Notes on Methods of Applied Mathematics
9/15/12
Contents
1 Ordinary Differential Equations
1.1 Superposition Principle . . . . . . . . . . . . . . . . . . .
1.2 Exactly Solvable Cases . . . . . . . . . . . . . . . . . . .
1.3 Integral Equations . . . . . . . . . . . . . . . . . . . . .
2 Partial Differential Equations
2.1 Method of Characteristics . . . .
2.2 d’Alambert’s Formula . . . . . . .
2.3 Method of Separation of Variables
2.3.1 Laplace’s Equation . . . .
2.3.2 Helmholtz Equation . . .
2.3.3 The Heat Equation . . . .
2.3.4 The Wave Equation . . . .
2.3.5 Schrödinger’s Equation . .
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3 Measure Theory and Function Spaces
3.1 Lebesgue Integration . . . . . . . . . .
3.2 Abstract Spaces . . . . . . . . . . . . .
3.2.1 Vector Space . . . . . . . . . .
3.2.2 Metric Space . . . . . . . . . .
3.2.3 Basic Topology . . . . . . . . .
3.2.4 Functions on Metric Spaces . .
3.2.5 Compactness . . . . . . . . . .
3.2.6 Contraction Mapping Theorem
3.2.7 Normed Spaces . . . . . . . . .
3.2.8 Linear Mappings . . . . . . . .
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93
4 Hilbert Spaces
4.1 Projections on Hilbert Spaces . . . . . . . . .
4.1.1 Gram-Schmidt Process . . . . . . . . .
4.2 Linear Functionals on Hilbert Spaces . . . . .
4.3 Fourier Series . . . . . . . . . . . . . . . . . .
4.3.1 Pointwise convergence of Fourier Series
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96
104
112
122
127
139
5 Distributions
5.1 Convergence in C08 pΩq . . . . . . . . . . . . . .
5.2 Dirac Delta distribution . . . . . . . . . . . . .
5.3 The Principal Value Distributions . . . . . . . .
5.4 Convergence in D 1 pΩq . . . . . . . . . . . . . . .
5.5 Algebra with Distributions . . . . . . . . . . . .
5.6 Calculus in D 1 pΩq . . . . . . . . . . . . . . . . .
5.6.1 The Heaviside function . . . . . . . . . .
5.6.2 The Derivative of the Delta Distribution
5.6.3 General definition of T 1 . . . . . . . . . .
5.6.4 Integration by Parts . . . . . . . . . . .
5.7 The Laplacian . . . . . . . . . . . . . . . . . . .
5.8 Convolution . . . . . . . . . . . . . . . . . . . .
5.9 Weak derivatives . . . . . . . . . . . . . . . . .
5.9.1 Strong vs Weak Derivatives . . . . . . .
5.10 Fourier Series in D 1 pΩq . . . . . . . . . . . . . .
5.10.1 Fourier coefficients of a distribution . . .
5.11 Fourier Transform . . . . . . . . . . . . . . . . .
5.12 Fourier Inversion Theorem . . . . . . . . . . . .
5.13 Properties . . . . . . . . . . . . . . . . . . . . .
5.14 Schwartz Space . . . . . . . . . . . . . . . . . .
5.15 Tempered Distribution . . . . . . . . . . . . . .
5.16 Fourier Transform on S 1 . . . . . . . . . . . . .
5.16.1 Some more properties . . . . . . . . . . .
5.17 Dual Spaces . . . . . . . . . . . . . . . . . . . .
5.18 Differential Equations in D 1 . . . . . . . . . . .
5.18.1 PDE . . . . . . . . . . . . . . . . . . . .
5.18.2 Fundamental Solutions . . . . . . . . . .
5.18.3 Some Examples . . . . . . . . . . . . . .
5.18.4 Some Remarks . . . . . . . . . . . . . .
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2
Chapter 1
Ordinary Differential
Equations
The nth order ODE is an equation of the following form
F pt, y, y 1 , y 2 , . . . , y pn´1q q “ 0
for unknown yptq. Usually it can be solved for the y pnq , so
y pnq “ f pt, y, y 1 , y 2 , . . . , y pn´1q q
It is linear if it can be written as,
n
ÿ
aj ptq y pjq ptq “ gptq.
j“0
If g “ 0, we say it is homogeneous.
We use the Operator notation D “ d{dt to mean
3
D2 y “ DpDyq “ Dpy 1 q “ y 2 , . . .
Dy “ y 1 ,
Above linear ODE is
Ly “
n
ÿ
aj ptqDj y “ g
j“0
where
L“
n
ÿ
aj ptqDj ,
j“0
We say L is a linear operator if
Lpc1 y1 ` c2 y2 q “ c1 Ly1 ` c2 L2
for any c1 , c2 P R.
1.1
Superposition Principle
Any linear combination c1 y1 ` c2 y2 of solutions of Ly “ 0 is again a
solution. The general solution of an ODE is the set of all solutions
Example 1.1.1. y 2 “ 0,
ùñ yptq “ C1 t ` C2 ,
is a general solution. 4
4
Example 1.1.2. y 2 “ 1
ùñ yptq “
t2
` C1 t ` C2 ,
2
4
Example 1.1.3. y 2 ` y “ 0
ùñ yptq “ C1 cosptq ` C2 sinptq
4
In general, the ODE Ly “ g, with L “
tion of the form
yptq “
n
ÿ
řn
j“1
aj ptqy pjq , has general solu-
cj yj ptq ` yp ptq,
j“1
where yp is any particular solution and y1 , . . . , yn are linearly independent
solutions of Ly “ 0, provided a0 , . . . , an , g are continuous functions.
In particular, the general solution depends on n parameters c1 , . . . , cn .
For a nonlinear ODE it is typically true that a general solution is an n
parameter family.
You should be able to specify then n side conditions. Common type of
side condition are y pjq paq “ γj , for j “ 0, . . . , n ´ 1.
ODE plus side conditions of this form make up an initial value problem,
(IVP).
#
Ly “ g
y pjq paq “ γj ,
j “ 0, . . . , n ´ 1.
5
All side conditions take place at t “ a. In other cases, we can have side
conditions at several points, and this is usually called a boundary value problem
(BVP).
Example 1.1.4.
y2 ` y “ 0
yp0q “ 5,
yp1q “ 3
4
Example 1.1.5.
y2 ` y “ 0
yp0q “ 5
y 1 p1q “ 3
4
Example 1.1.6.
y 2 ` y “ 0,
yp0q “ yp1q “ 2,
y 1 p0q ´ 3yp2q “ 5.
4
Definition 1.1.1. A side condition of the form
n´1
ÿ
ck y pkq paq “ γ,
k“0
is a separated bounded condition (BC).
Example 1.1.7. The conditions yp0q “ 5 and yp1q “ 3 are separated
but yp0q ` yp1q “ 2 is not.
4
6
Remark: Nonlinear (BC) are possible, e.g. yp1qyp2q “ 2, but we won’t
use them much.
In an ODE course, you prove early on a theorem about the IVP,
$
y pnq “ f pt, y, y 1 , y 2 , . . . , y pn´1q q
’
’
’
’
’
’
&ypaq “ γ0
y 1 paq “ γ1
’
..
’
’
’
.
’
’
% pnq
y paq “ γn´1
Theorem 1. If f is continuous in a nbhd of pa, γ0 , . . . , γn´1 q, then the
IVP has at least one solution on pa ´ , a ` q. for some ą 0 (a
local solution exists). If f is continuously differentiable in the nbhd,
there is at most on solution.
Remark: You can’t do better than local solution in general.
Example 1.1.8.
+
y1 “ y2
1
ùñ yptq “
1´t
yp0q “ 1
4
But if ODE is linear, then “ 8 works. (solution exists globally).
Recall that the IVP
#
y pnq “ f pt, y, y 1 , y 2 , . . . , y pn´1q q
y pjq paq “ γj ,
j “ 0, . . . , n ´ 1
has a unique solution under minimal assumptions on f , at least locally.
The BVP is a different story.
7
Example 1.1.9. Note,
#
y2 ` y “ 0
yp0q “ 0, ypπq “ 1
ùñ yptq “ c1 sinptq ` c2 cosptq
0 “ yp0q “ c1 ¨ 0 ` c2 ¨ 1 “ c2 ,
1 “ ypπq “ c1 ¨ 0 ` c2 ¨ p´1q “ ´c2 ,
So there is no solution.
4
Example 1.1.10. Note,
y2 ` y “ 0
yp0q “ 0, ypπq “ 0
ùñ yptq “ c1 sinptq ` c2 cosptq
0 “ c2 ,
0 “ c2 .
So yptq “ c1 sinptq is a solution of any c1 .
4
1.2
Exactly Solvable Cases
(1) Linear, homogeneous, constant coefficient equations.
Ly “
n
ÿ
j“0
8
aj D j y “ 0
for a0 , . . . , an constant. The technique is to guess a solution of the
form yptq “ eλt , where λ is to be found (“Ansatz”). Substitute into
ODE to find this works if and only if ppλq “ 0, where
ppλq “
n
ÿ
aj λj “ 0
j“0
(2) Euler’s equation:
Ly “
n
ÿ
aj pt ´ t0 qj Dj y “ 0.
j“0
for some constants a0 , . . . , an and t0 . The Ansatz is yptq “ pt´t0 qλ for
some λ. Again, this works if and only if λ is a root of the associated
polynomial.
1.3
Integral Equations
Let Rn “ n dimensional space and x “ px1 , . . . , xn q, with Ω Ď Rn . Let
ż
Lupxq “
Kpx, yqupyqdy
Ω
be a linear integral operator with kernel Kpx, yq. (To show linearity use
properties of the integral). The associated integral equations typically
have the form,
Lu “ λu ` g
9
for some constant λ and function g. In other words
ż
Kpx, yqupyq dy “ λupxq ` gpxq
Ω
Definition 1.3.1. If λ “ 0 it is an integral equation of the first kind,
otherwise it is second kind.
Example 1.3.1. Let Ω “ p0, 1q Ď R and Kpx, yq “ 1. The 1st kind is:
ż1
upyq dy “ gpxq
0
Solution is possible only if gpxq is a constant function, in which case
there will be infinitely many solutions. The 2nd kind is:
ż1
upyq dy “ λupxq ` gpxq
0
Any solution must have the form:
upxq “
c ´ gpxq
λ
for some constant c. Subsitute into equation:
10
ż1
c ´ gpyq
c ´ gpyq
dy “ λ
` gpxq
λ
λ
0
ż
c
1 1
ùñ
´
gpyq dy “ c
λ λ 0
˙
ˆ
ż
1 1
1
´1 “
gpyq dy
ùñ c
λ
λ 0
ş1
ş1
gpyq dy
gpyq dy
0`
˘ “ 0
ùñ c “
1
1´λ
λ λ ´1
If λ ‰ 1 then solution is
ş1
upxq “
gpyq dy gpxq
´
λp1 ´ λq
λ
0
If λ “ 1 then
ż1
gpyq dy
0“
0
This is a necessary condition on g in order that a solution exists. By
directly checking we see that
upxq “ c ´ gpxq
is a solution for any c. In summary,
λ “ 0 solution ðñ g is contant.
ż1
λ “ 1 solution ðñ 0 “
gpyq dy.
0
11
For λ ‰ 1, 0 we have a solution for any g. Conditions in λ “ 0, 1 cases
give a.e. solvability or consistency conditions.
4
Integral equations are classified in various wasps depending on K. For
example, consider Ω “ pa, bq Ď R and suppose that Kpx, yq “ 0 if y ą x.
The integral equation
żb
Kpx, yq upyq dy “ λupxq ` gpxq
a
becomes
żx
Kpx, yq upyq dy “ λupxq ` gpxq
a
This is said to be of Volterra type.
Example 1.3.2. Let Ω “ p0, 1q
#
1, y ă x
Kpx, yq “
0, y ą x
Then
żx
upyq dy “ λupxq ` gpxq
0
If λ “ 0 then
12
żx
upyq dy “ gpxq ðñ upxq “ g 1 pxq
0
provided that g 1 exists. Also, gp0q “ 0 must hold. This is the solvability
condition. On the other hand, if λ ‰ 0 then
żx
upyq dy “ λupxq ` gpxq
0
ùñ upxq “ λu1 pxq ` g 1 pxq
#
u1 pxq “ u´g
λ
ùñ
´gp0q
up0q “ λ
By ODE techniques (“integrating factor”) we have
´e´x{y gp0q 1
upxq “
´
λ
λ
żx
e
x´y
λ
g 1 pyq dy
0
is our solution.
4
Volterra equation often arise as equivalent forms of IVPs.
Example 1.3.3. Consider the ODE problem:
$
2
’
&u “ u
up0q “ γ0
’
% 1
u p0q “ γ1
13
Then by ODE techniques
żx
żx
2
1
upyq dy
u pyq dy “
0
żx
żxżz
upyq dy dz “ px ´ yupyq dy
upxq ´ γ0 ´ γ1 x “
0
0
0
żx
ùñ upxq “ px ´ yqupyq dy ` γ0 ` γ1
u pxq ´
0
0
The above are 2nd kind Volterra equation. 4
BVPs often correspond to non-Volterra integral equations. Other special
types of kernels:
• Kpx, yq “ Kpy, xq, symmetric type
• Kpx, yq “ Kpx ´ yq, convolution type.
• Kpx, yq Ñ 8 for some x, y are singular type.
Example 1.3.4. Let
żx
Lupxq “
0
upyq
?
dy
x´y
The above is called Abel integral operator. This is a Volterra, convolutional, singular type operator.
4
Other famous integral operators:
Example 1.3.5.
ż
e´xy upyq dy
Lupxq “
Rn
is the Laplace transform.
4
14
Example 1.3.6.
ż
e´ixy upyq dy
Lupxq “
Rn
is the Fourier transform.
4
Example 1.3.7.
1
Lupxq “
π
ż
R
upyq
dy
x´y
is the Fourier transform.
4
Nonlinear integral operators can also be studied. They are of the form:
ż
Kpx, y, upyqq dy
Lupxq “
Ω
15
Chapter 2
Partial Differential Equations
In a PDE upxq is the unknown quantity and assume x P Ω Ď Rn . We
use the multi index notation: α “ tαi uni“1 where each αi is a nonnegative
integer and
|α| “ α1 ` α2 ` ¨ ¨ ¨ ` αn
B |α| u
Bxα1 1 Bxα2 2 ¨ ¨ ¨ Bxαnn
ÿ
aα pxqDα u
Lupxq :“
Dα u :“
|α|ďm
where L is the linear partial differential operator of order m. We say
Lu “ f is an m-th order linear PDE. Nonlinear PDEs would look like
F px, tDα uu|α|ďm q “ 0
We study the following analytic techniques:
• 1st order: Method of characteristics.
• 2nd order: Separation of variables.
16
2.1
Method of Characteristics
we consider the case that m “ 1 and n “ 2. Let u “ upx, yq.
Example 2.1.1. Consider Bu
“ 0. This holds if and only if u “ f pxq for
By
some function f . The General solution involves an arbitrary function f .
4
Example 2.1.2. Consider
Bu Bu
`
“0
Bx By
Recall,
ˆ
~ “
∇u
Bu Bu
,
Bx By
˙
If θ~ “ pa, bq, then
~ ¨ θ~ “ a Bu ` b Bu
∇
Bx
By
is the derivative of u in the θ~ (unit vector) direction. In our case, with
θ~ “ p1, 1q, u has zero derivative in θ~ direction, so u is constant along any
line parallel to θ~
17
Then
ùñ upx, yq “ upx ´ y, 0q
ùñ u depends on x ´ y.
If f pxq “ upx, 0q, then upx, yq “ f px ´ yq. Conversely, if f is differentiable, it satisfies the PDE. Again we have a general solution depending
on an arbitrary function. The line parallel to p1, 1q, i.e. x “ y ` c, is a
characteristic for this PDE. Can think of it this way: solution u depends
only on c.
upx, yq “ f pcq “ f px ´ yq.
4
Now consider the PDE of the form
apx, yqux ` bpx, yquy “ 0
Consider the associated ODE system:
#
dx
dt
dy
dt
“ apx, yq
“ bpx, yq
for xptq and yptq. Solution is a curve in x-y plane.
18
Look at u along one such curve:
˘
d `
dx
dy
u xptq, yptq “ ux
` uy
dt
dt
dt
“ aux ` buy
“0
if u solves PDE then u is constant on this curve. These curves are the
characteristics for the PDE.
Example 2.1.3. Consider a “ b “ 1:
ux ` uy “ 0
dy
dx
“ 1,
“1
ùñ
dt
dt
ùñ xptq “ t ` c1 , yptq “ t ` c2
ùñ x ´ y “ c1 ´ c2 “ c
as before.
4
Example 2.1.4. Consider the problem with a “ x and b “ 1.
dy
dx
“ x,
“ 1,
dt
dt
ùñ xptq “ c1 et , yptq “ t ` c2
ùñ x “ c1 ey´c2 “ cey
So, x0 “ ce0 “ c.
19
The characteristic on the figure above is
x “ x0 ey ðñ x0 “ xe´y
So,
upx, yq “ upx0 , 0q “ upxe´y , 0q “ f pxe´y q,
if f pxq “ upx, 0q or upx, yq “ f pcq “ f pxe´y q. We can check any such
function is a solution for differentiable f . So this is a general solution.
4
Now consider the PDE of the form
apx, yqux ` bpx, yquy “ cpx, y, uq
Start by solving
dy
dx
“ apx, yq,
“ bpx, yq
dt
dt
Let xptq and yptq be one such curve, vptq :“ upxptq, yptqq, then
d
upxptq, yptqq
dt
“ a ¨ ux ` bu9 y
“ cpx, y, uq
“ Cpt, vptqq
ùñ v 1 “ Cpt, vptqq
v 1 ptq “
is an ODE for the solution u along a characteristic. In summary, the
PDE
20
apx, yqux ` bpx, yquy “ cpx, y, uq
known as “semi linear equation” has characterstic equations
#
dx
dt
dy
dt
“ apx, yq
“ bpx, yq.
Side conditions may be imposed to specify one solution, usually like by
specifying u on some curve γ in the xy plane
Let us study the following examples:
Example 2.1.5. Given the problem
#
ux ` uy “ 0
upx, 0q “ sinpxq
Here γ is the x-axis. Recall that the general solution is upx, yq “ f px´yq.
So since
sinpxq “ upx, 0q “ f px ´ 0q “ f pxq,
the solution is upx, yq “ sinpx ´ yq.
4
21
Example 2.1.6. Consider the problem
#
ux ` uy “ 0
upx, 0q “ sinpxq
with γ as the diagonal line:
Note that
sinpxq “ upx, 0q “ f px ´ xq “ f p0q,
is contradictory. Hence it is not possible to find such on f .
4
Problem: γ is itself a characteristic curve, and solution must be constant
along characteristic, so we cannot prescribe this as an addition condition.
We must require that γ be nowhere tangent to a characteristic.
Note: The tangent to a characteristic pxptq, yptqq is px1 ptq, y 1 ptqq “ pa, bq,
so pa, bq is a vector field on R2 , tangent to any characteristic. So, we
require γ to be nowhere tangent to pa, bq.
Example 2.1.7. Consider the problem
#
xux ` uy “ yu,
upx, 0q “ cospxq
22
This is the case that a “ x and b “ 1. The characteristic equations are
dx
dy
“ x,
“ 1, ùñ
dt
dt
ùñ x “ cey
xptq “ c1 et , yptq “ t ´ c2
as before.
Clearly γ is not tangent to any characteristic. Next we solve the required
ODE along each characteristic. Let characteristic pass through px0 , 0q
when t “ 0, x0 “ c1 e0 and 0 “ 0 ´ c2 . This implies
#
xptq “ x0 et ,
ypxq “ t
The ODE along characteristic is
d
dx
dy
upxptq, yptqq “ ux
` uy
dt
dt
dt
“ xux ` uy
“ yu
Let vptq :“ upxptq, yptqq, and so
v 1 “ tv
t2
ùñ vptq “ Ce 2
23
Now, vp0q “ upxp0q, yp0qq “ upx0 , 0q “ cospx0 q and so
C “ vp0q “ cospx0 q,
t2
ùñ vptq “ cospx0 qe 2
t2
ùñ upxptq, yptqq “ cospx0 qe 2
Express x0 and t in terms of x and y. That is t “ y and x0 “ xe´y .
Hence the solution is
y2
upx, yq “ cospxe´y qe 2
4
We summarize the steps in the above procedure:
(1) Find characteristic through each point of γ.
(2) Solve the required ODE along each characteristic, using side condition. You’ll end up with u as a function of the parameter t and point
on γ.
(3) Express u in terms of x and y.
Consider the following example
Example 2.1.8. Consider the equation
yux ´ xuy “ 0
The characteristic equations are
dx
dy
“ y,
“x
dt
dt
24
and so
d2 x
dy
“ ´x
“
2
dt
dt
d2 x
ùñ
`x“0
dt2
ùñ xptq “ c1 cosptq ` c2 sinptq
or simply xptq “ A sinpt ` δq. Hence
yptq “ x1 ptq “ A cospt ` δq
ùñ x2 ` y 2 “ A2
The characteristics are circles
Note that
d
upxptq, yptqq “ 0
dt
So u is constant on any circle. Then
a
upx, yq “ upx0 q “ up x2 ` y 2 , 0q
a
upx, yq “ f p x2 ` y 2 q, p or f px2 ` y 2 qq.
25
If you try to solve with side condition upx, 0q “ x you’ll see that it can’t
be done. Since
upx, 0q “ f px2 q “ f pp´xq2 q “ up´x, 0q
then u must be even in x. The problem here is p0, 0q; so xptq, yptq “ 0 is
a degenerate characteristic, i.e. it solves the characteristic equations but
isn’t really a curve. We’d have to regard γ as tangent to a characteristic
at p0, 0q. We can show that if pa, bq ‰ p0, 0q on γ and γ is nowhere
tangent to pa, bq, then a solution of the PDE problem exists in some
neighborhood of γ. In other words, we have a local solution and it can
be found by the method of characteristics.
4
Here are some generalizations:
1. More than 2 variables.
2. Fully nonlinear PDEs
F px1 , . . . , xn , u, ux1 , . . . , uxn q “ 0
The case in which n “ m “ 2 is for the PDE
Auxx ` Buxy ` Cuyy ` Dux ` Euy ` F u “ G
For now assume that A, B, . . . G are constants, and at least one of A, B, C ‰
0. We introduce canonical forms. Any such PDE can be transformed to
one of the following:
Elliptic Ñ uξξ ` uνν ` plower orderq “ 0
Hyperbolic Ñ uξξ ´ uνν ´ plower orderq “ 0
Parabolic Ñ uξξ ` plower orderq “ 0
26
A PDE is expressed with respect to a certain coordinate system. If you
make a change of variables the expression for the PDE changes and you
find it via the chain rule.
Example 2.1.9. Consider the problem
uxx ´ uyy “ 0
Let
#
ξ “ x ` y1
ν “x´y
ùñ
#
x“
y“
ξ`ν
2
ξ´ν
2
Then we obtain
ùñ
Bu
Bu Bξ Bu Bν
“
`
Bx
Bξ Bx Bν Bx
Bu Bu
`
“
Bξ Bν
or
B
B
B
“
`
Bx
Bξ Bν
Also,
27
ˆ
˙
B2
B Bu Bu
`
“
Bx2
Bx Bξ Bν
ˆ
˙ˆ
˙
B
B
Bu Bu
“
`
`
Bx Bν
Bx Bν
“ uξξ ` uξν ` uνξ ` uνν
“ uξξ ` 2uξν ` uνν
Similarly,
B2u
“ uξν ´ 2uξν ` uνν
By 2
So
uxx ´ uyy “ 4uξν
ùñ uxx ´ uyy “ 0
ðñ uξν “ 0
4
Now, consider
Auxx ` Buxy ` Cuyy “ 0
constants A, B, C at least one not zero. Consider a change of variables :
#
ξ “ αx ` βy
ν “ γx ` δy
28
where αδ ´ βγ ‰ 0. So
Bu
Bu Bξ Bu Bν
“
`
Bx
Bξ Bx Bν Bx
Bu
Bu
`γ
“α
Bξ
Bν
or
B
B
B
“α `γ
Bx
Bξ
Bν
So
˙ˆ
˙
ˆ
Bu
B2
B
B
Bu
α
“ α `γ
`γ
Bx2
Bξ
Bν
Bξ
Bν
2
2
“ α uξξ ` 2αγuξν ` γ uνν
Similarly, for uxy , uyy . Hence
0 “ Auxx ` Buxy ` Cuyy
“ Apα2 uξξ ` 2αγuξν ` γ 2 uνν
`Bp. . . q ` Cp. . . q
“ auξξ ` buξν ` cuνν ,
a “ α2 A ` αβB ` β 2 C,
b “ 2αγA ` pαδβγqB ` 2βδC,
c “ γ 2 A ` γδB ` δ 2 C.
The idea is pick α, β, γ and δ to achieve simple form of new PDE.
29
Example 2.1.10. Can we arrange a “ c “ 0?
ˆ ˙2
β
C“0
α
ˆ ˙2
β
β
c “ 0 ðñ A ` B `
C“0
γ
γ
β
a “ 0 ðñ A ` B `
α
This works if β{α, δ{γ are roots of
A ` Bλ ` Cλ2 “ 0
If this quadratic has two real, unequal roots, we can choose β{α ‰ δ{γ
to be these roots (so αδ ´ βγ ‰ 0, as needed).
4
If B 2 ´ 4AC ą 0, then α, β, γ, δ can be chosen so that new PDE is the
canonical form uξν “ 0 (or to uxx ´ uyy “ 0 by 1st example).
If B 2 ´ 4AC ă 0, then α, β, γ, δ can be chosen so that b “ 0, a “ c, so
we get
uξξ ` uνν “ 0
If B 2 ´ 4AC “ 0, then we can transform to b “ c “ 0, and get uξξ “ 0.
