MA3G8 Functional Analysis II
June 10, 2013
Contents
0 Introduction
1
1 Basics
1.1 Normed Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2 Separability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.3 Linear Maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
1
2
3
2 Duality and the separation theorems
2.1 1-Codimensional Subspaces . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2 Separation Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.3 Weak and Weak-? Topologies . . . . . . . . . . . . . . . . . . . . . . . . .
4
7
8
13
3 The
3.1
3.2
3.3
Category Theorems
Open Mapping Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Closed Graph Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Application to Fourier Analysis . . . . . . . . . . . . . . . . . . . . . . . .
14
15
16
17
4 Unbounded Operators
4.1 Closed Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.2 The Spectrum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
18
21
24
5 The Laplacian and Quantum Harmonic Oscillator.
5.1 Quantum Harmonic Oscillator . . . . . . . . . . . . . . . . . . . . . . . . .
25
26
These notes are based on the 2012 MA3F6 Functional Analysis II course, taught
by Keith Ball, typeset by Matthew Egginton.
No guarantee is given that they are accurate or applicable, but hopefully they will assist
your study.
Please report any errors, factual or typographical, to m.egginton@warwick.ac.uk
i
MA3G8 Functional Analysis II Lecture Notes Spring 2013
0
Introduction
At the beginning of the 20th century.
1
Basics
1.1
Normed Spaces
Definition 1.1 A norm || · || on a vector space V is a map x 7→ ||x|| ∈ R satisfying
1. ||x|| ≥ 0 for all x and ||x|| = 0 ⇐⇒ x = 0.
2. ||λx|| = |λ|||x|| for all x ∈ V and λ ∈ R
3. ||x + y|| ≤ ||x|| + ||y|| for x, y ∈ V .
A norm gives us a way to measure distance: d(x, y) = ||x − y||. The unit ball B plays an
important role (geometry of space is invariant under translation) and it is a convex set.
In other words, if x, y ∈ B and λ ∈ [0, 1] then (1 − λ)x + λy ∈ B
Lemma 1.2 If N : V → R satisfies
1. N (x) ≥ 0 for all x and N (x) = 0 ⇐⇒ x = 0
2. N (λx) = |λ|N (x) for λ ∈ R and x ∈ V
3. {x ∈ V : N (x) ≤ 1} is convex
then N is a norm.
The proof essentially uses a simple trick.
Proof We need to check the triangle inequality. Suppose x, y ∈ V . If N (x) = 0 then
x = 0 and then x + y = y and so N (x + y) = N (y) = N (x) + N (y). Thus suppose
a = N (x), b = N (y) and are strictly positive. Then N (a−1 x) = a−1 N (x) = 1 and so
a−1 x ∈ {x : N (x) ≤ 1} and the same for b−1 y ∈ {x : N (x) ≤ 1} and thus
a −1
b −1
x+y
a x+
b y=
∈ {x : N (x) ≤ 1}
a+b
a+b
a+b
by convexity. Thus
N (x + y) = (a + b)N (
x+y
) ≤ (a + b)1 = a + b = N (x) + N (y)
a+b
Q.E.D.
We shall consider a number of “standard” spaces.
P
p 1/p for 1 ≤ p < ∞
1. lp the space of sequences (xi ) with finite norm ||(xi || = ( ∞
i=1 |xi | )
and l∞ is the space of bounded sequences with the norm ||(xi )|| = supi |xi |.
2. Consider the finite dimensional versions lpn , the space of sequences of length n with
P
norm ||(xi )|| = ( ni=1 |xi |p )1/p .
3. Also c0 the space of sequences converging to 0 with norm ||(xi )|| = max |xi | =
sup |xi |. Note c0 ⊂ l∞ in the obvious way.
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MA3G8 Functional Analysis II Lecture Notes Spring 2013
4. C[0, 1] the space of continuous functions on [0, 1] with norm ||f || = sup |f (x)| =
max |f (x)|
R
1/p
1
5. Lp [0, 1] the space of measurable “functions” with ||f || = 0 |f (x)|p dx
< ∞.
We actually mean equivalence classes of functions, i.e. f ∼ g if f = g a.e.
6. L∞ [0, 1] is the space of measurable functions such that ||f || = sup |f (x)| < ∞.
The triangle inequality in Lp is often called the Minkowski inequality
Z
p
1/p
|f + g|
Z
≤
|f |
p
1/p
Z
p
1/p
|g|
+
R
R
We can see that the unit ball is convex as follows. If |f |p , |g|p ≤ 1 and λ ∈ [0, 1] then
for each x,
|(1 − λ)f (x) + λg(x)|p ≤ (1 − λ)|f (x)|p + λ|g(x)|p
since the map t 7→ |t|p is convex on R and so
Z
Z
Z
p
p
|(1 − λ)f (x) + λg(x)| ≤ (1 − λ) |f (x) + λ |g(x)|p ≤ 1
Lemma 1.3 (Hölder Inequality) If f, g are measurable, and
Z
Z
p
|f | < ∞,
then
Z
Z
fg ≤
|f |
p
1
p
+
1
q
= 1 and
|g|q < ∞
1/p Z
q
1/q
|g|
The Cauchy-Schwartz inequality is Hölder with p = 2 = q.
1.2
Separability
Definition 1.4 A subset U of a metric space X is said to be dense in X if for every
x ∈ X and ε > 0 there is a u ∈ U with d(x, u) < ε. A normed space is called separable
if it has a countable dense subset.
Lemma 1.5 For a normed space X, the following are equivalent
1. X is separable
2. SX = {x ∈ X : ||x|| = 1} is separable
3. X contains a sequence (x1 , x2 , ...) whose linear span is dense.
We prove the following before the proof of the above
Lemma 1.6 A subset of a separable metric space is separable
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MA3G8 Functional Analysis II Lecture Notes Spring 2013
Proof Let (x1 , x2 , ...) be a dense subset in X and Y ⊂ X. We need to find a dense
subset of Y .
For each n and k, if Y contains a point whose distance from xn is at most k1 call this
un,k ∈ Y . The set {un,k } is countable. I claim that it is dense in Y . If y ∈ Y and ε > 0
choose xn with d(xn , y) < k1 where k1 < 2ε . There is a point un,k since y would have been
a candidate. Then
d(y, un,k ) ≤ d(y, xn ) + d(xn , un,k ) <
1 1
+ <ε
k k
Q.E.D.
Proof [of lemma 1.5] 1) =⇒ 2) is clear from the previous lemma since SX ⊂ X.
2) =⇒ 3) Choose (ui ) dense in SX . This sequence suffices for 3). Even the multiples
x
x
∈ SX so we can find a ui with || ||x||
− ui || <
(λui ) are dense. If x ∈ X and x 6= 0 then ||x||
ε
||x|| and hence
||x − ||x||ui || < ε
3) =⇒ 1). Let (x1 , x2 , ...) be a sequence whose span is dense. The span is much
bigger than countable and so 3) is weaker than 1).PWe want to build a countable dense
ri xi for ri ∈ Q. Given x ∈ X, ε > 0
set in X. We take the rational linear combination n1P
we can find some n ∈ N and θ1 , ..., θn ∈ R with ||x − θi xi || < 2ε . Now choose rationals
ε
ri with |θi − ri | < 2nM
where M = 1 + max{||x1 ||, ..., ||xn ||} and then
X
X
X
θi x i −
ri xi = (θi − ri )xi X
≤
||(θi − ri )xi ||
X
=
|θi − ri |||xi ||
X ε
M
≤
2nM
ε
≤
2
and combining above inequalities gives the desired result.
Q.E.D.
The homework includes proving lp is separable, and this uses the above.
1.3
Linear Maps
A linear map T : X → Y between normed spaces may or may not be continuous. It is
continuous if and only if it is bounded, i.e. there exists K > 0 with
||T x|| ≤ K||x||
For such a map we define its norm as inf{K : ||T x|| ≤ K||x||∀x} = ||T || = sup{||T x|| :
||x|| ≤ 1}. This is in some sense the maximum scale factor.
The space of bounded linear maps form a normed vector space under this norm. This
space B(X, Y ) is complete if and only if Y is complete. A complete normed space is called
a Banach space. In finite dimensions, T : U → V then dim(ker T ) + dim(ImT ) = dim U .
Two Banach spaces X and Y are called isomorphic if there is an invertible linear map
T : X → Y with ||T || and ||T −1 || bounded. They are isometric if ||T || = 1 = ||T −1 ||.
There is a natural (algebraic) isomorphism between U/ ker T and ImT , by “ cutting
the electric cable”.
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MA3G8 Functional Analysis II Lecture Notes Spring 2013
We can define the quotient in a Banach space. If Y is a closed subspace of X we
define X/Y to be the space of translates {x + Y : x ∈ X} with x + Y = z + Y if x − z ∈ Y .
As in the finite dimensional case, addition and scalar multiplication make sense. We
define the norm
||x + Y || = inf{||x + y|| : y ∈ Y }
and in some sense this is the closest point of the translate to 0, but the closest point may
not exist.