If you include Dux ` Euy ` F u ´ G terms, these transform to duξ ` euν `
f u ´ g, i.e. lower order terms (l.o.t.).
We summarize.
• B 2 ´ 4AC ě 0 implies PDE may be transformed to the canonical
form
30
uξν ` l.o.t. “ 0
The above is the hyperbolic case.
• B 2 ´ 4AC ă 0 implies that
uξξ ` uνν ` l.o.t. “ 0
The above is the elliptic case.
• B 2 ´ 4AC “ 0 implies that
uξξ ` l.o.t. “ 0
The above is the parabolic case.
Example 2.1.11. Consider the PDE
2uxx ` 2uxy ` uyy ` 4ux ´ uy “ 0
and so
B 2 ´ 4AC “ 4 ´ 4p2qp1q ă 0
implies we have an elliptic PDE.
4
31
Example 2.1.12. Consider the PDE
uy ´ uxx “ 0
and so
B 2 ´ 4AC “ 0
implies we have an parabolic PDE.
4
If A, B, C, . . . , G are functions of px, yq, then we classify the type of PDE
by above test at each point.
Example 2.1.13. Consider Tricomi’s equations
uxx ´ xuyy “ 0
This is the case that A “ 1, B “ 0, C “ ´x implies B 2 ´ 4AC “ 4x.
Hence for x ą 0 we have the hyperbolic case and for x ă 0 we have the
elliptic case
4
Remark: Above discussion is special to n “ 2 case. For n ě 3 there is
no complete classification like this. Hyperbolic, elliptic, and parabolic
classifications still exist but there are PDEs which don’t belong to any
of these classes.
32
We can make use of canonical forms to solve PDEs. Consider uξν “ 0.
This implies
uξ “ f pξq
ùñ upξ, νq “ F pξq ` Gpνq
This is the general solution, depending on 2 arbitrary functions. If uxx ´
uyy “ 0, then by using the coordinate transformation from 1st example
we get
upx, yq “ F px ` yq ` Gpx ´ yq
since ξ “ x ` y and ν “ x ´ y. The above is called d’Alambert’s solution.
Ex: Consider the PDE
$
’
&uxx ´ uyy “ 0
upx, 0q “ f pxq
’
%
uy px, 0q “ gpxq
So f pxq “ F pxq ` Gpxq and gpxq “ F 1 pxq ´ G1 pxq.
The following are important model equations:
(1) utt ´ uxx “ 0 The wave equation is a hyperbolic PDE. Here u is
displacement of a vibrating string at location x and at time t.
(2) ut ´ uxx “ 0 The heat equation is a parabolic PDE. Here u is the
temperature in a bar at location x and at time t.
(3) utt ` uxx “ 0 Laplace’s equation is a elliptic PDE. This equation
describes potential due to some charge distribution at location px, yq.
We study each model equation.
33
2.2
d’Alambert’s Formula
Consider the wave equation.
utt ´ uxx “ 0
As discussed before, the general solution is given by
upx, yq “ F px ` yq ` Gpx ´ yq
We can set up the initial value problem
$
’
&utt ´ uxx “ 0
upx, 0q “ f pxq
’
%
ut px, 0q “ gpxq
As before
f pxq “ F pxq ` Gpxq
gpxq “ F 1 pxq ´ G1 pxq
żx
ùñ
gpsq ds “ F pxq ´ Gpxq ` C
0
and hence
34
żx
gpsq ds “ 2F pxq ` C
f pxq `
ż0x
gpsq ds “ 2Gpxq ´ C
f pxq ´
0
ˆ
˙
żx
1
C
ùñ F pxq “
f pxq `
gpsq ds ´
2
2
0
ˆ
˙
żx
1
C
Gpxq “
f pxq ´
gpsq ds `
2
2
0
upx, yq “ F px ` tq ` Gpx ´ tq
1
1
“ pf px ` tq ` f px ´ tqq `
2
2
ż x`t
gpsq ds
x´t
The above is known as d’Alambert’s solution to the wave equation.
Ex: Let f pxq “ x2 and gpxq ´ ex . Then the solution to the wave
equation IVP by d’Alambert’s solution is
˘ 1
1`
px ` tq2 ` px ´ tq2 `
upx, tq “
2
2
1
“ x2 ` t2 ` pex`t ´ ex´t q
2
ż x`t
es ds
x´t
Note that any time you can transform the PDE exactly to uξν “ 0, there
will be a similar formula, but even a lower order term invalidates this
formula. An example would be the PDE
utt ´ uxx ` u “ 0
ùñ 4uξν ` u “ 0
which can be solved so easily.
35
2.3
Method of Separation of Variables
This method is more widely useful. The typical case is the heat equation ut ´ uxx “ 0. Look for solutions of the form upx, tq “ φptqψpxq.
Substituting into the PDE we have
ut “ φ1 ptqψpxq
uxx “ φptqψ 1 pxq
Then
ùñ
ùñ
ùñ
0 “ ut ´ uxx “ φ1 ptqψpxq ´ φptqψ 1 pxq
φ1 ptqψpxq “ φptqψ 1 pxq
ψ 1 pxq
φ1 ptq
“
φptq
ψpxq
This works if and only if both sides equal some constant. So,
ψ 1 pxq
φ1 ptq
“
“ ´λ
φptq
ψpxq
for some constant λ to be determined. Then we have the ODE system
#
φ1 ` λφ “ 0
ψ 2 ` λψ “ 0
Now include some typical side conditions
$
’
0 ă x ă 1, t ą 0
&ut ´ uxx “ 0,
up0, tq “ up1, tq “ 0, t ą 0
’
%
upx, 0q “ f pxq,
0ăxă1
36
If upx, tq “ φptqψpxq works in 2nd condition then
0 “ φptqψp0q “ φptqψp1q
should hold. Then φptq ‰ 0 should hold and so ψp0q “ ψp1q “ 0. We
obtain
#
ψ 2 ` λψ “ 0,
0ăxă1
ψp0q “ ψp1q “ 0
We must consider several cases. If λ “ 0 then λ “ k 2 . Then
ψpxq “ C1 sinpkxq ` C2 cospkxq
0 “ ψp0q “ C2
ùñ ψpxq “ C sinpkxq
0 “ ψp1q “ C sinpkq
We know that C ‰ 0 must hold and thus sinpkq “ 0. This implies that
k “ nπ for n P N. So we have λ “ pnπq2 for n P N. Now,
φ1 ` λφ “ 0
ùñ φptq “ Ce´λt “ Ce´pnπq
So the function
37
2t
2t
upx, tq “ Ce´pnπq
sinpnπxq
satisfies the PDE and boundary conditions for all n P N. Now we consider
the case that λ ă 0. Let λ “ ´k 2 for some constant k to be determined.
In this case
ψ2 ´ k2ψ “ 0
ùñ ψpxq “ C1 ekx ` C2 e´kx
Using the boundary conditions we have
0 “ C1 ` C2
0 “ C1 ek ` C2 e´k
We note that this forms the matrix equation
ˆ
1
1
ek e´k
˙ˆ ˙ ˆ ˙
C1
0
“
C2
0
Note that the 2 ˆ 2 matrix above is nonsingular and hence C1 “ C2 “ 0.
This means no nonzero ψ is possible. In the case that λ “ 0 we also see
that no solution can be found. Therefore
!
)8
2
e´pnπq t sinpnπxq
n“1
are solutions. By linearity, any linear combination of these is also a
solution. We can try
38
upx, tq “
8
ÿ
2t
Cn e´pnπq
sinpnπxq
n“1
We need
f pxq “
8
ÿ
Cn sinpnπxq
n“1
It turns out, we must find constants C1 , C2 , . . . to make this work. This
touches on the theory of Fourier series which will be discussed later. We
show when and how this may be done.
Similar calculations can be done for the other basic problems.
Example 2.3.1. Consider the wave equation
uxx ´ utt “ 0
Look for solutions of the form upx, tq “ ψptqφpxq. Then
φ2 ψ ´ φψ 2 “ 0
φ2 ψ 2
´
“0
ùñ
φ
ψ
φ2
ψ2
ùñ
“
“ ´λ
φ
ψ
#
φ2 ` λφ “ 0,
ùñ
ψ 2 ` λψ “ 0,
By ODE techniques so wee that sinpkxq sinpktq is a solution.
44
39
If variable px, tq, look for product solution φptqψpxq. If you have side conditions, you may be able to impose them also. If superposition property
holds, the linear combinations of these are again solutions. This method
can be adapted to other coordinate systems. Consider the Laplace equation
#
uxx ` uyy “ 0
x2 ` y 2 ă 1
It is natural to use polar coordinates:
a
r “ x2 ` y 2 ,
θ “ arctanpy{xq
#
x “ r cos θ
y “ r sin θ
First we need to find an expression for PDE in these variables:
Bu Br Bu Bθ
Bu
“
`
Bx
Br Bx Bθ Bx
x
y
Bu
Bu
“a
´a
x2 ` y 2 Br
x2 ` y 2 Bθ
Bu sin θ Bu
“ cos θ
´
Br
r Bθ
Thus
B
B
sin θ B
“ cos θ ´
Bx
Br
r Bθ
etc . . . . The result is
40
1
1
uxx ` uyy “ urr ` ur ` 2 uθθ “ 0
r
r
Look for product solutions: RprqΘpθq. So
1
1
R2 Θ ` R1 Θ ` 2 RΘ2 “ 0
r
r
1
2
2
R
Θ
R
`
` 2 “0
R
rR r Θ
2
R
R1 Θ2
r2
`r `
“0
R
R
Θ
R2
R1
Θ2
r2
`r “´
“λ
R
Θ
#R
Θ2 ` λΘ “ 0
r2 R2 ` rR1 ´ λR “ 0
ùñ
ùñ
ùñ
ùñ
The “Hidden side conditions” for Θ are that Θ must be 2π periodic in
order that RprqΘpθq be a single valued function of px, yq. Consider the
following cases for the Θ equation.
If λ ą 0 take λ “ k 2 for some constant k. Then the solution to the Θ
ODE is
Θpθq “ C1 sinpkθq ` C2 cospkθq
This is 2π periodic if and only if k P N.
If λ ă 0 take λ “ ´k 2 for some constant k. Then the solution to the Θ
ODE is
Θpθq “ C1 ekθ ` C2 e´kθ
This is never 2π periodic.
41
If λ “ 0 then the solution to the Θ ODE is
Θpθq “ C1 ` C2 θ
This is periodic if and only if C2 “ 0. So
#
cospnθq, n “ 0, 1, 2, . . .
Θpθq “
sinpnθq, n “ 1, 2, . . .
satisfy the requirements. If follows that λ “ n2 .
Now let us consider the R equation:
r2 R2 ` rR1 ´ n2 R “ 0
is an Euler equation. The general solution is given by
#
C1 rn ` C2 r´n , n “ 1, 2, 3, . . .
Rprq “
C1 ` C2 log r, n “ 0
The “Hidden side conditions” for R are that Rp0q should be finite. This
implies that C2 “ 0. THus Rprq “ rn for n “ 0, 1, 2, 3, . . . . We have
found that
#
rn cospnθq, n “ 0, 1, 2, 3, . . .
rn sinpnθq, n “ 1, 2, 3, . . .
42
satisfy the PDE. If we add boundary conditions to this problem we obtain
#
uxx ` uyy “ 0,
x2 ` y 2 ă 1
upx, yq “ f px, yq, x2 ` y 2 “ 1
which is know as the “Dirichlet problem for the Laplacian”.
The standard PDE problems are setup as follows. We let Ω Ď Rn and
denote BΩ to be the boundary of Ω.
2.3.1
Laplace’s Equation
The Laplace equation is ∆u “ f for x P Ω where ∆ is the Laplacian
operator
∆u “
n
ÿ
B2u
,
2
Bx
j
j“1
The Laplace equation is also known as the Poisson equation. If f “ 0
then u is a harmonic function and we have an elliptic PDE. Here are
some typical side conditions:
(i) Dirichlet or first kind boundary condition is u “ g on BΩ.
(ii) The Neumann or 2nd kind boundary condition is
Recall that
Bu
~ ¨ ~n
“ ∇u
Bn
where ~n is the outward unit normal on BΩ.
43
Bu
Bn
“ g on BΩ.
(iii) The Robin or 3rd type boundary condition is
Bu
` σu “ g
Bn
on BΩ.
2.3.2
Helmholtz Equation
Helmholtz’s equation is given by
∆u ` λu “ f,
xPΩ
where λ is a constant. This equation is also elliptic and side conditions
are as in the discussion of the Laplace equation.
2.3.3
The Heat Equation
The general heat equation is
Bu
“ ∆u,
Bt
x P Ω, t ą 0
Here ∆ is the Laplacian only in x. This is a parabolic type PDE. Side
conditions are boundary conditions of type (i), (ii) and (iii). For example
upx, tq “ gpx, tq,
and initial condition upx, 0q “ u0 pxq.
44
x P BΩ, t ą 0
2.3.4
The Wave Equation
The general wave equation is
B2u
“ ∆u,
Bt2
x P Ω, t ą 0
This is a hyperbolic type PDE. Side conditions are boundary conditions
as same as for heat equation and initial conditions are
upx, 0q “ u0 pxq,
ut px, 0q “ u1 pxq
for x P Ω.
2.3.5
Schrödinger’s Equation
The Schrödinger’s equation is
Bu
“ i∆u,
Bt
x P Rn , t ą 0
This is a parabolic type PDE. THe initial condition is upx, 0q “ u0 pxq
for x P Rn .
Side conditions in each case are natural ones because they lead to “wellposed” problems. This is a somewhat vague concept. The meaning is
1. The problem has existence of solution.
2. The solution is unique.
3. The solution depends continuously on the data of the problem.
We say a problem is ill-posed if it is not well-posed. Let us consider an
ill-posed problem
45
Example 2.3.2. Consider the one-dimensional space Laplace equation
$
’
x P R, y ą 0
&uxx ` uyy “ 0,
upx, 0q “ 0,
’
%
uy px, 0q “ f pxq,
Note that by uy we mean the normal derivative on the boundary. This
PDE is ill-posed. If f pxq “ sinpkxq, then
upx, yq “
sinpkxqeky
k
is the solution. This is done by separation of variables. Notice that
|f | ď 1 for all x and k. However, as k Ñ 8 we have that u Ñ 8
exponentially. So the solution does not depend continuously on the data.
44
We must define and study some function spaces to precisely state properties as in the well-posedness.
46
Chapter 3
Measure Theory and
Function Spaces
We begin with some basic definitions.
ˇ
(
Definition 3.0.1. Let Bpx, rq “ y P Rn ˇ |y ´ x| ă r denote the inside of the sphere of radius r and center x.
We say Ω is open if for all x P Ω there is r ą 0 such that Bpx, rq Ď Ω. On
the other hand, Ω is closed if Ωc is open if an sonly if whenever txn u P Ω
with xn Ñ x then x P Ω.
The set Ω is bounded if Ω Ď Bp0, rq for some r ą 0.
We say BΩ is the boundary of Ω. Note x P BΩ if and only if for all r ą 0
we have Bpx, rq X Ω ‰ H and Bpx, rq X Ωc ‰ H.
The closure of Ω, denoted Ω is the smallest closed set containing Ω “
Ω Y BΩ.
The interior of Ω, denoted Ω0 is the largest open set in Ω.
We say that f is continuous in Ω if for all x P Ω and ą 0, there exists
δ ą 0 such that
47
|y ´ x| ă δ
|f pyq ´ f pxq| ă ùñ
and y P Ω. Let CpΩq be the set of all continuous functions on Ω. In
general,
C m pΩq “
C 8 pΩq “
ˇ
(
f ˇ Dα f P CpΩq, @ |α| ď m
ˇ
(
f ˇ Dα f P CpΩq, @α
An important function is space is that of Lp spaces defined by
"
p
L pΩq “
*
ż
ˇ
f ˇ
p
|f | dx ă 8
Ω
for 1 ď p ă 8.
3.1
Lebesgue Integration
To understand Lp spaces we must understand Lebesgue integration. Recall Riemann integration Theory. We say
żb
f pxq dx
a
is defined as a limit of Riemann sums, with some restrictions. We require
a and b to be finite, we need f to be bounded and piecewise continuous.
More generally, we require f to be continuous almost everywhere.
A set E Ď R is said to be of measure zero if there exists pan , bn q such
that
(i) E Ď
8
ď
pan , bn q.
n“1
48
(ii)
8
ÿ
pbn ´ an q ă .
n“1
Example 3.1.1. The set E “ Q is of measure zero. If Q “ tr1 , r2 . . . u
and take
´
¯
pan , bn q “ rn ´ n , rn ` n
2
2
then
EĎ
8
ď
pan , bn q
n“1
8
ÿ
pbn ´ an q “
n“1
8
´¯
ÿ
2 n “
2
n“1
shows that the set of rational numbers is of measure zero.
44
Definition 3.1.1. A property which holds except possibly on a set of
measure zero is said to hold almost everywhere. We usually abbreviate
a.e..
Example 3.1.2. Consider the function
#
0, x P Q
f pxq “
1, x R Q
This function is 1 almost everywhere since it is not 1 on a set of measure
zero. Note that f here is discontinuous at every point.
4
49
If a function f is bounded on ra, bs, then f is Riemann integrable if and
only if f is continuous a.e.. The Lebesgue Integration Theory amounts
to a major extension of the concept of the integral
żb
f pxq dx
a
We will be defining the above for “measurable” functions. Roughly, f is
measurable if and only if
f pxq “ lim fn pxq
nÑ8
where each fn is continuos. We say f is measurable if there exists a piecewise continuous function g such that f pxq “ gpxq almost everywhere.
Example 3.1.3. We said that the function
#
0, x P Q
f pxq “
1, x R Q
is 1 almost everywhere. This implies that f is measurable since 1 is a
continuous function.
4
If f is measurable and f ě 0, then
żb
f pxq dx
a
is defined as a finite number or as `8. We say that f is integrable if
the definite integral exists. That is
50
żb
f pxq dx ă 8
a
We will say that f is simply measurable if we can express f “ f` ´
f´ , where f` “ maxpf, 0q and f´ “ ´ minp0, f q. Note f` and f´ are
measurable and
żb
żb
f pxq dx “
a
żb
f` pxq dx ´
a
f´ pxq dx
a
provided at least one of these terms is finite. Next we discuss some
properties.
(1). The integral is unaffected by changing f on any set of measure zero.
Example 3.1.4. Consider once again the function:
#
0, x P Q
f pxq “
1, x R Q
Then
żb
żb
f pxq dx “
a
1 dx “ b ´ a
a
Since Q is a set of zero measure.
4
(2). If f is Riemann integrable, then the Lebesgue integral exists and
has same value.
51
(3). If f measurable and φ is continuous, then φ ˝ f is measurable.
Example 3.1.5. Let φpsq “ |s|p . For f measurable we have that
|f |p is measurable. So
żb
|f pxq|p dx
a
is defined. The spaces
żb
"
p
L pa, bq “
ˇ
f ˇ
*
p
|f pxq| dx ă 8
a
are of great interest.
4
(4). Let f and g be measurable. Then αf ` βg is also measurable since
żb
żb
żb
g dx
f dx ` β
αf ` βg dx “ α
a
a
a
for any constants α and β. It is a fact that f g is also measurable.
(5). Let E Ď R and consider the indicator function of E given by
#
1, x P E
χE pxq “
0, x R E
52
If χE is measurable, then E is a measurable set and Lebesgue measure of E is
ż
mpEq “
χE pxq dx
R
for any measurable function and measurable set.
(6). For E as in the previous property, it is a fact that
ˇż
ˇ ż
ˇ
ˇ
ˇ f pxq dxˇ ď
|f pxq| dx
ˇ
ˇ
E
E
Furthermore, if |f | ď M for some M ą 0, then
ˇż
ˇ
ˇ
ˇ
ˇ f pxq dxˇ ď M mpEq
ˇ
ˇ
E
(7). This theory generalizes to Rn . That is
ż
f pxq dx
E
is defined for measurable functions f and measurable sets E Ď Rn .
We say that E Ď Rn is of measure zero if for all ą 0 there exists
xk and rk such that
EĎ
8
ď
Bpxk , rk q,
k“1
8
ÿ
rk ă k“1
Example 3.1.6. A line in R2 has measure zero. 4
53
Convergence Theorems
Now we state without proof two important convergence theorems. Now
that even if fn Ñ f , it may not be the case that
ż
ż
fn Ñ
f
E
E
The following give conditions for this to occur.
Theorem 2. “Dominated Convergence Theorem” If fn is measurable
and for all n P N we have that fn Ñ f almost everywhere
with the
ş
property that there exists g such that |fn | ď g and E g ă 8, then we
have
ż
ż
fn Ñ
E
f
E
Theorem 3. “Monotone Convergence Theorem” If 0 ď fn ď fn`1 with
fn measurable and fn Ñ f almost everywhere, then
ż
ż
f
fn Ñ
E
E
3.2
Abstract Spaces
In this section we discuss Vector spaces (linear spaces), Metric spaces
and Normed spaces.
3.2.1
Vector Space
We say X is a vector space if for all x, y P X, the element αx ` βy also
belongs to X for any scalars α and β and such that
54
1) x ` y “ y ` x, for all x, y P X.
2) x ` py ` zq “ px ` yq ` z, for all x, y, z P X.
3) There exits 0 P X such that x ` 0 “ x for all x P X.
4) For all x P X there exits ´x P X such that x ` p´xq “ 0.
5) αpβxq “ pαβqx for all x P X and scalars α, β.
6) 1 ¨ x “ x for all x P X.
7) pα ` βqx “ αx ` βx for all x P X and scalars α, β.
8) αpx ` yq “ αx ` αy for all x P X and scalar α.
We give some examples of vector spaces.
Example 3.2.1. The sets
Rn “
Cn “
ˇ
(
x “ px1 , . . . , xn q ˇ xj P R, @j P t1, . . . , nu
ˇ
(
z “ pz1 , . . . , zn q ˇ zj P C, @j P t1, . . . , nu
Note that scalars are always R or C. It should be clear which one is
referred to by context.
4
Example 3.2.2. The set of all polynomials of degree less than or equal
to n forms a vector space.
4
Example 3.2.3. The set of all polynomials forms a vector space.
4
Example 3.2.4. The set of all functions forms a vector space.
4
Example 3.2.5. The sets CpΩq, C m pΩq, C 8 pΩq and Lp pΩq are all vector
spaces. For example, for f, g P CpΩq we have that αf ` βg belongs to
CpΩq by basic properties of continuous functions.
4
55
For Lp pΩq, we must show that if
ż
ż
p
|g|p dx ă 8
|f | dx,
Ω
Ω
then
ż
|αf ` βg|p dx
Ω
Let p “ 1. Then
ż
ż
|α||f | ` |β||g| dx
ż
ż
“ |α| |f | dx ` |β| |g| dx
|αf ` βg| dx ď
Ω
Ω
Ω
Ω
To show the case for 1 ă p ă 8 we need to prove a special inequality.
We do this later. For now let’s mention some important concepts.
ř
If x1 , x2 , . . . , xn P X, then n1 αj xj is also in X for scalars αj . We call
this sum a linear combination.
We say Ltx1 , x2 , . . . , xn u is the set of all linear combination of x1 , x2 , . . . , xn .
We also refer to it by the span of x1 , x2 , . . . , xn . Note that Ltx1 , x2 , . . . , xn u
is itself a vector space.
We say that x1 , x2 , . . . , xn are linearly independent if
n
ÿ
α j xj “ 0
ðñ
α1 “ α2 “ ¨ ¨ ¨ “ αn “ 0
j“1
56
Otherwise we say x1 , x2 , . . . , xn are linearly dependent, in which case at
least one of the xj s can be expressed as a linear combination of the others.
The set X has dimension n (denoted dimpXq “ n) if
1. There exits n linearly independent vectors in X.
2. Any n ` 1 vectors are linearly dependent.
The set X has infinite dimension if for all n there exists n linearly independent vectors.
Example 3.2.6. The spaces Rn and Cn have dimension n. Note that
e1 “ p1, 0, . . . q
e2 “ p0, 1, 0, . . . q
e3 “ p0, 0, 1, 0, . . . q
..
.
are linearly independent. These are the standard basis.
4
Example 3.2.7. If tx1 , x2 , . . . , xn u are linearly independent then
dimpLtx1 , x2 , . . . , xn uq “ n
The reason is that there are n linearly independent vectors by assumption. Suppose y1 , y2 , . . . yn`1 are in Ltx1 , x2 , . . . , xn u. Have
yj “
n
ÿ
ckj xk , j “ 1, 2, . . . , n ` 1
k“1
and so
57
n`1
ÿ
αj yj “
n`1
n
ÿÿ
αj ckj xk “
j“1 k“1
n`1
ÿ
ùñ
n n`1
ÿ
ÿ
αj ckj xk “ 0
k“1 j“1
αj ckj , k “ 1, 2, . . . , n
j“1
is a homogeneous system of n equations for n`1 unknowns. This implies
that at least one nonzero solution α1 , α2 , . . . , αn`1 . It then follows that
y1 , y2 , . . . yn are linearly independent.
4
Example 3.2.8. Consider the set of polynomials of degree at most n.
This set is denoted Pn . For f P Pn we can equivalently say
f ptq “
n
ÿ
aj tj
j“0
and hence Pn “ Lt1, t, . . . , tn u. Here 1, t, . . . , tn are linearly independent
and hence dimpPn q “ n ` 1.
4
Example 3.2.9. Consider the set of all polynomials which we denote
P . Then dimpP q “ 8 since P contains n linearly independent vectors
for any n.
4
Example 3.2.10. The sets Cpa, bq, C m pa, bq, C 8 pa, bq and Lp pa, bq are
all infinite dimensional since they also contain 1, t, t2 , . . . , tn . The same
is true with Ω Ď Rn instead of pa, bq.