1. The norm is well defined; i.e. ||x + Y || = ||z + Y || if x − z ∈ Y . This is because we
have seen a form of the norm without reference to a representative.
2. ||λ(x + Y )|| = |λ|||x + Y || is easy to show
3. We show the triangle inequality explicitly. Given ε > 0 choose yx ∈ Y and yz ∈ Y
with ||x + yx || < ||x + Y || + ε and ||x + yz || < ||z + Y || + ε and then
||x + z + Y || ≤ ||x + z + yx + yz || ≤ ||x + yx || + ||z + yz || ≤ ||x + Y || + ||z + Y || + 2ε
and taking the infimum over ε gives us our result.
4. Show ||x + Y || = 0 if and only if x + Y = 0, i.e. x ∈ Y .
If x ∈ Y then choose y = −x and so x + y = 0 and so ||x + Y || = 0. Conversely
suppose ||x + Y || = 0. Then we can find a sequence (yk ) in Y with ||x + yk || → 0,
i.e. x + yk → 0 or yk → −x, and since Y is closed, −x ∈ Y and so x ∈ Y .
5. X/Y is complete. We take a Cauchy sequence and then construct a subsequence
and use the completeness of X. We might not find a convergent sequence of
representatives. It suffices to find a convergent subsequence. Choose nk so that
||xnk + Y − xm + Y || ≤ 21k if m ≥ nk . Hence ||xnk + Y − xnk+1 + Y || ≤ 21k . Choose
ynk one by one so that
||xnk + ynk − (xnk+1 + ynk+1 )|| ≤
2
2k
and so xnk + ynk is Cauchy and has a limit u. Then xnk + Y → u + Y as required.
We have used:
Lemma 1.7 If (xk ) is a sequence in a complete metric space with d(xk , xk+1 ) ≤ εk and
P
∞
1 εk < ∞ then (xk ) is Cauchy.
2
Duality and the separation theorems
If X is a normed space then a linear functional on X is a linear map X → R (or the
underlying field). The space of bounded linear functionals is denoted X ? ; it is a Banach
space with the operator norm ||φ|| = sup{|φ(x)| : ||x|| ≤ 1}. If X is a Hilbert space then
every bounded linear functional is of the form
x 7→ hx, yi
for some y ∈ X and these maps are all in X ? . Thus we can identify X ? with X. What
about lp and Lp ?
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MA3G8 Functional Analysis II Lecture Notes Spring 2013


c0
Theorem 2.1 (Duality of lp ) If X = lp


l1
1
1
with p + q = 1.


l1
?
1 < p < ∞ , then X is isometric to lq


l∞
Remark If p = 2 then q = 2 and lp? = lp .
Proof For each y of the supposed dual (in each case) we can define φy : X → R by
φy (x) =
∞
X
yi x i
1
where x = (x1 , x2 , ...) and this makes sense by Holder:


∞
||x||∞ ||y||1
X
|yi xi | ≤ ||x||p ||y||q


1
||x||1 ||y||∞
and so the series converges absolutely.
φy is clearly linear and the map y 7→ φy (lq → X ? ) is linear
X
X
X
φy+z (x) =
(yi + zi )xi =
yi x i +
zi xi = φy (x) + φz (x)
We also have |φy (x)| ≤ ||x||p ||y||q and so on so the map has ||φy || ≤ ||y||. Note that
the map y 7→ φy is obviously 1-1 but we shall get this automatically when we prove
||y|| ≤ ||φy ||
It remains to show that each φ ∈ X ? is of the form φy for some y (in the supposed
dual) with ||y|| ≤ ||φy ||.
Let (ei ) be the standard unit vector basis of X. For each i set yi = φ(ei ). Then φ and
φy agree on the (ei ). The problem is the norm. We show that ||y|| ≤ ||φ|| as then φ is of
the form φy for that y. If 1 < p < ∞ then
m
X
|yi |q =
1
m
X
yi |yi |q−1 sgn(yi )
1
=
m
X
φ(ei )|yi |q−1 sgn(yi )
1
= φ(
m
X
|yi |q−1 sgn(yi )ei )
1
≤ ||φ||||
m
X
|yi |q−1 sgn(yi )ei ||p
1
= ||φ||
m
X
!1
p
|yi |(q−1)p
1
= ||φ||
m
X
!1
p
|yi |q
1
We have shown that
m
X
q
|yi | ≤ ||φ||
m
X
1
1
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!1
p
q
|yi |
MA3G8 Functional Analysis II Lecture Notes Spring 2013
and since p1 < 1 the estimate has somePcontent. If y = 0 then φ = 0 and there was nothing
to prove. Thus we may assume that
|yi | > 0 and then
!1
m
X
q
q
|yi |
≤ ||φ||
1
and so taking limits as m → ∞ we have
m
X
!1
q
|yi |
q
= ||y|| ≤ ||φ||
1
we know that φ and φy agree on (ei ) and are continuous linear functionals. In lp for
1 < p < ∞ or c0 or l1 the closed span of the (ei ) is X. Thus the functionals agree on X.