4
Example 3.2.11. Consider the ODE u2 ` u “ 0. The general solution
is
58
uptq “ C1 cosptq ` C2 sinptq
In other words, the solution is the 2 dimensional space Ltcosptq, sinptqu.
The general solution of any n-th order linear homogeneous ODE is an
n-dimensional vector space.
4
If X is a vector space, then E Ď X is a subspace of X if for all x, y P E
then αx ` βy P E and α, β scalars.
ˇ
(
Example 3.2.12. Let X “ Rn and consider E “ x P X ˇ x1 “ 0
where x1 is the
of x. Then E is a subspace. However
ˇ first component
(
E “ x P X ˇ x1 “ 1 is not a subspace!
4
Note that a subspace is itself a vector space.
Example 3.2.13. Observe that Pn is a subspace of P . The space C m pΩq
is a subspace of CpΩq. Lastly, the set
ˇ
(
f P Cr0, 1s ˇ f p0q “ 0
is a subspace of Cr0, 1s.
4
Example 3.2.14. If x1 , x2 , . . . , xn P X, Ltx1 , x2 , . . . , xn u is a subspace
of X, the subspace spanned by x1 , x2 , . . . , xn .
4
We say that S Ď X is a basis of X if for every x P X, the element x can
be expressed in exactly one way say a linear combination of elements of
S. That is
x“
n
ÿ
αj xj ,
j“1
59
xj P S
So x is a finite sum! There are other notions of basis that we will get to
when we can talk about infinite sums. This kind of basis (finite sum) is
a Hamel basis.
Theorem 4. Every vector space has a basis in this sense.
Theorem 5. S is a basis of X if and only if LpSq “ X and S is linearly
independent.
Proof. Note that LpSq “ X if and only if every x can be written at least
one way as a linear combination of elements of S. Furthermore, S is
linearly independent if and only if at most one way.
In summary: Suppose X is a vector space and S Ď X. If LpSq “ X, then
S is a spanning set for X. A basis is a linearly independent spanning set
and the dimension of X is the number of elements in any basis.
3.2.2
Metric Space
Let X be a set. A metric on X is a function d : X ˆ X Ñ R such that
1) dpx, yq ě 0 for all x, y P X.
2) dpx, yq “ 0 if and only if x “ y.
3) dpx, yq “ dpy, xq for all x, y P X.
4) dpx, yq ď dpx, zq ` dpz, yq for all x, y, z P X.
We think of dpx, yq as distance from x to y. A metric space is a pair
pX, dq where d satisfies the above properties. We can refer to X itself as
a matrix space if d is clear from context.
Example 3.2.15. Consider X “ Rn . Then the metric here is
g
fÿ
f n
dpx, yq “ e |xj ´ yj |2
j“1
The properties 1-3) are obvious. 4
60
!
)
ˇ řn
2
ˇ
Example 3.2.16. Consider X “ x
j“1 xj ă 1 . Same d satisfies
the metric properties. X could be any other subset of Rn . 4
Example 3.2.17. Consider X “ Rn with
˜
dpx, yq “
n
ÿ
¸1{p
|xj ´ yj |
p
j“1
We’ll see that d is a metric for any such p.
4
Example 3.2.18. Consider X “ Cra, bs with
dpf, gq “ max |f pxq ´ gpxq|
aďxďb
Since a continuos function on a compact interval achieves its maximum,
this is defined and finite. Metric properties are simple to check. The
space Cpa, bq however will not be a metric space with this d.
4
In general CpKq is a metric space for any K Ď Rn compact. Next we
list some useful inequalities.
(1) “The arithmetric-geometric mean” is given by
ab ď
a2 b 2
`
2
2
for all a, b P R. The reason is that
0 ď pa ´ bq2 “ a2 ´ 2ab ` b2
61
(2) For all a, b P R and ą 0
b2
ab ď a `
4
2
The reason is that
?
ab “ p 2q
ˆ
b
?
2
˙
ď a2 `
b2
4
using (1). A function φ on ra, bs is convex if
φpλx1 ` p1 ´ λqx2 q ď λφpx1 q ` p1 ´ λqφpx2 q
for all x1 , x2 P ra, bs with λ P r0, 1s.
For any c P pa, bq, there exists a liner (supporting line) such that
ψpxq “ mpx ´ cq ` φpcq
such that ψpxq ď φpxq on ra, bs. Note that m “ φ1 pcq if φ1 pcq exists.
62
(3) Let φ be convex on R, then
ˆ
φ
1
b´a
żb
˙
f ptq dt ď
a
1
b´a
żb
φpf ptqq dt
a
The above is called Jensen’s inequality.
Proof. Let
1
c“
b´a
żb
f ptq dt
a
and let ψ be a support line for φpxq at c. Then
φpxq ě mpx ´ cq ` φpcq
for all x. It follows that
φpf ptqq ě mpf ptq ´ cq ` φpcq
˙
ˆż b
żb
żb
ùñ
φpf ptqq dt ě m
f ptq dt ´ c dt ` φpcqpb ´ aq “ φpcqpb ´ aq
a
a
a
The above is equivalent to the given statement.
(4) For a, b ě 0 with q and p conjugate (or dual exponents) we have
63
ab ď
ap b q
`
p
q
The above is called Young’s inequality.
Proof. Consider the function
#
p logpaq, 0 ď t ď 1{p
q logpbq, 1{p ď t ď 1
Then for φpxq “ ex and a “ 0, b “ 1 we obtain
ż1
0
ˆ
˙
1
1
f ptq dt “ p logpaq ` q logpbq 1 ´
p
p
“ logpaq ` logpbq
and so
ˆż 1
φ
˙
f ptq dt “ ab
0
On the other hand,
ż1
φ pf ptqq dt “
0
ap b q
`
p
q
So Jensen’s inequality gives the proof of t.
64
(5) For a, b ě 0 with p and q conjugate (or dual exponents) we have
ab ď
ap ´q{p bq
`
p
q
for ą 0. The proof follows a similar argument to that of (2).
(6) For p and q conjugate (or dual exponents) we have
˙1{p ˆż
ˆż
ż
p
|upxqvpxq| dx ď
|upxq| dx
Ω
Ω
˙1{q
q
|vpxq| dx
Ω
The above is called Hölder’s inequality.
Proof. Consider
ż
|upxqvpxq| dx ď
p
Ω
ż
´q{p
|upxq| dx `
q
ż
p
Ω
Choose
ˆş q
˙1{q
|v| dx
“ ş p
|u| dx
Hence
65
|vpxq|q dx
Ω
˙1{q ˆż
˙1{p
p
|vpxq| dx
|upxq| dx
ˆż
ż
p
q
|upxq| dx “
˙1{q ˆż
˙1{p
p
|vpxq| dx
|upxq| dx
ˆż
q
“
ˆż
˙1{q ˆż
˙1{p
p
|upxq| dx
,
|vpxq| dx
q
´q{p
“
so
˙1{p ˆż
ˆż
ż
p
|upxqvpxq| dx ď
˙1{q ˆ
q
|u| dx
|v| dx
1 1
`
p q
˙
as needed. We use the notation
˙1{p
ˆż
p
|upxq| dx
||u||p “
Ω
for the Lp norm of u. If a is a finite or infinite sequence then
´ÿ
|a|p “
|ak |p
¯1{p
|a|8 “ supp|ak |q,
, for 1 ď p ă 8
for p “ 8
is the p norm of the sequence a. Hölders inequality amounts to say
||uv||L1 ď ||u||p ||v||q
This is valued for p “ 1 and q “ 8 as well.
66
(7) The triangle inequality
||u ` v||p ď ||u||p ` ||v||q
holds in Lp and it is known as Minkowski’s inequality.
Proof. Consider the case that a ă p ă 8. Then
pu ` vqp “ upu ` vqp´1 ` vpu ` vqp´1
˙1{p ˆż
ˆż
ż
|upu ` vq
ùñ
p´1
p
| dx ď
˙1{q
|u ` v|
|u|
pp´1qq
Ω
˙1´1{p
ˆż
p
|u ` v| dx
“ ||u||p
˙1{p ˆż
ˆż
ż
p´1
|vpu ` vq
| dx ď
|v|
p
˙1{q
|u ` v|
pp´1qq
Ω
˙1´1{p
ˆż
“ ||v||p
p
|u ` v| dx
So,
ż
|u ` v|p dx ď ||v ` u||pp´1 p||u||p ` ||v||q q
After dividing obtain the desired inequality. The case that p “ 1, 8
is easier.
67
(8) The sum analog of Hölder’s and Minkowski’s inequalities
8
ÿ
˜
|ak bk | ď
k“1
˜
8
ÿ
8
ÿ
¸1{p ˜
|ak |p
k“1
¸1{q
|bk |q
k“1
¸1{p
|ak ` bk |p
8
ÿ
˜
8
ÿ
ď
k“1
¸1{p
|ak |p
˜
8
ÿ
`
k“1
¸1{p
|bk |p
k“1
We now discuss some more examples of metric spaces
Example 3.2.19. Consider X “ Rn . Then
dp px, yq “ |x ´ y|p
˜
¸1{p
n
ÿ
“
|xj ´ yj |p
j“1
for 1 ď p ă 8. THe first three metric properties are clear. Note
dp px, yq “ |px ´ zq ` pz ´ yq|p
ď |x ´ z|p ` |z ´ y|p
“ dp px, zq ` dp pz, yq
by Minkowski’s inequality. Hence dp is a metric on X. The same works
for Cn .
4
Example 3.2.20. Consider X “ CpKq with K Ď Rn compact with
dpf, gq “ max |f pxq ´ gpxq|
xPK
68
Hence for all x P X
|f pxq ´ gpxq| ď |f pxq ´ hpxq| ` |hpxq ´ gpxq|
ùñ max |f pxq ´ gpxq| ď max |f pxq ´ hpxq| ` max |hpxq ´ gpxq|,
xPK
xPK
xPK
which is the triangle inequality.
4
Example 3.2.21. Consider X “ C m pKq with K Ď Rn compact with
dpf, gq “
ÿ
max |Dα f pxq ´ Dα gpxq|
xPK
|α|ďm
4
Example 3.2.22. Consider X “ C 8 pKq. It is not so clear since unite
sum may not converge.
4
Example 3.2.23. Consider X “ Lp pΩq with Ω Ď Rn compact with
dpf, gq “ ||f ´ g||Lp
Properties (1), (3) and (4) are clear but (2) is not. If dpf, gq “ 0 we
cannot conclude f “ g for all x P X. Only that f “ g almost everywhere.
4
We are obliged to make the following modification. If f “ g a.e. then
we regard them as identical elements of Lp . More precisely, we define an
equivalence relation f „ g if and only if f “ g almost everywhere. For
any f there is an equivalence class
ˇ
(
g P Lp ˇ g „ f
69
The space Lp must be regarded as a set of equivalence classes of functions.
To work with an element of Lp , pick any representative of the equivalence
class.
Convergence in metric spaces
We say xk converges to x, that is xk Ñ x, in X if dpxk , xq Ñ 0 as k Ñ 8.
Example 3.2.24. Consider X “ Rn , Cn with d “ dp . Then xk Ñ
x if and only if each component of xk converges to the corresponding
component of x. For example, for xk “ pxk1 , xk2 , . . . , xkn q
xk 1 Ñ x1
xk 2 Ñ x2
xk 3 Ñ x3
..
.
4
Example 3.2.25. Consider X “ CpKq with K Ď Rn compact. Then
fk Ñ f in X if and only if
max |fk pxq ´ f pxq| Ñ 0
xPK
which is the same as fk Ñ f uniformly on K. In C m pKq, we have
Dα fk Ñ Dα f uniformly for all |α| ď m.
4
Example 3.2.26. Consider X “ Lp pΩq. then fk Ñ f if and only if
ż
|fk ´ f |p dx Ñ 0
Ω
70
for 1 ď p ă 8. The case that p “ 2 is called the “root mean square
convergence”.
4
Definition 3.2.1. We say txk ukPN is a Cauchy sequence if for all ą 0
there is N P N such that
@m, k ě N
ùñ dpxm , xk q ă We claim that every convergent sequence is Cauchy. Let ą 0 be given.
Then there exists N P N such that for k ě N , we have dpxk , xq ă {2
where x is the limit of the sequence. If m, k ě N then
dpxm , xk q ď dpxm , xq ` dpx, xk q ă as needed. However, the converge is not true in general.
Example 3.2.27. Consider X “ Q which
? is a metric space with absolute
2 in R. Then tqk u is Cauchy, but
value metric. Pick qk P Q with
q
Ñ
k
?
not convergent in Q since 2 R Q.
4
Complete Metric Spaces
Define the set X to be complete if every Cauchy sequence in X is convergent in X.
Example 3.2.28. The set of rationals Q is not complete. The set of
real numbers R is complete.
Similarly Rn and Cn are complete.
4
Example 3.2.29. The set of continuous functions CpKq with supremum
norm is complete.
71
Proof. Fix ą 0. Then find N P N such that for all k, m ě N
max |fk pxq ´ fm pxq| ă xPK
Hence tfk pxqukPN is Cauchy in R for each fixed x. By completeness of
R, the limit limkÑ8 fk pxq exists for all x. Call the limit f pxq. That is
f pxq :“ lim fk pxq for each x. We claim that fk Ñ f in CpKq. If ą 0
there is N such that for x P K and k, m ě N we have
|fk pxq ´ f pxq| ď for x P K and k ě N . Thus dpfk , f q ď for k ě N . Then fk Ñ f
in CpKq. Finally, f P CpKq since f is a uniform limit of continuous
functions. Refer to Theorem 2.3 in the textbook.
4
Similarly, C m pKq is complete.
Example 3.2.30. The space Lp pΩq is complete for 1 ď p ă 8 and
Ω Ď Rn . This requires a certain amount of measure theory. See section
12.6.
4
Incomplete spaces can arise by choosing the wrong metric.
Example 3.2.31. Consider X “ Cr´1, 1s with
ż
dpf, gq “
|f pxq ´ gpxq| dx
r´1,1s
Then pX, dq is a metric spas but not a complete one. To show why,
we find a sequence of continuous functions on r´1, 1s whose limit is not
continuous. Let
72
$
’
&xk, |x| ă 1{k
fk pxq “ 1,
x ą 1{k
’
%
´1, x ă ´1{k
If is a Cauchy sequence since
ż
|fk pxq ´ fm pxq| dx ď
r´1,1s
4
N
if k, m ě N . But
$
’
x“0
&0,
lim fk pxq “ 1,
xą0
kÑ8
’
%
´1, x ă 0
is not in Cr´1, 1s.
4
An incomplete space can be complete. In other words, elements can be
added to the space to give a complete metric space. Refer to section 1.6
in the text. For example, R is a completion of Q.
73
3.2.3
Basic Topology
Denote the open ball of center a and radius r by
Bpa, rq :“
ˇ
(
x P X ˇ dpx, aq ă r
We say that G Ď X is open if for all x P G there is r ą 0 such that
Bpx, rq Ď G. The definitions of closet set, bounded set, closure, interior
and boundary of a set are exactly as in Rn case. However, the definition
of a compact set is not the same.
The union of opens sets is open. The finite union of closed sets is closed.
Refer to section 1.5.
Definition 3.2.2. A set G Ď X is dense if G “ X.
Definition 3.2.3. A set X is separable if it contains a countable dense
subset.
Example 3.2.32. The set of real numbers R is separable since Q is
countable and dense. Same goes for Cn and Rn .
4
Example 3.2.33. The sets CpKq and Lp pΩq for 1 ď p ă 8 are separable. However, the set L8 pΩq is not.
4
3.2.4
Functions on Metric Spaces
Let f : X Ñ Y be a function.
Definition 3.2.4. We say that f is continuous at x0 P X if for all ą 0
there is δ ą 0 such that
dX px, x0 q ă δ
dY pf pxq, f px0 qq ă ùñ
We say f is continuous on X if it is continuous at each x0 P X.
74
Subscript, X and Y are clear from context.
Definition 3.2.5. We say that f is uniformly continuous on X if for all
ą 0 there is δ ą 0 such that
dX px, yq ă δ
dY pf pxq, f pyqq ă ùñ
Definition 3.2.6. We say that f is Lipschitz continuous if there is ρ ą 0
such that
dY pf pxq, f pyqq ă ρdX px, yq
Definition 3.2.7. If ρ ă 1 we say that f is a contraction.
Note that
Contraction ùñ Lipschitz ùñ Unif. continuous ùñ continuous
Theorem 6. The function f is continuous on X if and only if f pxk q Ñ
f pxq whenever xk Ñ x. This is known as the “sequential definition of
continuity”. Refer to Prop 1.34 in the text.
If f : X Ñ Y and E Ď X then the image of E under f is
f pEq “
ˇ
(
y P Y ˇ y “ f pxq, for some x P E
If G Ď Y then
f ´1 pGq “
ˇ
(
x P X ˇ f pxq P G
is the inverse image of G under f .
Theorem 7. The function f : X Ñ Y is continuous if and only if f ´1 pGq
is open in X whenever G is open in Y . Refer to Prop 1.46 in the text.
75
3.2.5
Compactness
We say tEα uαPA is a collection of sets with A as the index set.
Example 3.2.34. Let Eα “ pα, α ´ 1q in R. So tEα uαPR is a collection
of intervals in R.
4
Definition 3.2.8. We say tEα uαPA is an open cover of K Ď X if
(i) Eα is open for all α P A.
ď
(ii) K Ď
Eα
αPA
Definition 3.2.9. The set K Ď XŤis compact if any open cover has
a finite subcover. That is, if K Ď αPA Eα , Eα is open then there is
α1 , . . . , αk P A such that
KĎ
n
ď
Eαj
j“1
Definition 3.2.10. The set K Ď X is sequentially compact if any sequence in K has a subsequence convergent to an element of K. That is
txn unPN Ď K, then there is nj Ñ 8, and x P K such that xnj Ñ x.
Theorem 8. In any metric space, K compact is equivalent to sequentially compact.
Proof. See Theorem 1.62 in text.
Definition 3.2.11. We say K is precompact (or relatively compact) if
K is compact.
Theorem 9. K compact implies K is close and bounded.
Theorem 10. (“Heine-Borel”) In Rn or Cn , K is compact if and only if
K is closed and bounded.
76
Note that K being compact is equivalent to K being closed and totally
bounded. See the text for definition.
Definition 3.2.12. If X itself is compact, we say it a compact metric space.
Theorem 11. A compact metric space is complete.
Proof. Let txn unPN be Cauchy in X. By compactness of X, a convergent
subsequence exists, say xnj Ñ x, for some x P X. We want to show
xn Ñ x. If ą 0 there exists N such that
dpxk , xm q ă
2
for k, m ě N . Fix j such that nj ą N and dpxnj , xq ă {2. It follows
that
dpxk , xq ď dpxk , xnj q ` dpxnj , xq
ă `
2 2
“
if k ě N .
The converse if false.
Theorem 12. Let X be a compact metric space with f : X Ñ R
continuous. Then maxxPX f pxq exists.
Proof. First claim that f is bounded on X. The function f is compact in
R (a hw exercise). So the range of f is bounded. Let m :“ supxPX f pxq,
then m is finite. There must exists a sequence txn u in X such that
f pxn q Ñ m. By compactness of X, there is xnj Ñ z P X and by
continuity
77
f pzq “ lim f pxnj q “ m
jÑ8
So the supremum is achieved so sup “ max. That is, maxxPX f pxq exists
in X. A similar argument follows for the min.
Denote
z :“ arg max f pxq
xPK
to be the point at which the maximum occurs.
Theorem 13. Let f : X Ñ R be continuous with X compact. Then f
is uniformly continuous on X.
Proof. Refer to Theorem 1.67 in the text.
If X is a compact metric space,
CpXq “
ˇ
(
f ˇ f : X Ñ R continuous
If dpf, gq “ maxxPX |f pxq ´ gpxq| then CpXq is a metric space with this
d by the earlier theorem, and similar arguments to the CpKq, K Ď Rn ,
compact space.
CpXq is not itself a compact metric space, but there are useful criteria
for when subsets of CpXq are compact.
This comes from the Arzela-Ascoli theorem.
Definition 3.2.13. Let F “ tfn unPN Ď CpXq be a family of functions.
We say it is uniformly bounded if there is M ă 8 such that |fn pxq| ď M
for all x P X and n P N (or f P F).
78
Definition 3.2.14. The family of functions F “ tfn unPN Ď CpXq is
equicontinuous if for all ą 0 there is δ ą 0 such that
|x ´ y| ă δ
ùñ
|fn pxq ´ fn pyq| ă for all n P N (or f P F).
Given any metric space, it is of interest to characterize compact subsets
of X. The Heine-Borel theorem does this for X “ Rn . However, in
general closed and bounded sets need not be compact. Refer to 2.13
in the text. Any time a theorem tells you some set has a convergent
subsequence, it is a (pre)compactness result.
Theorem 14. Any bounded sequence in Rn has a convergent subsequence.
The Arzela-Ascoli Theorem is a corresponding compactness result for
X “ CpKq, with K compact.
Theorem 15. (“Arzela-Ascoli”) If F Ď CpKq is uniformly bounded and
equicontinuous then F is precompact.
Proof. Refer to 2.1.2 in text.
In particular, if a sequence tfn unPN is uniformly bounded and equicontinuous, there is f P CpKq and a subsequence fnj Ñ f in CpKq.
Example 3.2.35. Let
F“
ˇ
(
f P Cpr0, 1s ˇ f p0q “ 0, |f 1 pxq| ď M
Then for f P F we have
żx
f pxq “ f p0q `
f 1 psq ds
0
żx
ùñ |f pxq| ď
M ds ď M
0
79
It follows that f is uniformly bounded. Similarly for f P F,
żx
f 1 psq ds
f pxq ´ f pyq “
y
ùñ |f pxq ´ f pyq| ď M |x ´ y|
Given ą 0, there is δ “ {M that works such that we obtain equicontinuity. So F is pre compact in Cr0, 1s. Note that however, it is not
compact.
4
3.2.6
Contraction Mapping Theorem
Let X be any metric space. Let T : X Ñ X (T is a mapping on X). T
is a contraction if
dpT x, T yq ď ρdpx, yq
for all x, y P X and some fixed ρ ă 1. We say x is a fixed point of T if
T x “ x.
Example 3.2.36. Let
ş
Tu “
Ω
Kpx, yqupyq dy ´ f pxq
λ
Then
ż
Kpx, yqupyq dy “ λupxq ` f pxq
Ω
is of the form T u “ u.
4
80
Example 3.2.37. Any equation f pxq “ y is equivalent to x “ x`f pxq´
y “ T x.
4
Theorem 16. If T is a contraction on a complete metric space X, then
it has a unique fixed point.
Proof. First we show the uniqueness result. If T x1 “ x1 and T x2 “ x2 ,
then
dpx1 , x2 q “ dpT x1 , T x2 q ă ρdpx1 , x2 q
If x1 ‰ x2 then dpx1 , x2 q ‰ 0. Division gives 1 ď ρ which is a contradiction.
Now for the existence part. Pick x1 P X and let xx`1 “ T xn . We claim
that txn u is Cauchy. Note
dpx3 , x2 q “ dpT x2 , T x1 q ď ρdpx2 , x1 q
dpx4 , x3 q “ dpT x3 , T x2 q ď ρdpx3 , x2 q ď ρ2 dpx2 , x1 q
..
.
dpxn`1 , xn q ď ρn´1 dpx2 , x1 q
If m ą k, then
81
dpxm , xk q ď
m´1
ÿ
dpxj`1 , xj q
j“k
m´1
ÿ
ď
ρj´1 dpx2 , x1 q
j“k
“ dpx2 , x1 q
m´1
ÿ
ρj´1
j“k
“ dpx2 , x1 qρk´1 p1 ` ρ ` ¨ ¨ ¨ ` ρm´k´1 q
ˆ
˙
1
k´1
ďρ
dpx2 , x1 q
1´ρ
by geometric series. If ą 0, let N be such that
ρN ´1
dpx2 , x1 q ă 1´ρ
Then for m ą k ě N , we have dpxk , xm q ă . Similarly if k ą m ě N .
By completeness, there is x P X such that xn Ñ x. Then the continuity
of T gives the result.
xn`1 “ T pxn q
Ó
Ó
x
“ Tx
as needed.
So we have shown that a contraction on a complete metric space has a
unique fixed point.
Example 3.2.38. Let Ω Ď Rn be compact and consider X “ CpΩq.
Then
82
ş
Ω
T upxq :“
Kpx, yqupyq dy ´ f pxq
λ
where f P X, K P CpΩ ˆ Ωq, so T u “ u if and only if
ż
kpx, yqupyq dy “ λupxq ` f pxq
Ω
for x P Ω.
4
We claim that T is a contraction for sufficiently large |λ|. Note there is
M such that |Kpx, yq| ď M for all x and y. THen
dpT u, T vq “ max |T puqpxq ´ T pvqpxq|
xPΩ
ˇ
ˇ ż
ˇ
ˇ1
ˇ
Kpx, yqpupyq ´ vpyqq dy ˇˇ
“ max ˇ
xPΩ λ Ω
ż
M
ď
|upyq ´ vpyq| dy
|λ| Ω
M
ď
mpΩq max |u ´ v|
|λ|
M
“
mpΩqdpu, vq
|λ|
So, |λ| ą M mpΩq. This is a contradiction. One other thing to check is
that T u P X if u P X. It is enough to show
ż
Kpx, yqupyq dy
Ω
83
is continuous in x, for u P X. Let xn Ñ x, must show that
ż
ż
Kpxn , yqupyq dy Ñ
Ω
Kpx, yqupyq dy
Ω
the continuity of K. This implies
Kpxn , yqupyq Ñ Kpx, yqupyq
for all y. Since
|Kpn , yqupyq| ď M max |u| “: M1 ,
and
ş
Ω
M1 ă 8, so the Dominated Convergence Theorem applies, so
ż
ż
Kpx, yqupyq dy
Kpxn , yqupyq dy Ñ
Ω
Ω
Note that the proof of the Contraction Mapping Theorem (CMT) shows
not only that fixed point exists, but a way to approximate it, namely,
pick any xn P X with xn`1 T xn and so xn Ñ x. This is known as the
fixed point iteration procedure.