If X = c0 then
m
X
|yi | =
1
=
m
X
1
m
X
yi sgn(yi )
φ(ei )sgn(yi )
1
=φ
m
X
!
sgn(yi )ei
1
≤ ||φ||||
m
X
sgn(yi )ei ||
1
= ||φ||
and if X = l1 then
|yi | = |φ(ei )| ≤ ||φ||||ei || = ||φ||
Q.E.D.
1. In each case we show that if y is in the suggested dual then it gives a φy ∈ X ? , by
using Holder. In particular each y ∈ l∞ acts on l1 and each y ∈ l1 acts on (l∞ and
hence) c0 .
2. For each case, given φ ∈ X ? there is a y in the supposed dual with φ(ei ) = φy (ei )
for each i. Since the closed span of the ei is the space required this gives φ = φy on
X.
? = l . Moreover l?
However, since l∞ is not separable, we cannot conclude that l∞
1
∞
is not isomorphic to l1 . Note that the above proof uses separability in the construction
ofthe basis.
The word duality suggests some kind of symmetry between X and X ? . In finite
dimensions, indeed (X ? )? ∼
= X in a canonical way: if x ∈ X and φ ∈ X ? then the action
of x on φ is the action of φ on x.
In Banach spaces, X still acts on X ? , so X ⊂ X ?? as for example c0 acts on l1 and
c0 ⊂ l∞ but X may not fill up X ⊂ X ?? , eg c0 6= l∞ (they are massively different).
Things in Lp are somewhat similar. There is no analogue of c0 and L?∞ is every bit as
1
1
? . C[0, 1] is a little bit like c . We have L? ∼ L
? ∼
nasty as l∞
0
1 = ∞ and Lp = Lq if p + q = 1 for
1 < p < ∞.
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MA3G8 Functional Analysis II Lecture Notes Spring 2013
The proof of L?p ∼
= Lq is very similar to the proof that lp? ∼
= lq . If g ∈ Lq we can
R1
define φg on Lp by φg (f ) = 0 f g and Hölder still works. On the other hand, if we have
a functional φ on Lp we have no “basis”
( vectors to apply φ to.
1 x∈A
We can apply φ instead to 1A (x) =
so we get numbers φ(1A ) and the map
0 x 6∈ A
A 7→ φ(1A ) is aR signed measure. So by the Radon-Nikodým
theorem there is a function g
R
with φ(1A ) = g1A for all A. So φ is given by f 7→ f g for f = 1A and by the standard
machine φ = φg on Lp .
We motivate the next steps we take. Suppose X is a Banach
P space, how might we
define a functional on X? Choose a basis {eα }α∈A and then φ( λα eα ) = λ1 is a linear
functional. This though will not in general be continuous, or how do we show so? We
need, ironically, the axiom of choice to prove that a continuous functional exists.
First though we consider RC[0, 1] and its dual space. If µ is a finite Borel measure
1
on [0, 1] then the map f 7→ 0 f dµ is a bounded linear functional on C[0, 1]. It turns
out that C[0, 1]? ∼
= {finite Borel signed measures} but this is aR subtle theorem, the Riesz
1
representation theorem. Newton showed that the map f 7→ 0 f dx is a bounded linear
functional. Thus Riesz says that the ordinary integral is given by a measure, so Riesz
implies the existence of the Lebesgue measure. Every linear functional on C[0, 1] is given
by a signed measure.
2.1
1-Codimensional Subspaces
Definition 2.2 A subspace Y of a Banach space X is called 1-codimensional if
1. Y 6= X
2. If Z is a subspace such that Y ⊂ Z ⊂ X then Z = Y or Z = X
Theorem 2.3 If Y is a subspace of a Banach space X then the following are equivalent.