Example 3.2.39. (Last ex.) Pick u1 P CpΩq with
un`1 pxq “ T pun qpxq
ş
Kpx, yqun pyq dy ´ f pxq
“ Ω
λ
Another way to express this is as follows. For simplicity take λ “ 1 and
let
84
ż
Spuqpxq “
Kpx, yqun pyq dy
Ω
So we have u “ T u “ Su ´ f and hence un`1 “ Spun q ´ f . Thus
u2 “ Spu1 q ´ f
u3 “ Spu2 q ´ f “ SpSpu1 q ´ f q ´ f “ S 2 pu1 q ´ Spf q ´ f
u4 “ Spu3 q ´ f “ S 3 pu1 q ´ S 2 pf q ´ Spf q ´ f
..
.
n´2
ÿ
un “ S n´1 pu1 q ´
S j pf q
j“0
Since S n´1 pu1 q Ñ 0 as n Ñ 8, we have
u“´
8
ÿ
S j pf q
j“0
which is the familiar “Neumann series formula”. Using just a few terms,
we obtain an approximation of u known as Born approximation.
4
In general, for λ ‰ 1 we have
u“´
8
ÿ
S j pf q
λj`1
j“0
Example 3.2.40. Consider X “ R with f : X Ñ X. Say
85
f pxq ´ f pyq “ f 1 pcqpx ´ yq
for c between x and y. So if
|f 1 pxq| ď ρ ă 1
for all x, then
|f pxq ´ f pyq| ď ρ|x ´ y|
implies that f is a contraction on X. If we had
|f pxq ´ f pyq| ď |x ´ y|
then f may not have a fixed point.
4
Example 3.2.41. Consider the set X “ r0, 8q with f pxq “
Then
?
x2 ` 1.
x
ă1
f 1 pxq “ ? 2
x `1
ùñ |f pxq ´ f pyq| ď |x ´ y|
?
holds for all x, y but f pxq “ x if and only if x2 ` 1 “ x which happens
if and only if x2 ` 1 “ x2 . This is impossible!
4
86
Variation of CMT
Theorem 17. Let X be a complete metric space and suppose T n is a
contraction on X, for some positive integer n. Then T has a unique fixed
point.
Proof. T n has a fixed point, T n x “ x, implies
T n`1 x “ T x
ùñ
T n pT xq “ T x
but fixed point of T n is unique implies T x “ x. Now the uniqueness
part: If x is a fixed point of T , T x “ x, then T 2 x “ T x “ x implies x is
a fixed point of T 3 , T 4 , etc. THus, 2 distinct fixed points of T will be 2
distinct fixed points of T n . This is impossible.
Example 3.2.42. Consider the ODE problem
#
u1 “ f pt, uq
upaq “ u0
This is equivalent to
żt
f ps, upsqq ds “: T puqptq
uptq “ u0 `
a
The solution is a fixed point of this mapping T .
4
Let X :“ Cra, bs for some b ą a. Suppose f P Cpra, bs, Rq and
|f ps, uq ´ f ps, vq| ď L|u ´ v|
87
for all s P ra, bs and u, v P R for some L ě 0. In other words, f is
Lipschitz in u, uniformly with respect to s. Then
ˇ
ˇż t
ˇ
ˇ
ˇ
|T uptq ´ T vptq| “ ˇ pf ps, upsqq ´ f ps, vpsqqq dsˇˇ
a
żt
ď |f ps, upsqq ´ f ps, vpsqq| ds
a
żt
ď L |upsq ´ vpsq| ds
a
ď Lpt ´ aq max |u ´ v|
ď Lpb ´ aq max |u ´ v|
If Lpb ´ aq ă 1, then T is a contraction on Cra, bs so has a unique fixed
point. So, solution exists at least on some interval starting at a. It can
be shown that for any b ą a, T n is a contraction, so we get a solution
for any interval ra, bs.
Example 3.2.43. Consider the ODE
#
u2 ` λ “ g,
0ăxă1
up0q “ up1q “ 0
From hw problem #5,
#
u2 “ f,
0ăxă1
up0q “ up1q “ 0
has solution
ż1
upxq “
Kpx, yqf pyq dy,
0
88
where
#
Kpx, yq “
ypx ´ 1q, 0 ď y ď x ď 1,
xpy ´ 1q, 0 ď x ď y ď 1,
For our problem, use f “ ´λu ` g in above formula and so
ż1
upxq “
Kpx, yqr´λupyq ` gpyqs dy,
ż1
“ ´λ Kpx, yqupyq dy ` Gpxq,
0
0
where
ż1
Gpxq “
Kpx, yqgpyq dy
0
So BVP is equivalent to T u “ u where
ż1
T puqpxq “ ´λ
Kpx, yqupyq dy ` Gpxq
0
This is a contraction if maxx,y |λKpx, yq| ă 1. Since
max |Kpx, yq| “
1
4
ùñ
|λ| ă 4
and so it is a contraction. We can shown that |λ| ă 8 works, with a
slightly refined argument.
4
89
3.2.7
Normed Spaces
Metrics often arise from norms. Let X be a vector space. A norm on X
is a mapping ||¨|| : X Ñ R such that
(1) ||x|| ě 0 for all x P X.
(2) ||x|| “ 0 if and only if x “ 0.
(3) ||αx|| “ α ||x|| for all x P X and α P R.
(4) ||x ` y|| ě ||x|| ` ||y|| for all x, y P X.
The pair pX, ||¨||q is a normed linear space.
Example 3.2.44. Let X “ Rn , Cn and
˜
n
ÿ
|x|p “
¸1{p
|xj |p
,
1ďpă8
j“1
|x|p “ max |xj |,
1ďjďn
p“8
Then ||¨|| “ | ¨ |p is a norm for any such p. Note
dp px, yq “ |x ´ y|p
was the corresponding metric.
4
Example 3.2.45. Let X “ CpKq with K compact. This is a NLS with
||f || :“ max |f pXq|
xPK
dpf, gq “ ||f ´ g||
4
90
Example 3.2.46. Let X “ Lp pΩq. This is a NLS with
˙1{p
ˆż
p
||u||Lp :“
|u| dx
, 1ďpă8
ˇ
(
“ inf M ˇ |upxq| ď M, a.e.
“ ess sup
Ω
||u||L8
ˇ
(
|upxq| ˇ x P Ω ,
p“8
4
When using CpKq, we use ||u|| “ ||u||L8 .
On any NLS we can define dpx, yq “ ||x ´ y|| and all axioms of metric
hold. If the resulting metric space is complete, then X is a Banach space.
Example 3.2.47. The spaces Rn , Cn , CpKq, and Lp pΩq are all Banach
spaces.
4
In a Banach spaces, infinite series can be considered, so
8
ÿ
xj “ x
j“1
if lim Sn “ x, where
Sn :“
n
ÿ
xj
j“1
We claim that if
ř8
j“1
||xj || converges then
ř8
ˇˇ
ˇˇ
8
8
ˇˇ ÿ
ˇˇ ÿ
ˇˇ
ˇˇ
||xj ||
ˇˇ x j ˇˇ ď
ˇˇj“1 ˇˇ j“1
91
j“1
xj converges and
Consider
ˇˇ
ˇˇ
m
ˇˇ ÿ
ˇˇ
ˇˇ
ˇˇ
||Sk ´ Sm || “ ˇˇ
x j ˇˇ
ˇˇj“k`1 ˇˇ
m
ÿ
ď
||xj ||
j“k`1
“ tm ´ tk
if m ą k where
tn “
n
ÿ
||xj ||
j“1
But ttn u is convergent, and hence it is Cauchy, and so tsk u is Cauchy.
Therefore it is also convergent as needed.
Definition 3.2.15. A sequence txn unPN in X is a Shauder basis for X
if for all x P X there is a unique tcj ujPN of scalars such that
x“
8
ÿ
cj x j
j“1
X must be separable if a basis exists. For example, finite linear combinations with rational coefficients are dense.
Example 3.2.48. The space L8 pΩq has no Shauder basis since it isn’t
separable.
4
The question to ask here is whether every separable space has a basis.
The answer is NO! However, “common” spaces like CpKq, Lp pΩq with
1 ď p ă 8 do.
92
3.2.8
Linear Mappings
Now we discuss Linear Mappings between Banach Spaces. Let X and Y
be Banach spaces. We say that T : X Ñ Y is linear if
T pc1 x1 ` c2 x2 q “ c1 T x1 ` c2 T x2
for all x1 , x2 P X and scalars c1 and c2 . The norm of the operator is
given by
||T || :“ sup
x‰0
||T x||
||x||
Example 3.2.49. Consider X “ CpΩq with Ω Ď Rn compact and
ż
T upxq “
Kpx, yqupyq dy
Ω
So
|T upxq| ď M max |upyq|
yPΩ
M :“
max |Kpx, yq|
px,yqPΩˆΩ
Assume that M is finite. Then
ùñ
ùñ
||T u|| ď M ||u||
||T u||
sup
ďM
||u||
||T || ď M
4
93
If ||T || ă 8 then we say that T is a bounded linear operator. Denote
the set of bounded linear operators by
BpX, Y q “
ˇ
(
T : X Ñ Y ˇ T is linear and bounded
This is also a vector space since for T, S P BpX, Y q
ptT ` sSqpxq “ tT x ` sSpxq
It is easy to check that tT ` sS is linear. Must also check that ||tT ` sS||
is finite. Note that
||T x|| ď ||T || ||x|| ,
@x P X
So
||ptT ` sSqpxq|| ď ||tT x|| ` ||sSx||
“ |t| ||T x|| ` |s| ||Sx||
“ |t| ||T || ||x|| ` |s| ||S|| ||x||
“ p|t| ||T || ` |s| ||S||q ||x||
THus
ùñ
||tT ` sS|| ď |t| ||T || ` |s| ||S|| ă 8
||T ` S|| ď ||T || ` ||S||
Can check that ||T || defines a norm on BpX, Y q. Other properties besides
the triangle inequality are easy to check. So, BpX, Y q is a NLS. We can
also show that this forms a Banach Space.
94
Note that ||T || is the Lipschitz constant for T : X Ñ Y . So
dpT x, T yq “ ||T x ´ T y||
“ ||T px ´ yq||
“ ||T px ´ yq||
ď ||T || ||x ´ y||
“ ||T || dpx, yq
Note that T is a contraction if ||T || ă 1 . If Y “ X the notation is BpXq.
In this case, note that
ˇˇ 2 ˇˇ
ˇˇT xˇˇ “ ||T ˝ T x||
“ ||T pT xq||
“ ||T || ||T xq||
“ ||T ||2 ||xq||
and hence ||T 2 || ď ||T ||2 . Similarly, ||T n || ď ||T ||n for n P N. So T a
contraction implies that T n Ñ 0 in the BpXq sense.
In earlier discussion of Neumann series, we said that ||S n´1 u1 || Ñ 0 and
that was true since ||S|| ă 1.
95
Chapter 4
Hilbert Spaces
Let X be a vector space. An inner product on X is a mapping x¨, ¨y :
X ˆ X Ñ C such that
(1) xx, xy ě 0 for all x P X.
(2) xx, xy “ 0 if and only if x “ 0.
(3) xx, yy “ xy, xy for all x, y P X.
(4) xx ` y, zy “ xx, zy ` xy, zy for all x, y, z P X.
(5) xαx, yy “ α xx, yy for all x, y P X and α P R.
Consider the following examples:
Example 4.0.50. Take X “ Rn . Then the natural inner product is
xx, yy “
n
ÿ
j“1
which is the dot product as in calculus.
4
96
xj y j
Example 4.0.51. Take X “ Cn . Then the natural inner product is
xx, yy “
n
ÿ
xj y j
j“1
4
Example 4.0.52. Take X “ L2 pΩq. Then the corresponding inner product is
ż
xu, vy “
upxqvpxq dx
Ω
This is finite since
ˇż
ˇ
ˇ
ˇ
ˇ upxqvpxq dxˇ ď ||u|| ||v||
2
2
ˇ
ˇ
Ω
“ ||u||2 ||v||2
ă8
4
We say that pX, x¨, ¨yq is an inner product space.
Theorem 18. The Cauchy-Schwarz Inequality is given by
| xx, yy | ď
a
a
xx, xy xy, yy
for all x, y P X.
97
Proof. Let x, z P X. Then
0 ď xx ´ z, x ´ zy
ùñ 0 ď xx, xy ´ xx, zy ´ xz, xy ` xz, zy
“ xx, xy ´ 2Re xx, zy ` xz, zy
1
1
ùñ Re xx, zy ď xx, xy ` xz, zy
2
2
The case that y “ 0 is obvious. Else take
z“
xx, yy
y,
xy, yy
and so
F˙
ˆB
xx, yy
y
Re xx, zy “ Re
x,
xy, yy
˜
¸
xx, yy
xx, yy
“ Re
xy, yy
“
| xx, yy |2
xy, yy
Also,
ˇ
ˇ
ˇ xx, yy ˇ2
1
1
ˇ xy, yys
rxx, xy ` xz, zys “ rxx, xy ` ˇˇ
xy, yy ˇ
2
2
1
|xx, yy|2
“ rxx, xy `
s
xy, yy
2
So,
98
ùñ
ùñ
1 |xx, yy|2
1
ď xx, xy
2 xy, yy
2
2
|xx, yy| ď xx, xy xy, yy
a
a
|xx, yy| ď xx, xy xy, yy
as needed.
Define ||x|| :“
a
xx, xy. So the C-S inequality is
| xx, yy | ď ||x|| ||y||
for all x, y P X.
Theorem 19. The function ||¨|| as defined above is a norm on the inner
product space of X.
Proof. It is clear that ||x|| ě 0 and equality is attained if and only if
x “ 0. Then note that
a
xαx, αxy
a
“ |α|2 xx, xy
a
“ |α| xx, xy
“ |α| ||x||
||αx|| “
Also,
99
||x ` y||2 “ xx ` y, x ` yy
“ xx, xy ` xx, yy ` xy, xy ` xy, yy
“ ||x||2 ` 2Re xx, yy ` ||y||2
ď ||x||2 ` 2 ||x|| ||y|| ` ||y||2
“ p||x|| ` ||y||q2
shows that the triangle inequality holds.
Any inner product space may be regarded as a normed linear space.
Definition 4.0.16. If X is complete with respect to this norm, it is a
Hilbert space.
Example 4.0.53. The spaces Rn and Cn with
g
fÿ
a
fn
||x|| xx, xy “ e |xj |2 “ |x|2
j“1
are Hilbert spaces.
4
Example 4.0.54. The space L2 pΩq with norm
dż
a
||u|| xu, uy “
|u|2 dx “ ||x||L2
Ω
is a Hilbert space.
4
100
Example 4.0.55. Consider the space `2 of infinite square summable
sequences x “ txn unPN .
#
`2 “
8
ˇ ÿ
ˇ
|xn |2 ă 8
txn unPN
+
n“1
Elements of `2 are added component-wise and same for scalar multiplication. Recalling the Minkowski inequality for sequences, `2 is a vector
space. We define
8
ÿ
xx, yy :“
xn y n ,
n“1
for x, y P `2 . The inner product axioms can be seen to hold. Then
g
f8
fÿ
|xn |2
||x|| “ e
n“1
is the norm. Completeness holds and thus `2 is a Hilbert space.
4
Other sequence spaces are possible. For example:
#
`2 pZq “
txn unPZ
8
ˇ ÿ
ˇ
|xn |2 ă 8
n“1
Recall that in R3
cos θ “
x¨y
||x|| ||y||
101
+
where θ is the angle between x and y. We can use this angle concept in
any inner product space:
cos θ “
xx, yy
||x|| ||y||
Definition 4.0.17. If xx, yy “ 0 we say that x and y are orthogonal (or
perpendicular) vectors, x K y.
If x K y then
||x ` y||2 “ xx, xy ` 2Re xx, yy ` xy, yy
“ ||x||2 ` ||y||2
gives the Pythagorean theorem. Also, calculation shows that
||x ` y||2 ` ||x ´ y||2 “ 2 ||x||2 ` 2 ||y||2
which is the Parallelogram law.
Example 4.0.56. Consider X “ L2 r0, 1s and take u “ sinpπxq and
v “ cospπxq. THen
ż1
xu, vy “
sinpπxq cospπxq dx
0
sinpπxq ˇˇ1
ˇ
2π 0
“0
“
shows that u K v.
4
102
Definition 4.0.18. If M Ď X and x K y for all y P M , we’ll say that
x K M.
MK “
ˇ
(
xPX ˇxKM
is the orthogonal complement of M.
Usually M is a subspace, but doesn’t have to be. It is easy to check that
M K is always a subspace. Note that M Y M K Ď t0u since x P M Y M K
implies that xx, xy “ 0 and thus x “ 0.
Example 4.0.57. Consider X “ R3 .
M“
ˇ
(
x P X ˇ x1 “ x3 “ 0 “ Lte3 u
implies that
MK “
ˇ
(
x P X ˇ xx, e3 y “ 0
ˇ
(
x P X ˇ x3 “ 0
“
“ Lte1 , e2 u
4
Example 4.0.58. Consider X “ L2 pΩq with M “ Lt1u, the constant
functions. Then
v P M K ðñ xv, 1y “ 0
ż
ùñ
vpxq dx “ 0
Ω
and hence M K is the set of functions of zero mean.
4
103
It can be shown that M K is always closed. We give a reason: Let xn P M K
such that xn Ñ x in X. For any y P M , we have xxn , yy “ 0 and
| xxn , yy ´ xx, yy | “ | xxn ´ x, yy |
ď ||xn ´ x|| ||y||
and this tends to zero as n Ñ 8. So 0 “ xxn , yy Ñ xx, yy implies that
xx, yy “ 0 and thus x P M K as needed to show the complement is closed.
4.1
Projections on Hilbert Spaces
Definition 4.1.1. If M Ď X then y “ projM x denotes the projection of x onto M .
This is the closest point in M to x. That is,
||y ´ x|| ď ||z ´ x||
for all z P M or y “ arg minzPM ||z ´ x||.
Note that y need not exists or be unique in general.
Example 4.1.1. Fix e P X nonzero with M “ Ltxu. Suppose x R M
and suppose that y P M is the closet point in M to x. Look at
φptq :“ ||y ´ x ` te||2 ,
tPR
Then φptq ě φp0q for all t P R. This implies that φ1 p0q “ 0 and so
φptq “ ||y ´ x||2 ` 2Re xx ´ y, tey ` t2 ||e||2
“ ||y ´ x||2 ` 2tRe xx ´ y, ey ` t2 ||e||2
and so φ1 p0q “ 2tRe xx ´ y, ey if such y exists. Similarly, look at
104
φptq :“ ||y ´ x ` ite||2 ,
tPR
to get
0 “ φ1 p0q “ 2Rep´i xy ´ x, eyq
“ 2Im xy ´ x, ey
and so xy ´ x, ey “ 0 or xy, ey “ xx, ey. But y “ αe for some α. So
α xe, ey “ xx, ey
xx, ey
ùñ y “
e “ projM pxq
xe, ey
Note that y ´ x K y since
xy ´ x, yy “ α xy ´ x, ey
and x “ y ` px ´ yq for y P M and x ´ y P M K .
4
Example 4.1.2. Let X “ R3 , e “ p1, 2, 3q and x “ p4, 0, 1q. Then
105
xx, ey
e
xe, ey
7
“ p1, 2, 3q
14
“ p1{2, 1, 3{2q
y“
4
Theorem 20. Let M Ď X a closed subspace. Then for all x P X there
is a unique y P M and z P M K such that x “ y ` z. Here
y “ projM x,
z “ projM x
Proof. First the uniqueness part. Suppose that x “ y1 ` z1 “ y2 ` z2 for
y1 , y2 P M and z1 , z2 P M K . Then
ùñ
y1 ´ y2 “ z2 ´ z1
ùñ
ùñ
y1 ´ y2 P M Y M K “ t0u
y1 “ y2 , z2 “ z1
as needed to obtain uniqueness. Now we show the existence part. Let
d “ inf ||x ´ u|| ě 0
uPM
So there is un P M such that
||x ´ un || “ dn Ñ d
106
Apply parallelogram law with x ´ un and x ´ um so that
||un ´ um ||2 “ ´ ||2x ´ un ´ um ||2 ` 2 ||x ´ un ||2 ` 2 ||x ´ um ||2
The first term on the right is equal to
ˇˇ
un ´ um ˇˇˇˇ2
ˇˇ
4 ˇˇx “
ˇˇ ě 4d2
2
since pun ´ um {2 P M . As n, m Ñ 8
||un ´ um ||2 ď ´4d2 ` 2d2n ` 2d2m Ñ 0
This implies that tun u is Cauchy, and so there is u P M such that un Ñ y,
since M is closed.
d ď ||x ´ y|| “ lim ||x ´ un || “ d
nÑ8
ùñ ||x ´ y|| “ d
ùñ y “ arg min ||x ´ z|| “ projM x
zPM
Now let z “ x ´ y and so x “ y ` z. We claim that z P M K . Note
||x ´ y||2 ď ||x ´ y ` λu||2
for any u P M and λ P C. Then the right-hand side above equals
||x ´ y||2 ` 2Re xx ´ y, λuy ` |λ|2 ||u||2
ùñ 2Repλ xy ´ x, uyq ď |λ|2 ||u||2
107
Choose
λ“
xy ´ x, uy
||u||2
and hence
2
| xy ´ x, uy |2
| xy ´ x, uy |2
ď
||u||2
||u||2
which occurs if xy ´ x, uy “ 0. Then xz, uy “ 0 gives our claim that
z P M K.
Finally, if u P M K
||x ´ u||2 “ ||x ´ z ` z ´ u||2
“ ||x ´ z||2 ` 2Re xx ´ z, z ´ uy ` ||z ´ u||2
ď ||x ´ z||2
ùñ z “ arg min ||x ´ u|| “ projM K pxq
uPM K
as needed.
If M, N Ď X then
M `N “
ˇ
(
z “ x ` y ˇ x P M, y P N
. If M Y N “ t0u, x, y are unique in which case
is the sum of M and NÀ
we use the notation M ÀN to denote the direct sum.
We have shown X “ M
M K (with M a closed subspace of X) to be
an orthogonal direct sum.
From now on, we write PM x “ projM x. We claim that PM : X Ñ X is
linear. If x “ c1 x1 ` c2 x2 then
108
x“y`z
x1 “ y1 ` z1
x2 “ y2 ` z2
as in theorem. Then
ùñ y ` z “ c1 py1 ` z1 q ` c2 py2 ` z2 q
ùñ y ´ c1 y1 ´ c2 y2 “ c1 z1 ` c2 z2 ´ z P M Y M K
This implies that both sides are zero. So
y “ PM x “ c1 PM x 1 ` c2 P M x 2 “ . . .
etc. We claim that PM is bounded. Note x “ y ` z implies
||x||2 “ ||y||2 ` ||z||2
2
(expand with inner product)
“ ||PM x|| ` ||PM K x||
2
ď ||PM x||2
This implies that ||PM || ď 1. For x P M we have PM x “ x and hence
||PM || “ 1 (unless M “ t0u in which case PM “ 0).
How to compute PM ? Suppose
M “ Lte1 , . . . , en u
xej , ek y “ 0, j ‰ k
109
We say te1 , . . . , en u is an orthogonal set, also if ||ej || “ 1 for all j it is
orthonormal.
Recall
PM x “ arg min ||x ´ z|| ,
xPM
so y “ PM x solves a “best approximation” problem. Suppose that M “
Lte1 , . . . , en u where te1 , . . . , en u is an orthogonal set. Then
PM x “
n
ÿ
xx, ej y
ej
xe
,
e
y
j
j
j“1
We did the n “ 1 case earlier, if Mj “ Ltej u, this says
PM x “
n
ÿ
PM j
j“1
Let
n
ÿ
xx, ej y
y“
ej
xej , ej y
j“1
So y P M . Pick z P M and so
||x ´ z||2 “ ||px ´ yq ` py ´ zq||2
“ ||x ´ y||2 ` 2Re xx ´ y, y ´ zy ` ||y ´ z||2
Note, that
110
n
ÿ
xx, ej y
xej , ek y
xx ´ y, ek y “ xx, ek y ´
xej , ej y
j“1
“ xx, ej y ´ xx, ek y
“0
This implies that x ´ y P M K and so x ´ y K y ´ z since y ´ z P M . Then
||x ´ z||2 “ ||x ´ y||2 ` 2p0q ` ||y ´ z||2
ď ||x ´ y||2
The above shows that y “ PM x as needed.
For te1 , . . . , en u orthonormal,
PM x “
n
ÿ
xx, ej y ej
j“1
and also in this case,
||PM x|| “
n
ÿ
| xx, ej y |2
j“1
since all cross terms are zero, or in general
ˇˇ
ˇˇ g
n
n
ˇˇ ÿ
ˇˇ f
ÿ
ˇˇ
ˇˇ f
e
|cj |2
ˇˇ cj ej ˇˇ “
ˇˇj“1
ˇˇ
j“1
111
4.1.1
Gram-Schmidt Process
What if M “ Ltx1 , . . . , xn u and x1 , . . . , xn aren’t orthogonal? In this
case we use Gram-Schmidt procedure to replace x1 , . . . , xn by an equivalent orthonormal set.