1. Y is 1-codimensional
2. Y 6= X but if x ∈ X \ Y then span{Y ∪ {x}} = X
3. Y = ker φ for some non-zero linear functional φ on X. (note φ is not necessarily
bounded)
Proof 1 ⇐⇒ 2 is easy. If Y is 1-codimensional and x ∈ X \ Y then Y ⊂ span{Y ∪ {x}}
but Y 6= span{Y ∪ {x}} and so span{Y ∪ {x}} = X. On the other hand if Y satisfies 2
and Y ⊂ Z ⊂ X and Z 6= Y then choose z ∈ Z \ Y . Then Z ⊃ span{Y ∪ {z}} = X.
1, 2 =⇒ 3. Assume Y is 1-codimensional and choose x ∈ X \ Y . Each element
z ∈ X can be written as y + λx for some y ∈ Y and λ ∈ R. The representation is unique
because if y1 + λ1 x = y2 + λ2 x then (λ1 − λ2 )x = y2 − y1 ∈ Y and so λ1 = λ2 and hence
y1 = y2 . Now define φ(y + λx) = λ. Clearly Y ⊂ ker φ and φ 6= 0. ker φ = Y because
Y ⊂ ker φ ( X and so must be Y .
3 =⇒ 2. Suppose Y = ker φ and φ 6= 0. Take any x such that φ(x) 6= 0. We want to
show that Y ∪ {x} spans X. Given z ∈ X we want to choose y, λ such that z = y + λx.
φ(z)
We do this as follows. Set λ = φ(x)
and y = z − λx. Then φ(y) = φ(z − λx) = 0 and so
y ∈ Y as required.
Q.E.D.
Lemma 2.4 If Y = ker φ is 1-codimensional in X then
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MA3G8 Functional Analysis II Lecture Notes Spring 2013
1. Y is closed if and only if φ is continuous
2. Y is dense if and only if φ is not continuous.
Proof Since Y 6= X it cannot be closed and dense. So it suffices to prove that Y is
closed if φ is continous and Y is dense if not.
If φ is continuous, then ker φ = φ−1 ({0}) and so is closed. We need to show if φ is not
continuous then ker φ is dense.
φ is unbounded (i.e. not continuous) so choose a sequence (xn ) in X with ||xn || ≤ 1
φ(x)
with |φ(xn )| → ∞. If x ∈ X choose yn = x − φ(x
xn and then φ(yn ) = 0 so yn ∈ ker φ.
n)
On the other hand
φ(x) φ(x)
||xn || → 0
||yn − x|| = || −
xn || = φ(xn )
φ(xn ) and so x ∈ closure(Y ).
2.2
Q.E.D.
Separation Theorems
We state the important theorem, give some applications and then prove it, as this way
you are motivated why the theorem is so important.
Theorem 2.5 (Hahn-Banach) If Y is a subspace of a real normed linear space X and
φ : Y → R is a continuous linear functional, then there is a continuous linear functional
φ̃ : X → R which extends φ, i.e. φ̃(y) = φ(y) if y ∈ Y and ||φ̃|| = ||φ||.
Corollary 2.6 If x ∈ X is a unit vector then there is a functional φ̃ ∈ X ? with ||φ̃|| ≤ 1
and φ̃(x) = 1.
The corollary thus says that there is a supporting hyperplane to the unit ball B at each
point on the boundary.
Proof Let Y = span{x} ⊂ X and define φ : Y → R by φ(λx) = λ. Then φ(x) = 1 and
|φ(λx)| = |λ|||x|| = ||λx|| and so ||φ|| = 1. Then by Hahn-Banach we can extend to φ̃
with ||φ̃|| = 1 and φ̃(x) = 1.
Q.E.D.
Corollary 2.7 (Closest point witnesses) Let Y be a closed subspace of X and x ∈ X.
Let d = inf{||x − y|| : y ∈ Y }. Then there is a linear functional φ on X with
1. φ(y) = 0 for all y ∈ Y
2. ||φ|| = 1
3. φ(x) = d
Thus φ witnesses the fact that x is at least distance d from Y , since if y ∈ Y then
||x − y|| = ||φ||||x − y|| ≥ φ(x − y) = φ(x) − φ(y) = d
Proof Let Z = span{Y ∪{x}} and define φ : Z → R by φ(y +λx) = dλ. It satisfies 1 and
3 and we want to show that |φ(y + λx)| ≤ ||y + λx||. But how do we understand the right
hand side of this. We know that ||x − y|| ≥ d for all y ∈ Y and so ||λx − λ( −y
λ )|| ≥ |λ|d
and so we have this bounded. Then we have |φ(y + λx)| = |λd| = |λ|d ≤ |λ|||x − −y
λ || =
||y+λx||.By the Hahn-Banach theorem there is a functional on X with the same properties.