Without loss of generality, let x1 , . . . , xn be linearly independent. So let
e1 “
x1
||x1 ||
Now,
ẽ2 :“ x2 ´ xx2 , e1 y e1
ẽ2
e2 :“
||ẽ2 ||
Note that ẽ2 K e1 , ẽ2 ‰ 0 and Lte1 , e2 u “ Ltx1 , x2 u. Next,
ẽ3 :“ x3 ´ xx3 , e1 y e1 ´ xx3 , e2 y e2
ẽ3
e3 :“
||ẽ3 ||
Note that ẽ3 K e1 , e2 , ẽ3 ‰ 0 and Lte1 , e2 , e3 u “ Ltx1 , x2 , x3 u. At each
step
K xk
ẽk :“ xk ´ PMk´1 xk “ PMk´1
Mk´1 “ Lte1 , . . . , ek´1 u
ẽk
ek :“
||ẽk ||
112
With the equivalent orthonormal basis of M we can use
PM x “
n
ÿ
xx, ej y ej
j“1
Infinite dimensional cases
Let txn unPN be a sequence
ř in a Hilbert space H. We have already discussed convergence of 8
j“1 cn xn . We say txn u is linearly independent
if
8
ÿ
cn x n “ 0
ðñ
cn “ 0, @n P N
j“1
Here txn u is a basis if they are linearly independent and Ltx1 , . . . u “ H.
We say txn u is an orthogonal basis if xj K xk “ 0 for j ‰ k and is a
basis. If
x“
8
ÿ
cj x j ,
SN :“
j“1
N
ÿ
cj x j ,
j“1
then SN Ñ x.
xSN , xk y “
N
ÿ
cj xxj , xk y “ ck xxk , xk y
j“1
if N ą K. Let N Ñ 8 and hence
113
ck “
ùñ x “
xx, xk y
xxk , xk y
8
ÿ
xx, xn y
xxn , xn y
n“1
xn “ lim PMN x
N Ñ8
where MN “ Ltx1 , . . . , xn u. Note that PMN Ñ I the identity operator.We say
xx, xn y
xxn , xn y
is the n-th generalized Fourier coefficient.
Let H be a Hilbert space and let ten unPN be orthonormal in H. That is
#
xej , ek y “
0, j ‰ k,
1, j “ k,
If
x“
8
ÿ
cn e n
j“1
then cn “ xx, en y. Formally this can be derived as follows:
114
x“
8
ÿ
cn e n
j“1
C
ùñ
xx, ek y “
8
ÿ
G
cn en , ek
j“1
8
ÿ
“
cn xen , ek y
j“1
“ ck
The second to the last equality needs justification. Similarly,
xx, yy “
8
ÿ
j“0
8
ÿ
“
cn xen , yy
xx, en y xen , yy
j“0
can be justified. In particular, choosing x “ 0 we get
0“
8
ÿ
cn e n
ðñ
cn “ 0, @n P N
n“1
So ten unPN is linearly independent.
Theorem 21. (Riesz-Fischer)
8
ÿ
cn e n ă 8
8
ÿ
ðñ
n“1
n“1
115
|cn |2 ă 8
Proof. Let
SN “
N
ÿ
cn e n ,
tN “
n“1
N
ÿ
|cn |2
n“1
Then
ˇˇ
ˇˇ 2
N
ˇˇ ÿ
ˇˇ
ˇˇ
ˇˇ
||SN ´ SM ||2 “ ˇˇ
cn en ˇˇ
ˇˇn“M `1
ˇˇ
N
ÿ
“
|cn |2
n“M `1
“ tN ´ tM
So tSN u is Cauchy in H if and only if ttN u is Cauchy in R. This implies
tSN u is convergent if and only if ttN u is convergent as needed.
Theorem 22. (Riemann-Lebesgue Lemma) If x P H then xx, en y Ñ 0
as n Ñ 8.
ř
Proof. Let x P H and so |cn |2 ă 8. It follows that ck Ñ 0 and this is
equivalent to xx, en y Ñ 0 as needed.
Another way to state the Riesz-Fischer theorem is that x P H if and only
if tcn u P `2 where cn “ xx, en y.
Letting SN “ PMN x and MN “ Lte1 , . . . , en u. Then
116
||PM x|| ď ||PM || ||x|| ď 1 ¨ ||x|| “ ||x||
ùñ ||SN || ď ||x||
N
ÿ
ùñ
n“1
8
ÿ
ùñ
|cn |2 ď ||x||2
| xx, en y |2 “
n“1
8
ÿ
|cn |2 ď ||x||2
n“1
The above is known as Bessel’s inequality.
Definition 4.1.2. Let txn unPN be a sequence in H. Then txn u is closed
in H if the set of all finite linear combinations of txn u is dense in H.
Definition 4.1.3. Let txn unPN be a sequence in H. Then txn u is complete
if there is no nonzero vector orthogonal to xn for all n. That is
xxn , xy “ 0, @n P N
ðñ
x“0
Definition 4.1.4. Let txn unPN be a sequence in H. If txn u is orthonormal, we say they are a maximal orthonormal set if it is not contained in
any strictly larger orthonormal set.
Theorem 23. Let ten unPN be orthonormal in H. Then the following are
equivalent:
(1) ten u is maximal orthonormal.
(2) ten u is closed.
(3) ten u is complete.
(4) ten u is a basis of H.
ř
2
(5) ||x||2 “ 8
n“1 | xx, en y | for all x P H.
Proof.
117
p1q ùñ p2q Suppose maximal, not closed. Let M be the set of finite
linear combinations of ten u. Then M ‰ M and so pick x P HzM , and
K
PM K x ‰ 0. So there is e P M , ||e|| “ 1 and xe, en y “ 0 for all n P N.
Thus ten u Y teu is a larger orthonormal set.
p2q ùñ p3q Suppose closed, not complete. So there is x P H, x ‰ 0,
xx, en y “ 0 for all n P N. Fix ą 0. Then there is c1 , . . . , cn such that
ùñ
ˇˇ
ˇˇ2
n
ˇˇ
ˇˇ
ÿ
ˇˇ
ˇˇ
c j e j ˇˇ ă ˇˇx ´
ˇˇ
ˇˇ
j“1
C
G
n
n
ÿ
ÿ
||x|| ´ 2Re x,
cj e j `
|cj |2 ă j“1
j“1
2
ùñ ||x|| ă @ ą 0 ùñ
ùñ x “ 0
||x|| “ 0
as needed.
p3q ùñ p4q Assume complete. We know ten u are linearly in depended
Pick x P H. We know
8
ÿ
| xx, en y |2 ď ||x||2
n“1
ùñ y :“
8
ÿ
xx, en y en
n“1
exists in H by Riesz-Fischer theorem. Must show y “ x. Then
xx ´ y, ek y “ xx, ek y ´ xy, ek y
“ xx, ek y ´ xx, ek y
“0
118
This implies that x ´ y “ 0 by completeness. Therefore
8
ÿ
x“y“
xx, en y en
n“1
and so ten u is a basis.
p4q ùñ p5q Pick x P H, so x “
ř8
n“1
xx, en y en . This implies that
ˇˇ2
ˇˇ
G
C
N
N
N
ˇˇ
ˇˇ
ÿ
ÿ
ÿ
ˇˇ
ˇˇ
2
| xx, en y |2
xx, en y en ˇˇ “ ||x|| ´ 2Re x,
xx, en y en `
ˇˇx ´
ˇ
ˇ
ˇˇ
n“1
n“1
n“1
Ñ0
Thus
N
ÿ
2
||x|| ´
ùñ
||x||2 “
n“1
8
ÿ
| xx, en y |2 Ñ 0
| xx, en y |2
n“1
p5q ùñ p1q If ten u is not maximal, there is e P H, ||e|| “ 1 and xe, en y “
0 for all n P N. By (5)
||e||2 “
8
ÿ
| xe, en y |2 “ 0
n“1
and so e “ 0 which is a contradiction.
119
A 6th equivalent condition,
8
ÿ
xx, en y xen , yy “ xx, yy
n“1
for all x, y P H. The above is known as Parseval’s equality. Refer to 6.28
in text.
H must be separable in order for an orthonormal basis to exists. There
do exist non separable Hilbert spaces. The above discussion can be
generalized to uncountable orthonormal sets, see text.
Theorem 24. Every Hilbert space has an orthonormal basis. Refer to
6.29 in text.
Example 4.1.3. Consider X “ Rn , Cn . Then
e1 “ p1, 0, . . . , 0q
e2 “ p0, 1, . . . , 0q
..
.
en “ p0, 0, . . . , 1q
is an orthonormal basis.
4
Example 4.1.4. Consider X “ L2 p0, 2πq and let un pxq “ einx with
n P Z. Then tun u is an orthogonal set since
ż 2π
einx e´imx dx
xun , um y “
ż02π
eipn´mqx dx
“
0
“0
120
given that n ‰ m and it is 2π if n “ m. Let
einx
?
en pxq :“
2π
ùñ
ten u is o.n.
For f P L2 p0, 2πq
einx
cn ?
2π
nPN
ż
1
f pxqeinx dx
ùñ cn “ xf, en y “
2π r0,2πs
ÿ
f pxq “
We’ll eventually show that ten u is an o.n. basis of L2 p0, 2πq.
4
In any separable Hilbert space you can find an o.n. basis as follows:
1) Take any countable dense set txn unPN .
2) Apply Gram-Schmidt procedure.
Example 4.1.5. In L2 p0, 2πq take un pxq :“ xn with n P N0 . You’ll get
orthogonal polynomials.
4
Theorem 25. (Weierstrass) Polynomials are dense in Cra, bs. Refer to
Theorem 2.9 in text.
Theorem 26. Polynomials are dense in L2 pa, bq.
Proof. If ą 0, f P Cra, bs, so there is p P P such that ||f ´ p||8 ă .
So
ż
||f ´
p||2L2
|f ´ p|2 dx ď pb ´ aq ||f ´ p||2L8
“
ra,bs
ùñ
||f ´ p||L2
?
ď b ´ a
121
Finally, if f P L2 pa, bq there is g P Cpa, bq such that ||f ´ g||L2 ă (Refer
to any measure theory book). So if p P P approximates g as above,
||f ´ p||L2 ď ||f ´ g||L2 ` ||g ´ p||L2
?
ă ` b ´ a
as needed.
So the set P of finite linear combinations of u0 , u1 , . . . is closed in L2 pa, bq.
If p0 , p1 , . . . are orthonormal polynomials obtained by Gram-Schmidt.
Thus tp0 , p1 , . . . u are closed, o.n. and hence a basis of L2 pa, bq.
4.2
Linear Functionals on Hilbert Spaces
Any ` : H Ñ R or C is a functional. If
`pax ` byq “ a`pxq ` b`pyq
it is a linear functional.
Example 4.2.1. The inner product `pxq “ xx, yy, for some fixed y P H
is a linear functional.
4
Example 4.2.2. Let H “ L2 p´1, 1q and `puq “ up0q. Then ` is not
well-defined on all of H, but is on subspaces like Cr´1, 1s
4
Example 4.2.3. Let `pxq “ ||x||. This is a nonlinear functional.
4
Definition 4.2.1. A linear functional ` is said to be bounded on H if
122
"
sup
|`pxq| ˇˇ
x‰0
||x||
*
“Că8
That is, ` P BpH, Cq. We could use R instead of C as well.
We denote H ˚ “ BpH, Cq to be the dual space of H.
Example 4.2.4. Let `pxq “ xx, yy. Then for x ‰ 0
||x|| ||y||
|`pxq|
ď
“ ||y|| ă 8
||x||
||x||
and hence ` P H ˚ .
4
Example 4.2.5. Let `pxq “ up0q as above. Then ` R H ˚ . (See HW
problem # 40)
4
Theorem 27. (Riesz Representation Theorem) If ` P H ˚ , then there is
a unique y P H such that `pxq “ xx, yy.
Proof. Let
M “ N p`q “
ˇ
(
x P H ˇ `pxq “ 0
(we call this the null space or kernel of `), which is a closed subspace of
H (Refer to text).If M “ H, y “ 0 works, else there is e P M K with
norm one.
Pick x P H and let z “ `pxqe ´ `peqx. Then
123
ùñ
ùñ
ùñ
ùñ
ùñ
`pzq “ `pxq`peq ´ `peq`pxq “ 0
zPM
zKe
x`pxqe ´ `peqx, ey “ 0
`pxq xe, ey “ `peq xx, ey
A
E
`pxq “ x, `peqe
and so y :“ `peqe works. If `pxq “ xx, y1 y “ xx, y2 y, then
xx, y1 ´ y2 y “ 0,
@x P H
Choose x “ y1 ´ y2 and hence
||y1 ´ y2 || “ 0 ùñ y1 “ y2
as needed to show that y P H is unique.
Note that ||y|| “ ||`||, where
||`|| “ sup
x‰
|`pxq|
||x||
The above is the norm in BpH, Cq. The reason is as follows: We have
already that ||`|| ď ||y||. But
124
`pyq “ xy, yy “ ||y||2
|`pxq|
“ ||y||
ùñ
||x||
ùñ ||`|| ě ||y||
as needed.
The map R : H Ñ H ˚ , namely Rpyq “ `, is linear, one to one, and
||Rpyq|| “ ||y||
The map R is known as Riesz map and it is an isometry of H, H ˚ .
Let ` : H Ñ C is a linear functional on the Hilbert space H. We don’t
always mean that ` is defined for all x P H. Instead, there exists a
domain of ` which we denote Dp`q and
ˇ
(
x P H ˇ `pxq is define
Dp`q “
and could write ` : Dp`q Ď H Ñ C. Note that Dp`q must be a subspace.
Also,
"
||`|| “ sup
|`pxq| ˇˇ
x P Dp`q, x ‰ 0
||x||
*
in this case. If ||`|| ă 8, then ` can be extended to Dp`q. THat is there
exists `1 a linear functional on H, with `1 pxq “ `pxq on Dp`q and `1
defined on Dp`q. Furthermore, ||`1 || “ ||`||.
125
The idea is to pick x P Dp`q, then there is xn P Dp`q with xn Ñ x. So
|`pxn q ´ `pxm q| “ |`pxn ´ xm q| ď ||`|| ||xn ´ xm || Ñ 0
as n, m Ñ 8. Hence t`pxn qu is Cauchy in C. It follows that
lim `pxn q
nÑ8
exists. Define it to be `1 pxq. Check definition of `1 is unique, that is is
linear, and that ||`1 || “ ||`||. Finally, we can define
`2 pxq “ `1 pP xq
where P is the projection onto Dp`q, and `2 pxq “ `pxqon Dp`q and `2 :
H Ñ C is linear. Also
||`2 || “ ||`1 || “ ||`||
So, any such ` can be extended to all of H. Any such bounded linear
functional may be assumed defined on all H.
If dimpDp`qq ă 8, then ||`|| ă 8 must hold. The reason as follows: Let
Dp`q “ Lte1 , . . . , en u be an o.n. set. Then using Bessel’s inequality
126
x“
n
ÿ
cj ej
j“1
ùñ `pxq “
n
ÿ
cj `pej q
j“1
g
g
fn
fÿ
f n 2f ÿ
|`pxq| ď e cj e `pej q2
j“1
j“1
“ C ||x||
where
g
fÿ
f n
C :“ e `pej q2 ă 8
j“1
Therefore, ||`|| ď C as needed.
4.3
Fourier Series
Let H “ L2 p0, 2πq. Define
un pxq “ einx ,
einx
en pxq “ ? ,
2π
tun unPZ are orthogonal in H
are o.n. in H.
Theorem 28. ten unPZ are an o.n. basis of H.
So, if f P L2 p0, 2πq then
127
f pxq “
ÿ
cn en pxq
nPZ
ż 2π
f pxqen pxq dx “ fpn
cn “ xf, en y “
0
Here, the partial sums are
N
ÿ
SN pxq “
cn en pxq
n“´N
and statement is that
lim ||SN ´ f ||L2 “ 0
N Ñ8
It does not say that SN Ñ f pointwise, but does imply a subsequence
coverless a.e.
Bessel’s inequality is given by
ż 2π
|f pxq|2 dx “
0
ÿ
|fpn |2 “
nPZ
ÿ
|cpn |2
nPZ
Keep in mind that the !goal )
is to show that teinx u is closed in L2 p0, 2πq.
einx
Follows from this that ?
is an o.n. basis of L2 p0, 2πq.
2π
Any finite linear combination
N
ÿ
cn einx
n“´N
128
is a trigonometric polynomial. So we are dense in L2 p0, 2πq. Note that
nay such function is defined on all of R, as a 2π periodic function. We
say f is 2π periodic if f px ` 2πq “ f pxq for all x.
If is common to regard f as a function on T “ unite circle, meaning
r0, 2πs with 0 and 2π regarded as the same point. The angle x is a
variable in this point of view.
Formally, T “ R mod 2πZ, i.e. any two x’s differing by 2πn are the
same.
Any function on T can be regarded as 2π periodic on R.
For periodic f
ż
ż
f pxq dx “
r0,2πs
f pxq dx,
@a
ra,a`2πs
T is a compact metric space:
dpx, yq “ shorter arc length from x to y
and so CpTq is a Banach space. If f P Cr0, 2πs then f P CrTs if and only
if f p0q “ f p2πq. Similarly, L2 p0, 2πq can be regarded as L2 pTq.
Theorem 29. CpTq is dense in L2 pTq.
Proof. It is enough to show that trig polynomials are dense in CpTq. If
f P L2 pTq, then for all ą 0 there is g P CpTq such that
||f ´ g||L2 ă There is h1 trig polynomial such that
129
||f ´ h||L8 ă This implies that
ż 2π
||g ´
h||2L2
|g ´ h|2 dx ă 2 p2πq
“
0
ùñ
||f ´ h||2L2 ď ||f ´ g||2L2 ` ||g ´ h||2L2 ă p1 `
?
2πq
Key device uses convolution of two functions
ż 2π
pf ˚ gqpxq “
f px ´ yq gpyq dy
0
The above is the convolution of functions f and g in CpTq. Also, by a
change of variables
ż a`2π
pf ˚ gqpxq “
f px ´ yq gpyq dy
ża2π
f pyq gpx ´ yq dy
“
0
“ pg ˚ f qpxq
Also, f ˚ g P CpTq. Let, φn pxq :“ cn p1 ` cos xqn , where cn is chosen such
that
ż 2π
φn pxq dx “ 1
0
If f P CpTq and hn :“ φn ˚ f , then we claim that
130
(i) hn is a trig polynomial.
(ii) hn Ñ f as n Ñ 8 in CpTq.
Proof.
(i) Consider,
ˆ
φn “ cn
eix ` e´ix
1`
2
n
ÿ
˙2
anj eijx
“
j“´n
The above shows that φn is a trig polynomial. Then
hn pxq “
ż 2π ÿ
n
0
anj eijpx´yq f pyq dy
j“´n
n
ÿ
“
j“´n
n
ÿ
“
ż 2π
ijx
anj e´ijy f pyq dy
e
0
cn eijx
j“´n
(ii) Some properties of φn ,
131
φn ě 0,
ż 2π
φn pxq dx “ 1
0
ż
lim
φn dx “ 0,
nÑ8 δď|x|ďπ
@δ ą 0
Reason:
żπ
1“2
φn pxq dx
ż0π
cn p1 ` cos xqn dx
0
żπ
ě 2cn p1 ` cos xqn sin x dx
“2
0
p1 ` cos xqn`1 ˇˇπ
“ ´2cn
ˇ
n`1
0
2n`2
“
cn
n`1
ùñ
cn ď
n`1
2n`2
So, for δ ă |x| ă π, and so
132
φn pxq ď φn pδq
“ cn p1 ` cos δqn
ˆ
˙n
n ` 1 1 ` cos δ
ď
4
2
Ñ 0,
as n Ñ 8 since 1 ` cos δ ă 2 and so
ż
n`1
φn pxq dx ď 2pπ ´ δq
4
δă|x|ăπ
ˆ
1 ` cos δ
2
˙n
Ñ0
ş
Now, using φn “ 1,
ż 2π
hn pxq ´ f pxq “
φn pyqf px ´ yq dy ´ f pxq
ż02π
φn pyqrf px ´ yq ´ f pxqs dy
“
0
Fix ą 0 and so there is δ ą 0 such that
|f px ` hq ´ f pxq| ă If |h| ă δ, so
ˇż
ˇ
|hn pxq ´ f pyq| ď ˇˇ
|y|ăδ
ˇ
ˇ
φn pyqrf px ´ yq ´ f pxqs dy ˇˇ
ˇż
ˇ
` ˇˇ
δă|y|ăπ
133
ˇ
ˇ
φn pyqrf px ´ yq ´ f pxqs dy ˇˇ
In first term,
|f px ´ yq ´ f pxq| ď and hence
ż
1st term ď |φn pyq| dy ď ż
2nd term ď 2 ||f ||L8
φn pyq dy Ñ 0
|y|ăδ
δă|y|ăπ
as n Ñ 8. Thus
lim sup |hn ´ f | ď , @ ą 0
nÑ8
ùñ
ùñ
lim sup |hn ´ f | “ 0
nÑ8
lim ||hn ´ f ||L8 “ 0
nÑ8
as needed.
In proving completeness of teinx unPZ , we made use of a sequence tφn u in
CpTq satisfying
(a) φn pxq ě 0, @x.
ż
(b)
φn pxq dx “ 1,
@n.
r´π,πs
ż
(c) lim
nÑ8 δă|x|ăπ
φn pxq dx “ 0,
@δ ą 0.
134
Theorem 30. If f P CpTq and tφn u satisfies (a),(b) and (c) above, then
φn ˚ f Ñ f in CpTq. We proved this and it is thm 7.2 in text.
Equivalently,
ż
Tn f :“ φn ˚ f “
φn px ´ yqf pyq dy,
r´π,πs
then Tn f Ñ f or Tn Ñ I. For this reason, tφn u is called an approximate identity.
Here is a picture of the proof:
ż
pφn ˚ f qpxq “
φn px ´ yqf pyq dy
r´π,πs
ż
φn px ´ yqf pxq dy
«
r´π,πs
ż
“ f pxq
φn px ´ yq dy
r´π,πs
“ f pxq
We say pφn ˚ f qpxq is the local average of f near x. Continuity of φn not
always needed.
Example 4.3.1. Consider
#
n, 0 ă x ă 1{n
φn pxq “
0, otherwise
135
ż
pφn ˚ f qpxq “
φn px ´ yqf pyq dy
żx
“n
f pyq dy
x´1{n
ÝÑ f pxq
This is the fundamental theorem of calculus:
d
dx
żx
f ptq dt “ f pxq
0
4
Now we discuss other examples of Fourier series:
Example 4.3.2. Let I “ pa, a ` 2πq, f P L2 pIq. Then
f pxq “
ÿ
cn einx
nPZ
ż a`2π
cn “
1
2π
e´inx f pxq dx
a
4
i2πnx
Example 4.3.3. Let I “ pa, a ` Lq, and t e ?LL unPZ is o.n. i nL2 pIq.
For f P L2 pIq,
136
f pxq “
cn “
1
L
ÿ
cn e
i2πnx
L
nPZ
ż a`L
f pxqe´
i2πnx
L
dx
a
4
Example 4.3.4. The Real Form of a Fourier series:
ÿ
nPZ
cn e
inx
ÿ
“
´
cn
¯
cospnxq ` i sinpnxq
nPZ
“ c0 `
´
ÿ
cn
¯ ÿ
´
¯
cospnxq ` i sinpnxq `
c´n cospnxq ´ i sinpnxq
nPN
“ c0 `
ÿ
nPN
pcn ` c´n q cospnxq `
“: a0 `
ipcn ´ c´n q sinpnxq
nPN
nPN
ÿ
ÿ
an cospnxq `
nPN
ÿ
bn sinpnxq
nPN
where a0 :“ c0 , an :“ cn ` c´n and bn :“ ipcn ´ c´n q. So
ż
ż
1 2π
1 2π
´inx
an “
f pxqe
dx `
f pxqeinx dx
2π 0
2π 0
ż
1 2π
“
f pxq cospnxq dx
π 0
bn “ . . .
ż
1 2π
“
f pxq sinpnxq dx
π 0
1
a0 “
2π
ż 2π
f pxq dx
0
137
Here t1, cospnxq, sinpnxqu is an orthogonal basis of L2 p0, 2πq or L2 pa, a `
2πq.
4
Example 4.3.5. The Sine and Cosine Fourier Series: Let f P L2 p0, πq.Extend
f to be even on p´π, πq.
We’ll then have bn “ 0 if we apply previous formula to extended f , using
p´π, πq in place of p0, 2πq. This implies that
f pxq “ a0 `
ÿ
an cospnxq
nPN
ż
1 π
f pxq cospnxq dx
an “
π ´π
ż
2 π
“
f pxq cospnxq dx
π 0
Similarly by making odd extensions
f pxq “
ÿ
bn sinpnxq
nPN
bn “ . . .
ż
2 π
“
f pxq sinpnxq dx
π 0
8
2
tcospnxqu8
n“0 and tsinpnxqun“1 are orthogonal bases of L p0, πq. This is
i2πnx
also true for te L unPZ .
4
138
4.3.1
Pointwise convergence of Fourier Series
We know snk Ñ f a.e. for some subsequence. For f more regular (i.e.
more differentiable) this can be improved. Suppose that f P L2 pTq is
piecewise C 1 , i.e. there is π “ x0 ă x1 ă ¨ ¨ ¨ ă xm “ π such that
f P C 1 rxj , xj`1 s for all j.
Let
#
f˜pxq “
f pxq,
x ‰ xj
1
rf pxj `q ` f pxj ´qs, x “ xj
2
Theorem 31. With these assumptions,
sN pxq Ñ f˜pxq,
@x,
in particular, if f P C 1 pTq, then sN pxq Ñ f pxq for all x.
If f P CpTq only, this may be false, piecewise C 1 condition is needed.
We showed φn ˚ f Ñ f , for f P CpTq, but φn ˚ f ‰ sn . Some trig
polynomials converge pointwise to f , but not necessarily sn .
Let f P L2 p0, 2πq, then
sN “ DN ˚ f,
139
where DN is known as Dirichlet Kernel. Refer to Homework. DN is
almost an approximate identity. For piecewise C 1 function f , the proof
that sN converges is fairly similar to the φn ˚ f discussed before. Precise
result is that sN Ñ f˜ pointwise.