Q.E.D.
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MA3G8 Functional Analysis II Lecture Notes Spring 2013
2.2.1
Reflexivity
We saw that if x ∈ X we can define an element of X ?? , x̂ by x̂(φ) = φ(x) for φ ∈ X ? . The
map x 7→ x̂ is linear and ||x̂|| ≤ ||x|| since |x̂(φ)| = |φ(x)| ≤ ||φ||||x||. In fact ||x̂|| = ||x||
and so the map x 7→ x̂ is an isometric embedding of X into X ?? , because, if ||x|| = 1 then
by the Hahn-Banach there is a φ ∈ X ? with ||φ|| = 1 and φ(x) = 1 and so x̂(φ) = φ(x) = 1
and ||φ|| = 1 and so ||x̂|| ≤ 1. If ||x|| =
6 1 we scale.
Definition 2.8 If the map x 7→ x̂ is surjective then we say that X is reflexive.
2.2.2
Adjoints
Definition 2.9 If T : X → Y is linear and bounded then the adjoint T ? : Y ? → X ? is
defined by
T ? (ψ)(x) = ψ(T x)
for ψ ∈ Y ? and x ∈ X.
Lemma 2.10 The following are properties of the adjoint
1. for each ψ, T ? ψ is linear
2. T ? is linear
3. ||T ? || = ||T ||
Proof
1. clear since T is linear.
2.
T ? (ψ + φ)(x) = (φ + ψ)(T x) = φ(T x) + ψ(T x) = T ? (φ)(x) + T ? (ψ)(x)
and so T ? (ψ + φ) = T ? (φ) + T ? (ψ).
3. We have that
|T ? (ψ)(x)| = |ψ(T x)| ≤ ||ψ||||T x|| ≤ ||ψ||||T ||||x||
and so ||T ? (ψ)|| ≤ ||T ||||ψ|| and thus ||T ? || ≤ ||T ||.
We claim that ||T ? || ≥ ||T ||. We can choose x ∈ X with ||x|| = 1 and ||T x|| ≥ ||T ||−ε
and then by Hahn-Banach there is a ψ ∈ Y ? with ||ψ|| = 1 and |ψ(T x)| ≥ ||T || − ε.
Then |T ? (ψ)(x)| = |ψ(T x)| ≥ ||T || − ε but ||x|| = 1 = ||ψ|| and so ||T ? || ≥ ||T || − ε
and let ε → 0.
Q.E.D.
H?
In a Hilbert space, if we identify
with H via the inner product then this adjoint
is the usual one in the real case.
In finite dimensions we can represent T as a matrix with respect to bases in X and
Y . Then the matrix of T ? is the transpose of that of T , if we use the dual bases for Y ?
and X ? .
Theorem 2.11 ( 1 -step extension) If Y is a 1-codimensional subspace of a normed
space X and φ : Y → R is a bounded linear functional then there is an extension φ̃ : X →
R of φ with ||φ̃|| ≤ ||φ||
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MA3G8 Functional Analysis II Lecture Notes Spring 2013
The inequality of the norms is crucial if we want to apply this infinitely often.
To extend into 1 dimension higher we will have a single degree of freedom.
Proof
By scaling we may assume that ||φ|| = 1. Choose x ∈ X \ Y . We can write every
point of X uniquely as y + λx for y ∈ Y and λ ∈ R. For each α define φα : X → R by
φα (y + λx) = φ(y) + λα
Our aim is to show that for an appropriate α, ||φα || ≤ 1, since each φα is clearly linear.
We need to check that
|φ(y) + λα| ≤ ||y + λx||
for all y ∈ Y and λ ∈ R. This is automatic if λ = 0 so by considering λ−1 y instead of y
it suffices to establish this for λ = 1, and all y ∈ Y .
We thus want
|φ(y) + α| ≤ ||y + x||
for all y ∈ Y . This is the same as
φ(y) + α ≤ ||y + x||
∀y ∈ Y
and
φ(z) + α ≥ −||z + x||
∀z ∈ Y
This means we need to choose α such that
−||z + x|| − φ(z) ≤ α ≤ ||y + x|| − φ(y)
∀y, z ∈ Y
such an α exists provided
−||z + x|| − φ(z) ≤ ||y + x|| − φ(y)
∀y, z ∈ Y
in other words we need
φ(y) − φ(z) ≤ ||y + x|| + ||z + x||
but
φ(y) − φ(z) = φ(y − z) ≤ ||y − z|| = ||y + x − (z + x)|| ≤ ||y + x|| + ||z + x||
as required.