Example 4.3.6. Consider the function
#
x ` 2, 0 ă x ă π
f pxq “
´1,
π ď x ď 2π
4
Example 4.3.7. Consider the function f pxq “ x ` 2 for 0 ă π
4
If sN converges uniformly to some function, then it must be to f .
Reason: Say sN Ñ g uniformly and sN Ñ f in L2 . But sN Ñ g uniformly
implies that sN Ñ g in L2 , so f “ g must hold.
How can we tell if the Fourier Series converges uniformly? Recall the
Weierstrass M-test.
140
Theorem ř
32. (Weierstrass M-test) If
n pxq| ď Mn for all x P Ω, for all
ř|h
8
8
n P N and n“1 Mn converges, then n“1 hn pxq converges uniformly on
Ω.
ř
inx
|
ď
|c
In
our
case,
|c
e
|.
So
if
n P Z|cn | ă 8, then it follows that
n
n
ř
inx
is uniformly convergent and the limit must be in CpTq.
nPZ cn e
Example 4.3.8. Consider the series
ÿ einx
n2 ` 1
nPZ
Since
ÿ
nPZ
n2
1
`1
converges, by the Weierstrass M-test we have that the original series is
uniformly convergent.
4
The general principle can be thought of as
More smoothness of f ðñ
faster decay of tcn u
By smoothness we mean “number of continuous derivatives”, and includes matching at zero and 2π.
C m pTq “
ˇ
(
f ˇ f pkq P CpTq, k ď m
For example, for f P C 1 pTq we have f 1 p0q “ f 1 p2πq. For f P C 2 pTq we
have f 2 p0q “ f 2 p2πq.
141
Claim: If f P C m pTq then |nm cn | Ñ 0 as n Ñ ˘8.
Reason: Take
1
cn “
2π
ż 2π
f pxqe´inx dx
0
and after integrating by parts we obtain
ˆ
˙
ż
1 f e´inx ˇˇ2π
1 2π 1
´inx
f pxqe
dx
cn “
ˇ `
2π
´n 0
in 0
ˆ
˙
ż
1 2π 1
1
´inx
0`
f pxqe
dx
“
2π
in 0
ùñ
1
ncn “
2πi
ż 2π
f 1 pxqe´inx dx ÝÑ 0
0
since there are Fourier Coefficients of f 1 P C p Tq so go to zero by RiemannLebesgue Lemma. For f P C 2 pTq, integrate by parts again, and so
1
n cn “
p2πiq2
ż 2π
2
f 2 pxqe´inx dx ÝÑ 0
0
again. Continue in the same way for any m. This proves our claim.
Claim: For some α ą 1
|nm cn | ď
C
nα
ùñ f P C m pTq
We should point a technicality: If f pxq “
142
ř
cn einx , then can we ask if
f 1 pxq “
ÿ
incn einx
can be implied. The answer is NOT in general. However, if
gpxq “
8
ÿ
gn P C 1 ra, bs
gn pxq,
n“1
and
ř8
n“1
gn1 pxq converges uniformly, then g P C 1 ra, bs and
g 1 pxq “
8
ÿ
gn1 pxq
n“1
Reason: In our case, if |cn | ď
C
,
nα
then it follows that
ÿ
cn einx
converges uniformly, since
|cn einx | ď
for α ą 1. If |cn | ď
C
,
nα
C
,
nα
ÿ 1
ă8
nα
look at
ÿ
incn einx ,
ˇ
ˇ
ˇincn einx ˇ ď C
nα
143
and hence
C 1 pTq.
ř
incn einx converges uniformly. It then follows that
ř
cn einx P
If m “ 2, then
ÿ
incn einx P C 1
by same argument. We carry on in the same way for larger m and this
proves the claim.
Example 4.3.9. Consider
ÿ einx
2|n|
nPZ
and note that it belongs to C 8 pTq since
ˇ
ˇ
ˇ 1 mˇ
ˇ
ˇ
ˇ 2|n| n ˇ ÝÑ 0
for all m.
4
144
Chapter 5
Distributions
We now introduce the theory of distributions or generalized functions.
Example 5.0.10. The Dirac delta function δ is a “function” with the
properties
δpxq “ 0, x ‰ 0
ż
δpxq dx “ 1
R
This is not an ordinary functions in the ordinary sense. How to give it a
precise sense?
4
Example 5.0.11. Consider the PDE
utt ´ uxx “ 0
We have seen that the general solution is
upx, tq “ F px ` tq ` Gpx ´ tq
145
if F, G P C 2 . So,
u “ p left moving waveq ` p right moving waveq
No evident physical reason to exclude F, G R C 2 .
4
We’ll see how u can be regarded as a solution for essentially arbitrary
F, G.
Definition 5.0.1. Let
ˇ
(
x ˇ f pxq ‰ 0
supp pf q “
By E ĂĂ Ω we mean that E is compact in Ω, meaning that E is bounded
and E Ă Ω.
Definition 5.0.2. Let Ω Ď Rn and define
DpΩq :“ C8
0 pΩq “
ˇ
(
f P C 8 ˇ supp pf q ĂĂ Ω
the space D is known as the space of test functions. If f P DpΩq then
f pxq “ 0 in some neighborhood of BΩ.
The space C08 pΩq is a vector space, but is there any function in it besides
0? No analytic function in C08 pΩq besides the zero functions.
Example 5.0.12. Let n “ 1 and
# 1
e x2 ´1 , |x| ă 1
φpxq “
0,
|x| ě 1
146
Note that supp pφq ĂĂ R and φ P C08 for |x| ă 1 and |x| ě 1. To show
that φ is continuous at x “ 1, we need to show
lim φpxq “ 0
xÑ1´
To show that φ1 is continuous at x “ 1, we need to show
1
´2 e x2 ´1 x
lim´ φ pxq “ lim´ 2
“0
xÑ1
xÑ1
px ´ 1q2
1
to be true. This follows from L’Hôpital’s rule. Similarly, for all higher
derivatives and at x “ ´1. For any n, φp|x|q P C08 pRn q.
4
We get lots more test function by
(1) scaling: φpxq ÞÑ αφpxq.
(2) dilation: φpxq ÞÑ φpαxq for α ą 0.
(3) translation: φpxq ÞÑ φpx ´ x0 q.
(4) differentiation: φpxq ÞÑ Dα φpxq (for α a multiindex).
(5) sums of such terms:
ÿ
cn Dαn φpβn px ´ xn qq
147
5.1
Convergence in C08pΩq
Let φn P C08 pΩq. We say that
φn ÝÑ 0
in C08 pΩq if
(1) There exists K ĂĂ Ω such that supp pφn q Ă K for all n P N.
(2) ||Dα φn ||L8 pΩq Ñ 0 for all α.
We say that φn Ñ φ if φn ´ φ Ñ 0 in the above sense.
Definition 5.1.1. A distribution is a continuous linear functional on
C08 pΩq. That is, T : C08 pΩq Ñ C, and
T paφ ` bψq “ aT φ ` bT ψ
φn Ñ φ in C08 pΩq ùñ T pφn q Ñ T pφq
We say that D 1 pΩq is the set of distribution on Ω. It is itself a vector
space.
Example 5.1.1. Let f P L1 pΩq and define
ż
T pφq “
f pxqφpxq dx
Ω
We claim that T P D 1 pΩq. Consider
ˇż
ˇ
ˇ
ˇ
|T pφq| “ ˇˇ f pxqφpxq dxˇˇ
Ω
ď ||f ||L1 ||φ||L8
ă8
148
φ is continuous on its support which is a compact set. Linearity is
obvious. If φn Ñ φ then
ˇż
ˇ
ˇ
ˇ
|T pφn q ´ T pφq| “ ˇˇ f pxqpφpxqn ´ φpxqq dxˇˇ
Ω
ď ||f ||L1 ||φn ´ φ||L8
ÝÑ 0
as needed.
4
Definition 5.1.2. Let L1loc pΩq denote the set of locally integrable functions.
For f P L1loc we still have that T P D 1 pΩq. If φ P C08 pΩq, let K “
supp pφq ĂĂ Ω, so
ˇż
ˇ
ˇ
ˇ
ˇ
|T pφq| “ ˇ f pxqφpxq dxˇˇ
K
ď ||f ||L1 pKq ||φ||L8 pKq
ă8
If φn Ñ φ then
ˇż
ˇ
ˇ
ˇ
|T pφn q ´ T pφq| “ ˇˇ f pxqpφpxqn ´ φpxqq dxˇˇ
K
ď ||f ||L1 pKq ||φn ´ φ||L8 pKq
ÝÑ 0
and hence T P D 1 pΩq.
We say T “ Tf for f P L1loc as above. We claim that Tf1 “ Tf2 if and only
if f1 “ f2 almost everywhere. The right to left implication is obvious.
Let Tf1 “ Tf2 . Then
149
ż
pf1 ´ f2 qpxqφpxq dx “ 0
Ω
for all φ P C08 pΩq. If f :“ f1 ´ f2 , need
ż
f pxqφpxq dx “ 0
Ω
for all φ P C08 pΩq and this
supp pf q and K1 “ x P K
that φn Ñ χK1 and |φn | ď 1.
that f “ 0 a.e.. Let K “
ˇwould imply
(
ˇ f ě 0 . We can find φn P C08 pΩq such
So
ż
0“
f φn dx ÝÑ 0
Ω
ùñ f “ 0 a.e. on K1
Similarly for f “ 0 on tf ď 0u implies that f “ 0 almost everywhere.
This proves our claim.
In summary:
1. C08 pΩq “ DpΩq “ space of test functions.
2. D 1 pΩq “ set of continuous linear functional on C08 pΩq.
If f P L1loc pΩq then Tf P D 1 pΩq is defined by
ż
Tf pφq “
f pxqφpxq dx
Ω
The set
150
ˇ
(
T P D1 pΩq ˇ T “ Tf , for some f P L1loc pΩq
is a subspace of D1 pΩq in 1-1 correspondence with L1loc pΩq. We regard
L1loc pΩq Ă D 1 pΩq.
The traditional view of a function is that of x Ñ f pxq. The D 1 pΩq view
of a function is as
ż
f pxqφpxq dx
φÑ
Ω
We now study an important distribution.
5.2
Dirac Delta distribution
Pick x0 P Ω and T pφq “ φpx0 q. We claim that T P D1 pΩq. Note T :
C08 pΩq Ñ C is linear. If φn Ñ φ in C08 pΩq,
|T pφn q ´ T pφq| “ |φn px0 q ´ φpx0 q| ÝÑ 0
The functional T is the Dirac delta distribution denoted δx0 . That is
δx0 pφq “ φpx0 q,
δ0 “ δ
Since this is an important distribution, let us study its properties. If δ
were a function δpxq, and supp pφq Ă tx ‰ 0u, we’d have
ż
δpxqφpxq dx :“ δpφq “ φp0q “ 0
Ω
loooooooomoooooooon
Notation
151
and this implies δpxq “ 0 for x ‰ 0. If φ P C08 pΩq with φp0q “ 1 then
ż
ż
δpxq dx “
Ω
δpxqφpxq dx
Ω
“ δpφq
“1
These were the two “properties” of δ. It is common to use function-like
notation for δ (i.e. δpxq) even though it is not one.
ż
δpxqφpxq dx “ φp0q
Ω
In this notation δx0 ÝÑ δpx ´ x0 q. In other words
ż
δpx ´ x0 qφpxq dx :“ δx0 pφq :“ φpx0 q
Ω
Example 5.2.1. Pick x0 P Ω, α any multi-index and set
T pφq “ pDα φq px0 q
As in last example, we can show that T P D 1 pΩq. In particular, T pφq “
φ1 p0q in R is the dipole distribution.
4
Example 5.2.2. Let γ Ď R2 be a curve. Set
ż
T pφq “
φ ds
γ
We can also check that T P D1 pR2 q.
4
152
5.3
The Principal Value Distributions
Let
ż8
T pφq “
´8
φpxq
dx
x
This is of the form Tf with f “ x1 . The problem here is that f R L1loc pRq.
Some explanation is needed. Below is a more precise definition
ż
φpxq
dx
R x
ż
φpxq
“ lim`
dx
Ñ0
|x|ą x
T pφq “ pv
The above is known as the “Cauchy principal value” integral. Note that
ż
pv
R
ż
φpxq
dx ‰
x
R
φpxq
dx
x
For example,
ż1
´1
dx
x
is undefined in Lebesgue or improper Riemann sense.
ż1
pv
´1
dx
“ lim`
Ñ0
x
ż
153
ă|x|ă1
dx
“0
x
We claim that T P D 1 pΩq. Let supp pφq Ă r´M, M s
„ż ´
T pφq “ lim`
Ñ0
» ´M
φpxq
dx `
x
żM

φpxq
dx
x
fi
żM
ż ´
żM
—ż ´ φpxq ´ φp0q
φpxq ´ φp0q
φp0q
φp0q ffi
—
ffi
“ lim` —
dx `
dx `
dx `
dxffi
Ñ0 – ´M
fl
x
x
x
x
´M
looooooooooooooooomooooooooooooooooon
cancel
ż
ψpxq dx
“
ă|x|ăM
where ψpxq “ pφpxq ´ φp0qq{x. This implies that
|ψpxq| “ |φ1 pcx q|
ď ||φ1 ||L8
ùñ
for some cx P p0, xq by MVT
|T pφq| ď 2M ||φ1 ||L8 ă 8
and hence T : C08 pΩq ÝÑ C. Linearity can be shown. If φn Ñ φ then
|T pφn q ´ T pφq| ď 2M ||ψn ´ ψ||L8
ď 2M ||φ1n ´ φ1 ||L8
ÝÑ 0
We denote T “ pv x1 or pf x1 .
5.4
Convergence in D 1pΩq
We say Tn Ñ T in D 1 pΩq if Tn pΩq Ñ T pΩq for all φ P C08 pΩq.
154
Example 5.4.1. If fn Ñ f in L1 pΩq then we claim that Tfn Ñ Tf in
D 1 pΩq. Consider
ˇż
ˇ
ˇ
ˇ
|Tfn pφq ´ Tf pφq| “ ˇˇ pfn ´ f qpxqφpxq dxˇˇ
Ω
ď ||fn ´ f ||L1 ||φ||L8
ÝÑ 0
Same result if fn Ñ f in Lp pΩq for 1 ď 0 ă 8.
4
Example 5.4.2. Let fn “ einx in D 1 p0, 2πq. We want to show that
fn Ñ 0 in the sense of distributions i.e. in D 1 p0, 2πq. Consider
ż 2π
φpxqeinx dx “ c´n
Tfn pφq “
0
where c´n is the ´n-th Fourier Coefficient for φ. By the RiemannLebesgue Lemma we have that
c´n ÝÑ 0,
as n ÝÑ ˘8
Similarly, cospnxq, sinpnxq Ñ 0 in the sense of distributions.
4
Example 5.4.3. Let fn pxq “ nχr0, 1 s pxq. Then fn Ñ δ in D 1 pRq since
n
ż 1{n
Tfn pφq “ n
φpxq dx ÝÑ φp0q
0
by the Fundamental Theorem of Calculus.
4
155
Suppose tfn unPN satisfies
ş
(a) Rm fn pxq dx “ 1 for all n P N.
(b) There exists C such that ||fn ||L1 pRm q ď C for all n P N.
ż
(c) lim
|fn pxq| dx “ 0 for all δ ą 0.
nÑ8 |x|ąδ
Then fn Ñ δ in D 1 pRm q. That is for all φ P D 1 pRm q,
ż
fn pxqφpxq dx ÝÑ φp0q
The proof is similar to the CpTq cases. We call fn an approximate identity
again or a delta sequence. A
ş common way to construct such a sequence
1
m
is to pick g P L pR q with Rm g dx “ 1 and set
fn pxq “ nm gpnxq
For example, gpxq “ χr0,1s pxq then, fn pxq “ nχr0,1{ns pxq. Note that nm is
chosen such that
ż
ż
nm gpnxq dx
fn pxq dx “
ż
“
“1
156
gpyq dx
5.5
Algebra with Distributions
Addition and multiplication are clear. Product is not defined in general.
For example, δ 2 has no meaning.
If a P C 8 pΩq and T P D 1 pΩq, we define aT by
paT qpφq “ T paφq,
@φ P DpΩq
The reason is that if T “ Tf then aT should correspond to Taf , namely
paT qpφq “ TaT pφq
ż
“ apxqf pxqφpxq dx
Ω
“ Tf paφq
Note that aφ P C08 pΩq, so definition always makes sense, and is easy to
check that aT P D 1 pΩq.
Example 5.5.1. If T “ δ, then aδpφq “ δpaφq “ ap0qφp0q. In particular,
aδ “ ap0qδ
An example would be
ex δ “ δ
xδ “ 0
4
157
5.6
Calculus in D 1pΩq
If f P C 2 pa, bq, and T “ Tf then
żb
żb
f pxqφ1 pxq dx
1
f pxqφpxq dx “ ´
a
a
or Tf 1 pφq “ ´Tf pφq. Note that the RHS above is defined, even if f R C 1 .
This suggests that we define T 1 by
T 1 pφq “ ´T pφ1 q
for any T P D 1 pa, bq.
Theorem 33. We claim that if T P D 1 pa, bq then T 1 P D 1 pa, bq.
Proof. It is enough to show the continuity property. Let φn Ñ φ in
C08 pa, bq. Then φ1n Ñ φ1 and so
T 1 pφn q “ ´T pφ1n q ÝÑ ´T pφ1 q “ T 1 pφq
Hence T 1 P D 1 pa, bq as needed.
5.6.1
The Heaviside function
The Heaviside function is
#
1,
Hpxq “
0,
158
if x ą 0
if x ă 0
Classically, H 1 pxq “ 0 for x ‰ 0, and H 1 p0q is undefined. In the sense of
distributions
H 1 pφq “ ´Hpφ1 q
ż8
φ1 pxq dx
“´
0
“ loφp0q
omoon
δpφq
that is, H 1 “ δ in the sense of distributions.
5.6.2
The Derivative of the Delta Distribution
The Dirac Delta Distribution T “ δ must also have a derivative:
δ 1 pφq “ ´δpφ1 q “ ´φ1 p0q
which was referred to as the dipole distribution.
5.6.3
General definition of T 1
The general definition distribution derivative is given as follows. If T P
DpΩq then
BT
pφq “ ´T
Bxi
ˆ
Bφ
Bxi
˙
For any multi-index α
Dα T pφq “ p´1q|α| T pDα φq
Here are some facts:
159
1) Dα T P D 1 pΩq.
2)
B2T
B2T
“
for all i, j.
Bxi Bxj
Bxj Bxi
3) Tn Ñ T in D 1 pΩq implies Dα Tn Ñ Dα T in D 1 pΩq.
4)
BT
Ba
B
paT q “ a
`
T for all a P C 8 .
Bxj
Bxj Bxj
The notation xT, φy may be used instead of T pφq. For example
xδ, φy “ φp0q
xDα T, φy “ p´1q|α| xT, Dα φy
Let f P C 8 pRq except at x “ 0 and
lim f k pxq
xÑ0˘
exists for all k P N0 . We want to compute f 1 in D 1 pRq sense. That is,
we want to compute pTf q1 .
f 1 pφq “ ´f pφ1 q
ż
“ ´ f pxqφ1 pxq dx
żR0
ż8
1
“´
f pxqφ pxq dx ´
f pxqφ1 pxq dx
´8
0
ż0
ˇ0
ˇ
“
f 1 pxqφpxq dx ´ f pxqφpxqˇ
´8
´8
ż8
ˇ8
ˇ
1
f pxqφpxq dx ´ f pxqφpxqˇ
0
0
160
We introduce the following notation
#
f 1 pxq,
x‰0
rf 1 s :“
undefined, x “ 0
Consider the example of the Heaviside function H. Then
#
0,
x‰0
rH 1 s :“
undefined, x “ 0
So, with this notation:
ż
1
rf 1 sφ dx ` rf p0`q ´ f p0´qsφp0q
f pφq “
R
ùñ f 1 “ rf 1 s ` ∆f δ
where ∆f “ f p0`q ´ f p0´q, is the jump in f at x “ 0.
Example 5.6.1. Once again the Heaviside function H has derivative
H 1 “ rH 1 s ` ∆Hδ
“0`1¨δ
“δ
4
In particular, if f is continuous at 0, then f 1 “ rf 1 s.
161
Example 5.6.2. Consider the absolute value function f pxq “ |x|. This
function is continuous at zero.
#
1,
xą0
f 1 “ rf 1 s “ signpxq “
´1, x ă 0
4
The second derivative becomes
f 2 “ rf 1 s1 ` ∆f δ 1
“ rf 2 s ` `∆f 1 δ ` ∆f δ 1
Example 5.6.3. Consider the function
#
f pxq “
cospxq, x ą 0
2x,
xă0
Then
#
´ sinpxq, x ą 0
rf 1 s “
2,
xă0
#
´ cospxq, x ą 0
rf 2 s “
0,
xă0
and hence
162
f 1 “ rf 1 s ` δ
ùñ f 2 “ rf 2 s ` ´2δ ` δ 1
since ∆f “ 1 and ∆f 1 “ ´2.
4
This can be generalized to any order derivative and jumps at any other
point. For example, we would like to have a jump at x “ a. Then
f 1 “ rf 1 s ` ∆a f δa
etcetera. We could even have jumps at more than one point.
Example 5.6.4. Consider the 2π periodic extension of x on r´π, pis
Then rf 1 s “ 1 and so
f 1 “ rf 1 s `
8
ÿ
∆p2j`1qπ f δp2j`1qπ
j“´8
8
ÿ
“ 1 ` ´2π
δp2j`1qπ
j“´8
4
Example 5.6.5. Consider the delta sequence fn pxq “ nχr0,1{ns pxq.
163
Then
fn1 “ nδ ´ nδ 1 ÝÑ δ 1 pxq
n
or
fn1 pxq “
δpxq ´ δpx ´ n1 q
1
n
ÝÑ δ 1 pxq
4
Example 5.6.6. Consider the function
#
logpxq, x ą 0
f pxq “
“ Hpxq logpxq
0,
xă0
Note f P L1loc and
#
1
rf s “
1
,
x
0,
164
xą0
xă0
However, this can be f 1 , since rf 1 s R L1loc . So
f 1 pφq “ ´f pφ1 q
ż8
logpxqφ1 pxq dx
“´
0
ż8
logpxqφ1 pxq dx
“ ´ lim`
Ñ0
„ż 8
ˇ8 
φpxq
ˇ
“ lim`
dx ´ logpxqφpxqˇ
Ñ0
x
„ż8

φpxq
“ lim`
dx ` logpqφpq
Ñ0
x
Note that
pφpq ´ φp0qq logpq ÝÑ 0
as Ñ 0. Its equal to
φ1 pa q logpq ÝÑ 0
So
„ż 8
1
f pφq “ lim
0
`
´
This distribution is denoted p f
If

φpxq
` φp0q logpq
x
Hpxq
x
¯
for pseudo function or finite part.
#
0,
xą0
f pxq “
“ Hp´xq logp´xq
logp´xq, x ă 0
165
Then
„

ż ´
φpxq
f “ lim ´φp0q logpq `
dx
Ñ0`
´8 x
1
If f pxq “ log |x| is the sum of the above two, then
ż
1
f pφq “ lim
Ñ0` |x|ą
φpxq
dx “ pv
x
Check hw for the second derivative of f .
4
Example 5.6.7. Consider
#
1, x ą 0
f px, yq “
0, x ă 0
This function looks like f px, yq “ Hpxq and so
Bf
“ δpxq,
Bx
What is the meaning of δpxq here?
166
Bf
“0
By
ˆ ˙
1
pφq
x
ˆ ˙
Bf
Bφ
pφq “ ´f
Bx
Bx
ż ż8
Bφ
“´
dx dy
Bx
0
R
ż
ˇ8
ˇ
“ ´ φpx, yqˇ
dy
x“0
R
ż
“ φp0, yq dy
R
“ line integral of φ on tx “ 0u axis.
Check
4
Bf
By
“ 0 in a similar way.
Example 5.6.8. Let F P L1loc pRq and upx, tq “ F px, `tq. Then we want
to show that
utt ´ uxx “ 0,
in D 1 pR2 q. We must show that
putt ´ uxx qpφq “ 0
for all φ P C08 pRq. In other words
upφtt ´ φxx q “ 0
167
In other words
ż ż
F px ` tqpφtt px, tq ´ φxx px, tqq dx dt “ 0
R
R
Change variables
ξ “ x ` t,
η “ x ´ t,
x“
x“
ξ`η
2
ξ´η
2
and
ˇ
ˇ
ˇ Bpx, tq ˇ
ˇ dξ dη
dx dt “ ˇˇ
Bpξ, ηq ˇ
ˇ
„
ˇ
ˇ
xξ xη ˇˇ
ˇ
“ ˇdet
dξ dη
tξ tη ˇ
ˇ
„ 1 1 ˇ
ˇ
ˇ
ˇ
“ ˇdet 21 21 ˇˇ
´2
ˇ ˇ 2
ˇ 1ˇ
“ ˇˇ´ ˇˇ dξ dη
2
From earlier work
φtt ´ φxx “ ´4φξη
so integral becomes
ż ż
´2
ˆż
ż
F pξqφξη pξ, ηq dξ dη “ ´2
F pξq
R
and
168
˙
φξη dη
R
dξ
ż
ˇ8
ˇ
φξη dη “ φξ pξ, ηqˇ “ 0
´8
R
as needed. It goes similarly for Gpx ´ tq. We remark that F , and G need
not be even functions. For example, δpx ˘ tq is a solution.
4
Example 5.6.9. Consider the function
upxq “
´1
4π|x|
in R3 . Note that u P L1loc pR3 q since
ż 2π ż π ż R
ż
|upxq| dx “
|x|ăR
0
0
0
1 2
r sin θ dr dθ dφ
4πr
which is finite. We would like to claim that ∆u “ δ in D 1 pR3 q. However,
we need to review the concept of integration by parts first.
4
5.6.4
Integration by Parts
Let Ω Ď Rn bounded with open, smooth boundary. Let ~npxq denote the
unit outward normal on BΩ.
The basic formula is
ż
Bg
dx “ ´
f
Ω Bxi
ż
Ω
Bf
g dx `
Bxi
Replace by gi , sum on i
169
ż
f gni dS
BΩ
ÿż
ÿ
Bgi
f
dx “ ´
Ω Bxi
i
i
ż
Ω
ÿ
Bf
gi dx `
Bxi
i
ż
f gi ni dS
BΩ
or
ż
ż
~ ¨ ~g dx “ ´
f ∇
Ω
ż
~ ¨ ~g dx `
∇f
Ω
BΩ
f ~loomoon
g ¨ ~npxq dS
gn :“
where gn “ ~g ¨ ~n is the normal component of ~g . If f pxq “ 1, we get
ż
ż
~ ¨ ~g dx “
∇
gn dS
Ω
BΩ
~ for some scalar function h. Note:
Choose ~g “ ∇h,
~ ¨ ∇h
~ “ ∆h
∇
~ ¨ ~n “
p∇hqn “ ∇h
Bh
B~n
So
ż
ż
∆h dx “
Ω
BΩ
Bh
dS
B~n
~ in previous formula,
Using ~g “ ∇h
ż
ż
f ∆h dx “ ´
ż
∇g ¨ ∇h dx `
Ω
f
BΩ
170
Bh
dS
Bn
Exchange f and h and subtract. Hence
ż
ˆ
ż
pf ∆h ´ h∆f q dx “
Ω
BΩ
Bf
Bh
´h
f
Bn
Bn
˙
dS
The above is known as Green’s Formula.
5.7
The Laplacian
Now we discuss the Laplacian in spherical coordinates pr, θ, φq.
ı
2
1”
1
∆u “ urr ` ur ` 2 uθθ ` cotpθquθ `
u
φφ
r
r
sin2 pθq
If uprq “
´1
,
4πr
and so
2
∆u “ urr ` ur “ 0
r
for r ‰ 0. Now back to claim made in Example 5.6.9. of section 5.6.3.:
We want to show ∆u “ δ in D 1 pR3 q. We need
p∆uqpφq “ up∆φq
“ δpφq
“ φp0q
That is
¡
φp0q “
´1
∆φ dx
4πr
R3
171
To show this:
¡
´1
∆φ dx “ lim`
Ñ0
4πr
R3
¡
´1
∆φ dx
4πr
M ą|x|ą
Apply Green’s Formula with Ω “ tM ą |x| ą u,
¡
´1
∆φ dx “
4πr
M ą|x|ą
ˆ
¡
M ą|x|ą
˙
´1
∆
φ dx
4πr
loooomoooon
“0
ˆ
˙˙
¡ ˆ
B
1
´1 Bφ
`
´φ
´
dS
4πr Bn
Bn
4πr
|x|“
ˆ
˙˙
¡ ˆ
1
´1 Bφ
B
´
dS
`
´φ
4πr Bn
Bn
4πr
looooooooooooooooomooooooooooooooooon
|x|“M
1
“´
4π
¡
Bφ
dS `
Bn
“0
¡
1
φ dS
4π2
|x|“
where in the last equality we used the fact that B{Bn “ ´B{Br. For the
first term, note that
ˇ ˇ
ˇ Bφ ˇ
ˇ ˇďK
ˇ Bn ˇ
for some K. Then
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ 1 ij Bφ
1
ˇ
ˇ
dS ˇ ď
K4π2 ÝÑ 0
ˇ
ˇ 4π
ˇ 4π
Bn
ˇ
ˇ
|x|“
172
as Œ 0. For the second term, note it is the average of φ over BBp0, q.
So should tend to φp0q as Ñ 0` . In detail
1
4π2
ij
1
φpxq dx ´ φp0q “
4π2
|x|“
ij ´
¯
φpxq ´ φp0q dx
|x|“
“♣
If η ą 0, there is ą 0 such that
|x| ă ùñ |φpxq ´ φp0q| ă η
For such 1
η4π2 “ η
4π2
|♣| ď
So
1
lim
Œ0 4π2
ij
φpxq dx “ φp0q
|x|“
as needed.
5.8
Convolution
Now we discuss convolution in R3 .
ż
pf ˚ gqpxq “
f pyqgpx ´ yq dy
3
żR
f px ´ yqgpyq dy
“
R3
173
Theorem 34. (Young’s Inequality) Suppose f P Lp pR3 q and g P Lq pR3 q
and
1 1
1
` “ `1
p q
r
with 1 ď p, q, r ď 8. Then
||f ˚ g||Lr ď ||f ||Lp ||g||Lq
An example: ||f ˚ g||p ď ||f ||p ||g||1 is the case that r “ p.
Formally
ż
f ˚ δpxq “
f pyqδpx ´ yq dy “ f pxq
In other words, δ ˚ f “ f . So δ is the convolution identity. If gn Ñ δ we
expect
f ˚ gn ÝÑ f ˚ δ “ f
This is an important technique for approximation. It will even work if
the function f is replaced by T P D 1 pRn q
We ask: How to define T ˚ g?
Definition 5.8.1. Define the translation operator τh ¨ by
τh φpxq “ φpx ´ hq
174
Define the reflection operator q̈ by
q
φpxq
“ φp´xq
Observe that
q
q ´ xq “ φpx ´ yq
pτx φqpyq
“ φpy
If φ, ψ P C08 pRn q, then
q
pφ ˚ ψqpxq “ Tψ pτx φq
ż
q
“ ψpyqτx φpyq
dy
ż
“ ψpyqφpx ´ yq dy
Definition 5.8.2. If T P D 1 pRn q and φ P C08 pRn q, we may define the
convolution in the sense of distributions by
q
pT ˚ φqpxq “ T pτx φq
The above definition makes sense since τx φq P C08 pRn q. We may also say
that pT ˚ φqpxq is defined for all x P Rn .
Example 5.8.1. Consider the delta distribution T “ δ. Then
175
q
pδ ˚ φqpxq “ δpτx φq
ˇ
ˇ
“ τx φqˇ
y“0
q
“ τx φp0q
q ´ xq
“ φp0
q
“ φp´xq
“ φpxq
If Tn Ñ δ, then
q
pTn ˚ φqpxq “ Tn pτx φq
q
ÝÑ δpτx φq
“ φpxq
Notice that this is only pointwise convergence!
4
It is a fact that
T ˚ φ P C 8 pRn q
q
Dα T ˚ φ “ Dα T pτx φq
“ p´1q|α| pDyα φpx ´ yqq
“ T pDα φpx ´ yqq
“ T pτx pDα φqqq
“ T ˚ Dα φ
Let ψk P C08 pRq such that ψk Ñ δ in D 1 pRn q, and Tk “ T ˚ ψk . SO
Tk P C 8 . We claim that Tk Ñ T in D 1 pRn q. This would show that
distributions may be approximated by C 8 functions.
176
Reason: Pick φ P C08 pRn q and show Tk pφq Ñ T pφq. Note
q
Tk pφq “ pTk ˚ φqp0q
q
“ ppT ˚ ψk q ˚ φqp0q
q
“ pT ˚ pψk ˚ φqqp0q
explanation needed
q qq
“ T ppψk ˚ φq
However, since ψk Ñ δ in C08 pRn q, it follows that
q q ÝÑ φq
q“ φ
pψk ˚ φq
Hence
Tk pφq ÝÑ T pφq
as needed.
If T has better properties, then stronger convergence results can be
proved.
Example 5.8.2. Let T “ Tg with g P CpRn q and supp pgq ĂĂ Rn . Then
u
gk “ g ˚ ψk ÝÑ g
(The above limit is uniform). If g P Lp pRn q with 1 ď p ă 8, then gk Ñ g
in Lp pRn q sense.
4
177
5.9
Weak derivatives
If f P Lp pΩq, then Dα f P D1 pΩq. If there is g P Lq pΩq, such that
Dα f “ g, then we say f has the weak α derivative in Lq pΩq. That is
ż
ż
α
|α|
f pxqD φpxq dx “ p´1q
Ω
gpxqφpxq dx
Ω
Example 5.9.1. Consider the absolute value function f pxq “ |x|. Note
that f P L8 p´1, 1q. By previous discussion (Example 5.6.2.), we know
that f 1 “ signpxq. This implies that f P Lq p´1, 1q for all q.
4
Example 5.9.2. Consider Heaviside function f pxq “ Hpxq. Note that
f P L8 p´1, 1q. By previous discussion (section 5.6.1), we have that
f 1 “ δ which is not in Lq p´1, 1q for any q.
4
Definition 5.9.1. Define the set
W k,p pΩq “
ˇ
(
f P D 1 pΩq ˇ Dα f P Lp pΩq, |α| ď k
The meaning of Dα f P Lp is in the weak sense. The above set is known
as a Sobolov Space.
Some examples are
W 0,p pΩq “ Lp pΩq
W k,2 pΩq “ H k pΩq
Example 5.9.3. The absolute function f pxq “ |x| belongs to W 1,q p´1, 1q
for all q.
4
178
The Sobolov Space W k,p pΩq is a Banach space with norm
˛1{p
¨
ÿ
||f ||W k,p pΩq :“ ˝
||Dα f ||pLp pΩq ‚
|α|ďk
The norm for p “ 8 is suitably modified. In particular, H k pΩq is a
Hilbert space with
xf, gyH k pΩq :“
ÿ ż
|α|ďk
Dα f pxqDα gpxq dx
Ω
as the inner product.
5.9.1
Strong vs Weak Derivatives
A related concept is that of strong derivatives. The Lp pΩq function f has
strong α-derivative g in Lq pΩq if there is a sequence fn P C 8 pΩq such
that
fn ÝÑ f,
Dα fn ÝÑ g,
in Lp pΩq
in Lq pΩq
The key fact is the “Weak equals Strong”. Due to Friedrichs an MeyersSerrin. The implication Strong ùñ Weak is easy. The reverse implication is proven via convolution approximation. For example, in the Rn
case, if Dα f “ g in the weak sense, let fn “ ψn ˚ f , with ψn P C 8 and
ψn Ñ δ. Then
fn ÝÑ f,
in Lp pΩq
Dα fn “ ψn ˚ Dα f “ ψn ˚ g ÝÑ g,
Some more steps are needed if Ω ‰ Rn .
179
in Lq pΩq
5.10
Fourier Series in D 1pΩq
Consider the delta distribution T “ δ in D 1 p´π, πq. Formally,
8
ÿ
δ“
cn einx
n“´8
seems to imply that
żπ
δpxqe´inx dx “
cn “
´π
1
2π
for all n P Z. Is
8
1 ÿ inx
δ“
e
2π n“´8
Recall the Dirichlet Kernel (Section ?),
N
1 ÿ
DN pxq “
n “ ´N einx
2π
and that DN Ñ δ in D 1 p´π, πq, since
żπ
żπ
DN pyqφpyq dy “
´π
DN p´yqφpyq dy
´π
“ pDN φqp0q
“ SN p0q
ÝÑ φp0q
180
pSN for φq
by earlier discussion, sinceφ P C 1 pTq. So DN Ñ δ.
8
1 ÿ inx
δ“
e
2π n“´8
in D 1 p´π, πq. The 2π periodic extension of δ to R is
δ“
8
1 ÿ
δpx ´ 2πnq
2π n“´8
A Fourier series
8
ÿ
cn einx
n“´8
can be defined in D 1 p´π, πq or D 1 pRq, provided |cn | ď C|n|m , for some
C and m. Look at
8
ÿ
n“´8
ùñ
cn inx
e
m`2
n
ˇ c ˇ
C
ˇ n ˇ
ˇ m`2 ˇ ď 2
n
n
8
ÿ
cn inx
e “ f pxq
m`2
n
n“´8
for f P CpTq. So,
181
8
N
ÿ
dm`2 ÿ cn inx
dm`2
cn inx
e “ m`2 lim
e
m`2
m`2
m`2
N
Ñ8
dx
n
dx
n
n“´8
n“´N
N
dm`2 ÿ cn inx
“ lim
e
N Ñ8 dxm`2
nm`2
n“´N
N
ÿ
m`2
“ lim p´1q
N Ñ8
cn einx
n“´N
“ some distribution
ř
The series nPZ cn einx must be 2π periodic in some sense. Two ways to
define this:
(i) Define τh T , where T P D 1 pRq and τh is the translation operator
(refer to hw12). Then 2π if and only if τ2π T “ T .
(ii) Note D 1 pTq is the set of continuous linear functionals on C 8 pTq.
The set T is already compact and so C08 pTq “ C 8 pTq.
Example 5.10.1. Consider the delta distribution on the torus δ P
D 1 pTq. We know δpφq “ φp0q. Here T P D 1 pTq must be regarded as
a 2π periodic distribution.
4
5.10.1
Fourier coefficients of a distribution
We raise a question: If T P D 1 pTq, what are its Fourier coefficients?
If T “ Tf , then
ż
1 π
cn “
f pxqe´inx dx
2π ´π
1
“
Tf pe´inx q
2π
For any T P D 1 pTq, we have cn :“
e´inx P C 8 pTq.
1
T pe´inx q.
2π
182
This is well defined since
Example 5.10.2. Consider the delta distribution T “ δ. Then
cn “
1
1
δpe´inx q “
2π
2π
as above.
4
ř
Example 5.10.3. Consider ně0 einx . Then cn “ 1 for all n ě 0 and
cn “ 0 for all n ă 0. What is the limit?
N
ÿ
einx “
n“0
1 ´ eipN `1qx
“ SN
1 ´ eix
and this implies
żπ
SN pφq “
´π
żπ
“
´π
1 ´ eipN `1qx
φpxq dx
1 ´ eix
żπ
1 ´ eipN `1qx
1 ´ eipN `1qx
pφpxq
´
φp0qq
dx
`
φp0q
dx
1 ´ eix
1 ´ eix
´π
looooooooooomooooooooooon
“2π
The
the last term equals 2π is that the integral is the same as
řN şreason
inx
dx “ 2π. Note that the first term on the left
0 Te
żπ
φpxq ´ φp0q ipN `1qx
e
dx ÝÑ 0
1 ´ eix
´π loooooomoooooon
PL8
as N Ñ 8 by the Riemann-Lebesgue Lemma. Furthermore,
183
żπ
´π
φpxq ´ φp0q
dx “ . . .
1 ´ eix
ż
ż
1 π
1
φpxq dx ´ πφ p0q ` lim`
φpxq cotpx{2q dx
“
Ñ0
2 ´π
ă|x|ăπ
Hence,
1
SN pφq Ñ πφp0q `
2
żπ
1
φpxq dx ` pv
2
´π
ˆż π
˙
cotpx{2qφpxq dx
´π
Therefore,
8
ÿ
n“0
einx “ πδ `
¯
1 1 ´
` pv cotpx{2q
2 2
4
5.11
Fourier Transform
The Fourier transform of f at k P Rn is given by
1
fppkq “
p2πqn{2
ż
f pxqe´ik¨x dx
Rn
This is defined, e.g. for f P L1 pRn q since
1
|fppkq| ď
p2πqn{2
ż
|f pxq| dx ă 8
Rn
184
and hence fp P L8 pRn q with
ˇˇ ˇˇ
ˇˇ pˇˇ
ˇˇf ˇˇ
L8
ď
1
||f ||L1
p2πqn{2
Let F denote the mapping Fpf q “ fp. Then F is linear and ||F|| ď
p2πq´n{2 , as an operator from L1 to L8 .
Example 5.11.1. Consider n “ 1 with characteristic function f pxq “
χr0,1s pxq. Then its Fourier Transform is
ż1
1
e´ikx dx
p2πqn{2 0
1 e´ikx ˇˇ1
“´ ? ˇ
ik 2π 0
1 1 ´ e´ik
“?
2π ik
fppkq “
4
Theorem 35. The Fourier transform F is a bounded linear operator
from L1 pRn q into C0 pRn q.
Here, C0 pRn q is the set of continuous functions in CpRn q with lim|x|Ñ8 f pxq “
0. This is a closed subspace of L8 pRn q. So it is a Banach space with L8
norm.
Proof. First we show that fp P CpRn q. If km Ñ k, then
f pxqe´ikm ¨x ÝÑ f pxqe´ik¨x
and
185
a.e.
ˇ
ˇ
ˇf pxqe´ikm ¨x ˇ ď |f pxq| P L1
So
ż
1
lim fppkm q “ lim
f pxqe´ikm ¨x dx
mÑ8
mÑ8 p2πqn{2 Rn
ż
1
f pxqe´ik¨x dx
“
p2πqn{2 Rn
“ fppkq
To show fp Ñ 0, first suppose f P C 1 pRn q of compact support. This
implies that
´1
fppkq “
p2πqn{2
C
ùñ |fppkq| ď
|kj |2
2
ż
Rn
ˆ
´1
ikj
˙
e´ik¨x dx
ˇˇ
ˇˇ
ˇˇ Bf ˇˇ2
ˇˇ
ˇˇ
ˇˇ Bxj ˇˇ 1
L
«
C
ùñ |fppkq|2 ď min
j
kj 2
ùñ
Bf
Bxj
ˇˇ
ˇˇ ff
ˇˇ Bf ˇˇ2
C
ˇˇ
ˇˇ
ď
ˇˇ Bxj ˇˇ 1
|k|2
L
lim |fppkq| “ 0
|k|Ñ8
Such functions f are dense in L1 pRn q, so given f P L1 pRn q, pick g as
above with
||g ´ f ||L1 ă 186
Then
g pkq|
|fppkq| ď |fppkq ´ gppkq| ` |p
ď ||f ´ g||L1 ` |p
g pkq|
Since gp Ñ 0 as |k| Ñ 8, it follows that
lim |fppkq| ď |k|Ñ8
and hence
lim |fppkq| “ 0 ùñ fp P C0 pRn q
|k|Ñ8
We had shown that
ˇˇ ˇˇ
ˇˇ pˇˇ
ˇˇf ˇˇ
L8
ď
1
||f ||L1
p2πqn{2
as needed.
Example 5.11.2. Consider the Gaussians:
2
f pxq “ e´α|x| ,
Consider the case α “ 1, and n “ 1.
187
αą0
ż8
1
2
fppkq “
e´x e´ikx dx
2π ´8
ż
1 8 ´px2 `ikxq
e
dx
“
2π ´8
ż
1 8 ´px2 `ikx´ k2 q ´ k2
4 e
4 dx
“
e
2π ´8
k2 ż
e´ 4 8 ´px` ik q2
2
“?
e
dx
2π ´8
k2 ż
e´ 4 8 ´x2
e
dx
“?
2π ´8
k2
e´ 4
“?
2π
where complex variables techniques are used in the fourth equality.
2
If f pxq “ e´αx , then
k2
1
fppkq “ ? e´ 4α
2α
As α Ñ 0` , f pxq Ñ move spread out, fppkq Ñ more concentrated at zero.
Can show,
lim fppkq “ δ
αÑ0`
188
If α “ 1{2 then f “ fp. Now consider the case n “ 2. So
2
2
2
e´α|x| “ e´αx1 e´αx2
Then
ij
1
2
2
p
e´αx1 e´αx2 e´ix1 k1 e´ix2 k2 dx1 dx2
f pkq “
2π
ż
ż
1
2
´αx21 ´ix1 k1
“
e
e
dx1 e´αx2 e´ix2 k2 dx2
2π
1 ´ k12 ´ k22
e 4α e 4α
“
2α
1 ´ |k|2
“
e 4α
2α
In the case that n is arbitrary,
|k|2
f pxq “ e
´α|x|2
e´ 4α
ùñ fppkq “
p2αqn{2
4
5.12
Fourier Inversion Theorem
Theorem 36. If f, fp P L1 pRn q, then
1
f pxq “
p2πqn{2
ż Rn
fppkqeik¨x dk
q
p
p
We remark that this implies f “ fp or fq “ fp. This also means that if
R=reflexion, then Rf pxq “ f p´xq and hence F2 “ R.
189
Proof. Note
1
p2πqn{2
ż
fppkqeik¨x dk
ż
1
ppkqeik¨x e´α|k|2 dk
“ lim`
f
αÑ0 p2πqn{2
ij
1
2
“ lim`
f pyqe´ik¨y eix¨k e´α|k| dydk
n
αÑ0 p2πq
ˆż
˙
ż
1
´α|k|2 ´ik¨py´xq
“
lim f pyq
e
e
dk dy
p2πqn αÑ0`
|y´x|2
ż
e´ 4α
dy
“ lim` f pyq
αÑ0
p4παqn{2
“ lim` pf ˚ Gα qpxq
αÑ0
and hence
|y´x|2
e´ 4α
Gα pxq “
p4παqn{2
and Gα Ñ δ as α Ñ 0` since
1
Gα pxq “ ? n G
p αq
ˆ
x
?
α
with G “ G1 . SO
f ˚ Gα pxq ÝÑ f pxq
as α Ñ 0` as needed.
190
˙
,
Note that
fppkq “ pFf qpkq
ż
1
“
f pxqe´ix¨k dx
p2πqn{2 Rn
We can summarize. Recall that F : L1 pRn q Ñ C0 pRn q. If f, fp P L1 pRn q
q
p
then f “ fp or F´1 “ RF where R “ q̈ is the reflection.
5.13
Properties
(1) Note
1
p2πqn{2
ż
´ix¨k
f pxqe
Rn
ż
1
dx “
p´ixj f pxqqe´ix¨k dx
p2πqn{2 Rn
“ p´ixj f qp
Correct at least if xj f P L1 . In general
Dα fppkq “ pp´ixqα f qppkq
xα “ xα1 1 xα2 2 ¨ ¨ ¨ xαnn
if xα f P L1 .
(2) Note
191
ˆ
Bf
Bxj
˙p
ż
1
Bf ´ix¨k
e
dx
pkq “
n{2
p2πq
Rn Bxj
ż
´1
“
f pxqp´ikj qe´ix¨k dx
p2πqn{2 Rn
“ ikj fppkq
if f P C 1 , rapidly enough decaying at 8. In general
pDα f qppkq “ pikqα fppkq
for suitable functional f .
5.14
Schwartz Space
A class of functions for which the properties in the previous section work
is the following. Define
S“
ˇ
(
φ P C 8 pRn q ˇ xα Dβ φ P L8 pRn q, @α, β
The above space is known as the Schwartz space. Observe that xα Dβ φ P
L8 pRn q for all α and beta implies that
|Dβ φpxq| ď
C
,
p1 ` |x|2 qN
for all N ě 0. Any derivative of φ decays more rapidly than any reciprocal power. We say that is φ P S then φ is “rapidly decaying” (or rapidly
decreasing).
192
Example 5.14.1. Note that C08 pRn q Ď S and
2
e´α|x| P S
for all α ą 0.
4
Note that S Ă L1 pRn q.
Theorem 37. If φ P S then φp P S.
Proof. Note
´
¯p
p
“ pikqα p´ixqβ φ pkq
pikqα Dβ φpkq
´
¯p
“ Dα p´ixqβ φ pkq
where the properties mentioned before where used. So if φ P S, then
p´ixqβ φ P S. Hence
´
¯
Dα p´ixqβ φ P S Ď L1
´
¯p
ùñ
Dα p´ixqβ φ P L8 pΩq
ùñ pikqα Dβ φp P L8 pRn q
for all α and β. Similarly, φp P C 8 pRn q using property (1). So φp P S as
needed.
Theorem 38. (Corollary) The Fourier Transform F : S Ñ S is one-toone and onto.
Proof. If φ P S with φp “ 0 then
193
q
p q q
0“0“0
φ “ φp “ p
q
Therefore F is one-to-one. If ψ P S, let φ “ ψp and hence φ P S. So φ P S
and
p
p
q q
φp “ ψp “ ψp “ ψ
implies F is onto as needed.
Theorem 39. (“Plancherel’s Identity”) If φ, ψ P S then
ż
ż
p
φpxqψpxq
dx “
Rn
p
φpxqψpxq
dx
Rn
Proof. Note
ż
ij
1
φpxqψpxqe´ix¨y dy dx
p2πqn{2
ż
ż
1
“
ψpxq φpxqe´ix¨y dx dy
p2πqn{2
ż
p
“
φpxqψpxq
dx
p
φpxqψpxq
dx “
Rn
Rn
as needed.
ˇˇ ˇˇ
ˇˇ ˇˇ
Theorem 40. If ψ P S then ||φ||L2 “ ˇˇφpˇˇ .
L2
194
Proof. Pick ψ such that ψp “ φ, then
q
p q
p “ φp
ψ “ ψp “ φ
and hence
ż
ż
2
|φ| “
p2
|φ|
as needed.
We review the notation. In text,
φq “ F´1 φ,
S˚ “ S1
Recall that
S “ SpRn q “
ˇ
(
φ P C 8 ˇ xα Dβ φ P L8 pRn q, @α, β
We have shown that F : S Ñ S is one-to-one and onto and
ˇˇ ˇˇ
ˇˇ ˇˇ
||φ||L2 “ ˇˇφpˇˇ
L2
for all φ P S.
It can be shown that S is dense in Lp pRn q for 1 ď p ă 8. So it follows
that F extends to all of L2 pRn q. If f P L2 , pick φn P S with φn Ñ f in
L2 and so
195
ˇˇ
ˇˇ
ˇˇ
ˇˇ p
p
ˇˇφn ´ φm ˇˇ
L2
“ ||φn ´ φm ||L2 ÝÑ 0
and so the limit of φpn exists in L2 . We denote this limit by fp. We can
check fp is independent of the choice of φn , and fp coincides with the
original definition if f P L1 .