Q.E.D.
Definition 2.12 A partially ordered set (Ω, ≤) is a set Ω equipped with a relation ≤
satisfying
1. x ≤ x for all x ∈ Ω
2. If x ≤ y and y ≤ z then x ≤ z.
3. For every x, y ∈ Ω, if x ≤ y and y ≤ x then x = y
An example is to take Ω to be the set of all subsets of {1, 2, 3, 4} and A ≤ B means
A ⊆ B. Then {1} ≤ {1, 2} and ∅ ≤ {3} but {1, 2} is not related to {3}.
Definition 2.13 A chain in a partially ordered set is a subset in which every pair of
elements is related.
An upper bound for a chain C ⊂ Ω is an element u ∈ Ω with c ≤ u for all c ∈ C
A maximal element m of Ω is an element for which there is no v with m ≤ v but
v 6= m.
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MA3G8 Functional Analysis II Lecture Notes Spring 2013
Note that this doesnt mean that m ≥ u for all u ∈ Ω.
An example of such is Ω = {∅, {1}, {2}, {3}, {1, 2}} ordered by inclusion. Then the
maximal elements are {1, 2} and {3}.
Theorem 2.14 (Zorn’s Lemma) If (Ω, ≤) is a non-empty partially ordered set in which
every chain has an upper bound, then Ω contains at least one maximal element
If you have a 1-step extension then a maximal element must be the top, i.e. in our context
the Banach space.
Proof (Hahn Banach) Let Ω be the partially ordered set of pairs (Z, φZ ) where Z is
a subspace of X containing Y and φZ : Z → R is a linear functional extending φ and
satisfying ||φZ || ≤ ||φ||, with the ordering (Z1 , φZ1 ) ≤ (Z2 , φZ2 ) if Z1 ⊂ Z2 and φZ1 and
φZ2 agree on Z1 . It is clear that this is a partial order and Ω is non empty since (Y, φ) ∈ Ω.
Suppose we have a chain {(Zα , φZα )}α∈A in Ω. Let W = ∪A Zα and define φW : W →
R as follows: If w ∈ W then w ∈ Zα for some α and we set φW (w) = φZα (w) for that
α. This is well defined since if w ∈ Zα and w ∈ Zβ we have Zα ⊂ Zβ , say, and then φZβ
extends φZα because we are in a chain. W is a subspace since if u, v ∈ W then u ∈ Zα
and v ∈ Zβ say and then Zα ⊂ Zβ so u, v ∈ Zβ so u + v ∈ Zβ ⊂ W , and similarly if
Zα ⊂ Zβ . In the same way φW is linear.
Moreover, if w ∈ W with w ∈ Zα then
|φW (w)| = |φZα (w)| ≤ ||φ||||w||
and since this works for all w, ||φW || ≤ ||φ||. Thus (W, φW ) ∈ Ω and clearly this is an
upper bound for the chain. By Zorn’s lemma, Ω contains a maximal element (Z, φZ ). If
Z 6= X then we can find an x ∈ X \ Z and by the 1-step extension theorem we can extend
φZ to span{Z ∪ {x}}, contradicting the maximality of (Z, φZ ).
Q.E.D.
Homework 4 asked to use Hahn-Banach to build a functional φ on l∞ which is bounded
such that if (xi ) is convergent then φ((xi )) = limn→∞ xn . This is not unique though, and
is rather strange.
2.2.3
Complex Hahn Banach
If X is a complex normed space then we can regard it as a real normed space, using only
the real numbers as the scalars.
Lemma 2.15 If X is a complex normed space and φ : X → C is linear then there is a
real linear functional u : X → R with
φ(x) = u(x) − iu(ix)
If φ is bounded then ||φ|| = ||u||. Conversely, given u we can build φ and ||φ|| = ||u||.
This is saying that for a linear functional, the real and imaginary parts are related.
An example of this is with X = C and φ(z) = z. Thus φ(x + iy) = x + iy and so we
have u(x + iy) = x. Then the image under u of the unit disc is very small compared to
the image of it under φ. However, it still has norm 1.