Furthermore,
ˇˇ ˇˇ
ˇˇ pˇˇ
ˇˇf ˇˇ
L2
ˇˇ ˇˇ
ˇˇ ˇˇ
“ lim ˇˇφpn ˇˇ
nÑ8
“ lim ||φn ||L2
nÑ8
“ ||f ||L2
This implies that F is an isometry on L2 .
Ş
Ş
The subset L1 L2 is dense in L2 . Thus, if fn P L1 L2 , with fn Ñ f ,
for some f P L2 , then
fpn ÝÑ fp,
as n Ñ 8
since
ˇˇ
ˇˇ
ˇˇ p
ˇˇ
p
ˇˇf n ´ f ˇˇ
L2
“ ||fn ´ f ||L2 ÝÑ 0
A common choice for a sequence is
#
f pxq, |x| ă n
fn pxq “ f pxqχBp0,nq pxq “
0,
|x| ě n
196
So fn P L1
Ş
L2 and fn Ñ f in L2 , then fpn Ñ fp in L2 . That is
ż
fppkq “ lim
nÑ8 |x|ăn
f pxqeix¨k dx
in the sense of L2 convergence.
It can be shown that the inversion theorem holds in L2 . That is, if
f P L2 pRn q then
1
f pxq “ lim
mÑ8 p2πqn{2
ż
fppkqeix¨k dk
|k|ăm
Example 5.14.2. Consider the function f pxq “ χr0,1s pxq. Then f P
L1 pRq and so
ż1
1
e´ixk dx
fppkq “ ?
2π 0
ˆ
˙
1
1 ´ e´ik
“?
ik
2π
Therefore fp R L1 but fp P L2 .
1
f pxq “ lim ?
M Ñ8
2π
żM
´M
1
?
2π
ˆ
1 ´ e´ik
ik
4
5.15
Tempered Distribution
Definition 5.15.1. We say φm Ñ φ in S if
197
˙
eikx dk
ˇˇ
ˇˇ
lim ˇˇxα Dβ pφm ´ φqˇˇL8 “ 0
mÑ8
for all α and β.
The set of continuous linear functionals on S, denoted S 1 is called the
set of tempered distributions. We say that T P S 1 if T : S Ñ C, T is
linear and
φm ÝÑ φ in S ùñ T φm ÝÑ T φ
We claim that S 1 Ď D 1 pRn q. Pick T P S 1 . We need to shown that it
belongs to D 1 pRn q. If φ P C08 pRn q then φ P S, so T φ is well defined.
Linearity is clear. If φn Ñ φ in C08 pRn q then it also converges in the S
sense since
ˇˇ α β
ˇˇ
ˇˇx D pφn ´ φqˇˇ
L8
ˇˇ
ˇˇ
ď C ˇˇDβ pφn ´ φqˇˇ
for some C depending on supp pφn φq (the set containing all supports).
However,
ˇˇ β
ˇˇ
ˇˇD pφn ´ φqˇˇ ÝÑ 0
So T φn Ñ T φ. Next, we want to show that S 1 ‰ D 1 pRn q. Consider
2
T “ Tf where f pxq “ ex . Then f P L1loc and thus Tf P D 1 pRn q. However,
Tf R S 1 since otherwise
ż
2
ex φpxq dx
Tf pφq “
R
198
2
would be defined for φpxq “ e´x P S which isn’t true. This concludes
our claim that the dual of the Schwartz functions is a proper subset of
the set of distributions.
What kind of functions have the property that Tf P S 1 ?
Definition 5.15.2. We say f has slow growth if
|f pxq|
“ 0,
|x|Ñ8 |x|N
lim
for some N .
In this case, Tf P S 1 . For example
ˇż
ˇ
|Tf pφq| “ ˇˇ
ˇ
ˇ
f pxqφpxq dxˇˇ
Rn
ż
|x|N |φpxq| dx
ďC
Rn
which is finite for φ P S.
Example 5.15.1. Any polynomial is in S 1 .
4
5.16
Fourier Transform on S 1
If φ, ψ P S recall that
ż
ż
p
φpxqψpxq
dx “
or
199
p
φpxqψpxq
dx
p
Tψppφq “ Tψ pφq
Definition 5.16.1. If T P S 1 then
p
Tppφq “ T pφq
for all φ P S.
This definition makes sense since φ P S implies φp P S. So Tp : S Ñ C, Tp
is linear, and if φn Ñ φ in S, can show φpn Ñ φp in S, so
p “ Tppφq
Tppφn q “ T pφpn q ÝÑ T pφq
implies Tp P S 1 .
Example 5.16.1. Consider T “ δ. Formally,
1
p2πqn{2
1
“
p2πqn{2
ż
δpxqe´ix¨k dx
p “
δpkq
Rn
Rigorously,
p
p
δpφq
“ δpφq
p
“ φp0q
1
“
p2πqn{2
200
ż
φpxq dx
Rn
which implies
δp “
1
p2πqn{2
4
Example 5.16.2. Consider T “ Tf where f pxq “ 1. Note that f R L1 .
Then
1
fppkq “
p2πqn{2
ż
e´ix¨k dx “?
But Tf P S 1 , so Tpf must exists. Then
p
Tpf pφq “ Tf pφq
ż
i0¨x
p
φpxqe
dx
“
Rn
p
p
“ p2πqn{2 φp0q
q
“ p2πqn{2 φp0q
“ p2πqn{2 φp0q
“ p2πqn{2 δpφq
Hence p
1 “ p2πqn{2 δ.
4
Theorem 41. (Inverse Theorem) Let T P S 1 . So we have shown that
p for all φ P S implies Tp P S 1 . Then
Tppφq “ T pφq
q
p
T “ Tp
201
We say that T is even if Tq “ T . For example, δ is even since
Proof. We prove the Inversion theorem. Note
q
p
p q
Tppφq “ Tppφq
p
p
q
“ T pφq
q
p
p
“ T pφq
since the inversion formula holds on S.
Example 5.16.3. Note that
ˆ
q
p
δp “
1
p2πqn{2
˙qp
“ δq
“δ
4
5.16.1
Some more properties
(1) Let T P S 1 . Then
´
¯p
α
p
D T “ p´ixq T
α
The reason is as follows:
202
Dα Tp “ p´1q|α| TppDα φq
α φq
z
“ p´1q|α| T pD
¯
´
“ p´1q|α| T p´ixqα φp
p
“ p´ixqα T pφq
“ pp´ixqα T qppφq
(2) For T P S 1
pDα T qp “ pikqα T,
The reason is similar to property (1).
Example 5.16.4. Consider T “ δ 1 . Then
ik
p
Tp “ pδ 1 q “ ik δp “ ?
2π
using the above properties.
4
Example 5.16.5. Consider the Heaviside function T “ Hpxq. Then
203
p
p
Hpφq
“ Hpφq
ż8
p
“
φpkq
dk
0
ż8ż8
1
φpxqe´ikx dx dk
“?
2π 0 ´8
żRż8
1
“?
φpxqe´ikx dx dk
lim
RÑ8
2π
0
´8
„ż R

ż8
1
´ikx
“?
lim
φpxq
e
dk dx
2π RÑ8 ´8
0
„

ż8
1
1 ´ e´ixR
φpxq
dx
“?
lim
ix
2π RÑ8 ´8
„

ż8
ż8
sinpRxq
1
cospRxq ´ 1
lim
“?
φpxq dx ` i
φpxq dx
x
x
2π RÑ8 ´8
´8
It can be shown that
sinpRxq
ÝÑ πδ
x
ˆ ˙
cospRxq ´ 1
1
ÝÑ pv
x
x
One gets
c
p“
H
π
i
δ ´ ? pv
2
2π
ˆ ˙
1
x
4
Example 5.16.6. Consider the principal value distribution T “ pv
From above
204
`1˘
x
.
c
p
p“
H
i
pi 1
? ` ? pv
2 2π
2π
ˆ ˙p
1
x
and hence
#
´ 21 , k ă 0
1
,
ką0
2
q´1
“H
2
ˆ ˙p
i
1
“ ? pv
x
2π
1
´ signpkq “
2
and therefore
ˆ ˙p c
1
π
pv
“
isignpkq
x
2
4
Example 5.16.7. Consider the delta distribution δn pxq “ δpx ´ nq.
Then
p
δpn pφq “ δn pφq
p
“ φpnq
ż
e´inx
“ φpxq ?
dx
2π
R
implies that
205
e´inx
δpn “ ?
2π
If
T “
8
ÿ
δn
n´8
then
Tp “
8
ÿ
δpn
n´8
8
ÿ
e´inx
?
“
2π
n´8
8
1 ÿ inx
“?
e
2π n´8
8
? ÿ
δpx ´ 2nπq
“ 2π
n´8
p we conclude that
Since Tppφq “ T pφq,
8
8
ÿ
? ÿ
p
2π
φp2nπq “
φpnq
n´8
n´8
The above is known as the Poisson Summation Formula.
4
206
5.17
Dual Spaces
If X is a vector space, which there is a definition of convergence
xn ÝÑ x,
in X,
then we definite X 1 to be the set of continuous linear functionals on X.
We call this set the dual space of X.
Suppose X Ď Y , where Y is another such space, and convergence in X
implies convergence in Y .
Example 5.17.1. The spaces X “ C08 pRn q, and Y “ SpRn q.
4
Example 5.17.2. The spaces X “ L2 pΩq, and Y “ L1 pΩq where Ω is
of finite measure.
4
We’ll call this a “topological” inclusion. We then have Y 1 Ď X 1 . Since
if T P Y 1 , so T pxq is defined on X, linearity is inherited. If xn Ñ x in X
then xn Ñ x in Y . So
T pxn q Ñ T pxq
implies T P X 1 .
Example 5.17.3. Consider C08 pRn q Ď SpRn q. Then S 1 pRn q Ď D 1 pRn q.
4
Example 5.17.4. Let Ξ “ C 8 pRn q and φn Ñ φ in Ξ. If
||Dα pφn ´ φq||L8 pKq ÝÑ 0
207
for all α and all K ĂĂ Rn . Then
C08 pRn q Ď SpRn q Ď C 8 pRn q “ ΞpRn q
This implies that
Ξ1 pRn q Ď S 1 pRn q Ď D 1 pRn q
4
It can be shown that Ξ1 is the set of distributions with compact support.
Definition 5.17.1. We say that T P D 1 pRn q with x P supp pT q if and
only if given ą 0 there is φ P C08 pBpx, qq such that T pφq ‰ 0. In other
words
T P Ξ1
ðñ
supp pT q is compact in pRn q
Example 5.17.5. Consider the delta distribution T “ δ. Then supp pδq “
t0u.
4
5.18
Differential Equations in D 1
Consider the problem T 1 “ 0 on pa, bq Ď R. Classically we know only
constants are solutions. However, is it still true if we allow distributions?
Theorem 42. If T P D 1 pa, bq and T 1 “ 0, then T is constant.
Proof. Choose φ0 P C08 pa, bq such that
żb
φ0 pxq dx “ 1
a
208
If φ P C08 pa, bq, let
ˆż b
˙
φpyq dy ¨ φ0 pxq
ψpxq “ φpxq ´
a
So
żb
ψpxq dx “ 0
a
Let
żx
ξpxq “
ψpsq ds
a
and so ξ 1 “ ψ and so ξ P C 8 pa, bq with ξpaq “ ξpbq “ 0. Note that ξ 1 “ 0
near a or b since ψ P C08 pa, bq. Thus ξ P C08 pa, bq. Then
0 “ T 1 pξq “ ´T pξ 1 q
“ ´T pψq
żb
φpyq dy ¨ T pφ0 q
“ T pφq ´
a
which implies
żb
T pφq “ C
φpyq dy,
a
and therefore T “ C as needed.
209
C “ T pφ0 q
Now look at T 1 “ f with f P L1loc pa, bq. Let
żx
f psq ds
gpxq :“
a
From analysis g 1 “ f almost everywhere. We claim that g 1 “ f in
D 1 pa, bq. Note
g 1 pφq “ ´gpφ1 q
żb
“ ´ gpxqφ1 pxq dx
a
żbżx
f psq dsφ1 pxq dx
“´
a a
żbżb
“´
f psqφ1 pxq dx ds
a s
żb
żb
“ ´ f psq φ1 pxq dx ds
s
a
żb
ˇb
ˇ
“ ´ f psqφpxqˇ ds
s
a
żb
f psqφpsq ds
“
a
“ f pφq
It follows that the general solution of T 1 “ f is
żx
f psq ds ` C
a
5.18.1
PDE
Now we discuss PDEs. Let Ω Ď Rn and
210
ÿ
Lu “
aα pxqDα u
|α|ďm
If aα pxq P C 8 pΩq for all α, then u P D 1 pΩq. This implies that Lu P D 1 pΩq.
We say Lu “ f in the classical sense if u P C m pΩq and Lu “ f at every
point.
We say Lu “ f in the weak sense if u P L1loc pΩq and Lu “ f in Lu “ f
in D 1 pΩq.
We say Lu “ f in the distributional sense if u P D 1 pΩq and Lu “ f in
Lu “ f in D 1 pΩq.
We study the regularity of solutions. If u is a weak or distributional
solution, is it actually classical?
Example 5.18.1. Consider the problem u1 “ 0. We have seen that this
only has constant solutions. In this case the answer is “yes”.
4
Example 5.18.2. Consider the problem xu1 “ 0. Here u “ δ is a
solutions.
xδ 1 pφq “ ´xδpφ1 q
“ ´δpxφ1 q
“ ´0 ¨ φ1 p0q
“0
But the solution is not classical.
4
Example 5.18.3. Consider the problem Lu “ utt ´ uxx “ 0. We have
seen that
211
upx, tq “ F px ` tq ` Gpx ´ tq
with F, G P C 2 is a classical solutions. However, F, G P L1loc gives a weak
solution.
4
Example 5.18.4. Consider the problem Lu “ ∆u “ 0. It turns out
that any weak or distributional solution is C 8 .
4
5.18.2
Fundamental Solutions
Now we discuss Fundamental Solutions. Consider the function
#
upxq “
1
log |x|,
2π
1
4π|x|
in R2
in R3
We have seen that ∆u “ δ in D 1 . In general, if u P D 1 pΩq and Lu “ δξ ,
for some ξ P Ω, then u is a fundamental solution for L with singularity ξ.
If ξ “ 0 we just say fundamental solution.
We consider the constant coefficient case:
Lu “
ÿ
aα D α u
|α|ďm
and suppose LE “ δ with E P D 1 pRn q. So E is a fundamental solution.
If f P C08 pRn q and u “ E ˚ f then
Lu “ LpE ˚ f q “ pLEq ˚ f “ δ ˚ f “ f
In other words, we get a solution of Lu “ f .
212
Example 5.18.5. Consider L “ ∆ in R3 . Then
1
˚f
4π|x|
ż
1
f pyq
“´
dy
4π R3 |x ´ y|
u“´
solves ∆u “ f in R3 . Here we say u is the Newtonian potential of f .
4
Example 5.18.6. Consider
Lu “ utt ´ uxx
in R2 . It is a homework problem to show that
1
Epx, tq “ Hpt ´ |x|q
2
is a fundamental solution.
In this case
213
u“E˚f
ij
“
Epx ´ y, t ´ sqf py, sq dy ds
ij
1
“
Hpt ´ s ´ |x ´ y|qf py, sq dy ds
2
ż ż x`t´s
1 t
f py, sq dy ds
“
2 ´8 x´t`s
over the set |x ´ y| ă t ´ s
which is half the integral of f over the “backward-light cone” or “backward characteristic triangle” with vertex px, tq.
4
It can be shown that f P Cc1 pRn q implies that u P C 2 pR2 q which is a
classical solution. If f P L1 pR2 q then u is a weak solution of
utt ´ uxx “ f
Typically, f px, tq “ 0 for t ă 0, in which case the integral with respect
to s goes from 0 to t.
Another way E arises: If g “ gpxq then upx, tq “ E ˚x gpx, tq (convolution
wrt x) then
214
ż
upx, tq “
Epx ´ y, tqgpyq dy
ż
1 8
“
Hpt ´ |x ´ y|qgpyq dy
2 ´8
ż
1 x`t
gpyq dy
“
2 x´t
R
which equals one term in solution of utt ´ uxx “ 0 with upx, 0q “ hpxq
and ut px, 0q “ gpxq. The other term is
B
1
phpx ` tq ` hpx ´ tqq “ pE ˚x hq
2
Bt
(again convolution wrt to x). So d’Alembert’s solution amounts to
upx, tq “ E ˚x g `
B
pE ˚x hq
Bt
How to find fundamental solutions? We discuss the general strategy.
Consider LE “ δ and take the Fourier transform
´ÿ
α
¯p
aα D E “ δp
“
which implies
215
1
p2πqn{2
ÿ
aα pDα Eqp “
1
p2πqn{2
p“
aα pikqα E
1
p2πqn{2
|α|ďm
ÿ
ùñ
|α|ďm
If
ppkq :“ p2πqn{2
ÿ
pikqα
|α|ďm
and hence
p“
E
1
ppkq
1
ùñ Epxq “
p2πqn{2
ż
Rn
1 ix¨k
e
dk?
ppkq
where p is a polynomial of degree m. We say that p is the symbol of L.
Is this really valid? There are some obstacles. There is no reason why
p should exists. E may not be tempered.
E
From another point of view, roots of p may prevent 1{p from being in
S 1.
Example 5.18.7. Consider the operator Lu “ ∆u and so
ppkq “ ´p2πqn{2 |k|2
We use
216
B2u
p “ ´k12 u
p
“ p´ik1 q2 u
Bx21
4
Example 5.18.8. Consider the operator
utt ´ uxx
and so the symbol is
ppkq “ 2πp´k22 ` k12 q
4
Note that the ∆ case, we only have a root of p at k “ 0. In wave operator
case, we have a root of p if k1 “ ˘k2 .
Turns out you can work around this difficulty. This is studied in the
PDE class.
Theorem 43. (Malgrange-Ehrenpreis) If L ‰ 0 then a fundamental
solution exists.
It’s easier when p has fewer roots.
Elliptic PDE
Definition 5.18.1. The principal symbol of L is
pm pkq :“
ÿ
p2πqn{2 aα pikqα
|α|“m
217
Definition 5.18.2. Whenever pm pkq “ 0 for k P Rn if and only if k “ 0,
we say L is elliptic.
Example 5.18.9. The operator L “ ∆ is elliptic whereas the wave
equation isn’t.
4
Parabolic PDE
The heat equation in Rn`1 is given by
Lu “ ut ´ ∆u
First consider
#
ut ´ ∆u “ 0, t ą 0, x P Rn
upx, 0q “ f pxq
and apply Fourier Transform in x variables,
1
ppk, tq “
u
p2πqn{2
ż
upx, tqe´ix¨k dx
Rn
Formally put qp “ pp
uqt . So the Fourier Transform of the equation is
#
u|t ` |k|2 u
p“0
|p
ppk, 0q “ fppkq
u
For each k P Rn this is an ODE in time t, so
2t
u
ppk, tq “ fppkqe´|k|
218
Another Fourier Transform property is
{
pf
˚ gq “ p2πqn{2 fpgp
Refer to Theorem 11.35 in the text. Furthermore,
|x|2
´|k|2 t
gp “ e
e´ 4t
ðñ g “
p2tqn{2
The above is known as the Fourier Transform of Gaussian Formula. This
implies
˜
2
e´|k| t fppkq “ gppkqfppkq “
2
´ |x|
4t
e
˚f
p4πtqn{2
¸y
and thus
|x|2
e´ 4t
upx, tq “
˚f
p4πtqn{2
ż
2
1
´ |x´y|
4t
“
f pyq dy
e
p4πtqn{2 Rn
So we say
$
&
Epx, tq “
|x|2
e´ 4t
,
p4πtqn{2
%0,
219
tą0
t“0
is a fundamental solution.
(Note go to next day. Check if 4 or 2 in formula of E)
In summary, let
$
&
Epx, tq “
|x|2
e´ 4t
,
p2πtqn{2
%0,
tą0
t“0
Then the solution to the IVP
#
ut ´ ∆u “ 0,
x P Rn , t ą 0
upx, 0q “ f pxq, x P Rn
is given by
ż
Epx ´ y, tqf pyq dy “ E ˚x f
upx, tq “
Rn
Duhamel’s principle connects IVP to the inhomogeneous problem
ut ´ ∆u “ h
Let vpx, t, sq satisfy
#
vt ´ ∆v “ 0,
x P Rn , t ą 0
vpx, 0, sq “ hpx, sq, x P Rn
220
and consider
żt
vpx, t ´ s, sq ds,
upx, tq “
0
Then
żt
ut “ vpx, 0, tq ` vt px, t ´ s, sq ds
żt 0
“ hpx, tq ` ∆upx, t ´ s, sq ds
0
żt
“ hpx, tq ` ∆ vpx, t ´ s, sq ds
0
loooooooooomoooooooooon
“: u
and so u satisfies ut ´ ∆u “ h. So,
ż
Epx ´ y, tqhpy, sq dy
vpx, t, sq “
Rn
żtż
Epx ´ y, t ´ sqhpy, sq dy ds “ E ˚ h
ùñ upx, tq “
0
Rn
provided that hpx, tq “ 0 for t ă 0. We can verify directly that
Et ´ ∆E “ δ
in D 1 pRn`1 q. Note that since upx, 0q “ 0 then
upx, tq “ E ˚ h ` E ˚x f
221
and so
#
ut ´ ∆u “ h,
x P Rn , t ą 0
upx, 0q “ f pxq, x P Rn
Furthermore, if
|x|2
e´ 4
F pxq :“
p2πqn{2
then
ˆ
Epx, tq :“
1
?
t
˙n
ˆ
F
x
?
t
˙
Recall,
ż
F pxq dx “ 1
Rn
n
ùñ k F pkxq ÝÑ δ, as k Ñ 8
So if k “ t´1{2 then k Ñ 8 as t Ñ 0` . Thus
lim Epx, tq “ δpxq,
tÑ0`
in D 1 pRn q, so E may be regarded as a solution of
#
Et ´ ∆E “ 0, t ą 0
Epx, 0q “ δ
222
5.18.3
Some Examples
The following are some well-known Fundamental Solutions.
The Laplacian
The Laplacian is L “ ∆. Then
$ |x|
’
n“1
&2,
1
Epxq “ 2π log |x|, n “ 2
’
% Cn
,
ně3
|x|n´2
where
Cn “ ´
1
,
npn ´ 2qan
where an is the volume of Bp0, 1q.
The Heat Equation
The Heat equation comes from Lu “ ut ´ ∆u. Then
Epx, tq “
Hptq ´ |x|2
e 4
p2πtqn{2
is the fundamental solution.
The Wave Equation
The Wave equation comes from lu “ utt ´ ∆u. Then
$
1
’
´ |x|q, n “ 1
’
& 2 Hpt
Hpt´|x|q
1 ?
n“2
Epx, tq “ 2π t2 ´|x|2 ,
’
’
% δpt´|x| ,
n“3
4π|x|
223
is the fundamental solution. More complicated formulas are known for
n ě 4.
Example 5.18.10. What does E ˚ f mean in n “ 3? Consider
ż
R4
δps ´ |y|
f px ´ y, t ´ sq ds dy “
4π|y|
ż
R3
f px ´ y, t ´ |y|q
dy
4π|y|
If f px, tq “ 0 for t ă 0 then
ż
f px ´ y, t ´ |y|q
dy
4π|y|
“
Bp0,tq or |y|ăt
ż
f py, t ´ |x ´ y|q
dy
4π|x ´ y|
“
Bpx,tq or |x´y|ăt
4
The Schrödinger Equation
The Schrödinger Equation comes from Lu “ ut ´ i∆u. Then
π
|x|2
Hptqeipn´2q 4 e´ 4it
Epx, tq “
p2πtqn{2
is the fundamental solution.
The Helmholtz Equation
The Helmholtz Equation comes from Lu “ ∆u ´ k 2 u. Then
#
Epxq “
1
K pk|x|q,
2π 0
´k|x|
´ e4π|x| ,
224
n“2
n“3
is the fundamental solution where K0 is the modified Bessel function of
order zero.
The Klein-Gordon Equation
The Klein-Gordon Equation comes from
Lu “ utt ´ uxx ´ u
in R2 . Then
´?
¯
1
Epx, tq “ Hpt ´ |x|qJ0
t2 ´ x2
2
is the fundamental solution.
The Biharmonic Equation
The Biharmonic Equation comes from Lu “ ∆∆u “ ∆2 u. Then
Epxq “ |x|2 log |x|, with n “ 2
5.18.4
Some Remarks
Some last thoughts. The solution operator f ÞÑ E ˚ f is translation
invariant. That is, commutes with translations. It is an indication that
original PDE problem has translation invariance: Lu “ f . So
(translate, solve)
ðñ
(solve, translate)
This is true because L has constant coefficients:
225
ÿ
Lu “
aα D α u
|α|ďm
Note
ÿ
τh pLuq “
aα τ h D α u
|α|ďm
ÿ
“
aα Dα pτh puqq
|α|ďm
“ Lpτh uq
If
ÿ
Lu “
aα pxqDα u
|α|ďm
then solution of Lu “ f cannot be expected to be of the form E ˚ f .
Instead, try to solve
LE “ δy
ùñ E “ Epx, yq “ Fundamental solution with singularity at y
and thus
ż
upxq “
Epx, yqf pyq dy
Rn
and hence
226
ż
Lu “
LEpx, yqf pyq dy
n
żR
δpx ´ yqf pyq dy
“
Rn
“ f pxq
We can allow Ω ‰ Rn , in this case.
Example 5.18.11. Refer to homework problem. Consider
#
Kpx, yq “
ypx ´ 1q, 0 ď y ď x ď 1
xpy ´ 1q, 0 ď x ď y ď 1
If
ż1
Kpx, yqupyq dy
upxq “
0
then u2 “ f . We could check that Kxx “ δpx ´ yq. This is a fundamental
solution of this generalized type for Ω “ p0, 1q.
4
227