Proof Let u and v be the real and imaginary parts of φ so that
φ(x) = u(x) + iv(x)
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u, v : X → R and are real linear since for λ ∈ R,
u(λx) + iv(λx) = φ(λx) = λφ(x) = λu(x) + iλv(x)
Also we have that u(ix) + iv(ix) = φ(ix) = iφ(x) = iu(x) − v(x) and equating real and
imaginary parts gives
u(ix) = −v(x)
u(x) = v(ix)
and so we have φ(x) = u(x) − iu(ix).
p
|φ(x)| = u(x)2 + u(ix)2 ≥ |u(x)|
and so ||φ|| ≥ ||u||. On the other hand for any x we can write |φ(x)| = eiθ φ(x) = φ(eiθ x) =
u(eiθ x) − iu(ieiθ x) = u(eiθ x) for some θ ∈ R. Since |φ(x)| is real we have
|φ(x)| = u(eiθ x) ≤ ||u||||x||
and so ||φ|| ≤ ||u||.
Conversely, if we build φ as φ(x) = u(x) − iu(ix) then φ is linear and its real part is
u so ||φ|| = ||u||.
Q.E.D.
Theorem 2.16 (Complex Hahn-Banach) If Y is a subspace of a complex Banach
space X and φ : Y → C is linear then there is an extension φ̃ of φ with ||φ̃|| ≤ ||φ||.
Proof Write φ(x) = u(x) − iu(ix) for a real linear functional u on Y . Extend u to
ũ : X → R with ||ũ|| ≤ ||u|| by the real Hahn-Banach theorem. Set φ̃(x) = ũ(x) − iũ(ix)
for x ∈ X. Then φ̃ is complex linear and extends φ. Furthermore
||φ̃|| ≤ ||ũ|| ≤ ||u|| = ||φ||
Q.E.D.
Corollary 2.17 If X is a real Banach space then X is reflexive if and only if X ? is
reflexive.
Without the Hahn-Banach theorem, we cannot prove the ⇐= direction.
Proof We first prove the =⇒ direction. We want to show that each element F of X ???
is φ̂ for some φ ∈ X ? . Define φ : X → R by φ(x) = F (x̂). If x ∈ X, then
φ̂(x̂) := x̂(φ) := φ(x) := F (x̂)
and so φ̂ = F on all elements of X ?? of the form x̂. If X is reflexive, then this is all of
X ?? .
Now the other direction. Suppose that X is not reflexive. Then the image of X under
∧ map is a closed subspace of X ?? since it is complete. Since we are assuming that it
is not the whole of X ?? we can find F ∈ X ??? which is not 0 but with F (x̂) = 0 for all
x ∈ X. By assumption F = φ̂ for some φ ∈ X ? . But if x ∈ X then
φ(x) = x̂(φ) = φ̂(x̂) = 0
This contradicts φ 6= 0
Q.E.D.
Corollary 2.18 l1 and l∞ are not reflexive.
Proof c0 is not reflexive, and l1 ∼
= c?0 and l∞ ∼
= l1?
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Q.E.D.
MA3G8 Functional Analysis II Lecture Notes Spring 2013
2.3
Weak and Weak-? Topologies
Definition 2.19 Let X be a Banach space. The weak topology on X consists of all
possible unions of sets of the form
{x ∈ X : φ1 (x) < α1 , ..., φn (x) < αn }
for some finite sequence φ1 , ..., φn ∈ X ? and αi ∈ R.
In the same way, the weak-? topology on X ? consists of unions of sets of the form
{φ ∈ X ? : φ(x1 ) < α1 , ..., φ(xn ) < αn }
for x1 , ..., xn ∈ X and αi ∈ R.
The sets of the above form can be called a “ basis of neighbourhoods” for the open
sets.
If X is reflexive, then the weak and weak-? topologies on X ? are the same.
The weak topology on X makes each φ ∈ X ? continuous, because φ−1 (U ) is a union
of neighbourhoods if U is open in R.
U = ∪α (aα , bα ) = ∪α ({xα < bα } ∩ {−xα < −aα })
and thus
φ−1 (U ) = ∪α {x : φ(x) < bα , −φ(x) < −aα }
The weak topology is the weakest topology which makes each element of X ? continuous.
If X is infinite dimensional then the weak and weak-? topologies are not metrisable, so
compactness is not the same as sequentially compactness.
Theorem 2.20 (Banach-Alaoglu) The unit ball BX ? is compact in the weak-? topology
Q The proof uses Tychonov’s theorem. If (Xα ) are topological spaces then the product
α Xα is the Cartesian product {(xα ) : xα ∈ Xα ∀α} with the topology consisting of
unions of sets {(xα ) : xα1 ∈ Uα1 , ..., xαn ∈ Uαn } where Uαi is open in Xαi for each i. Then
Tychonov’s theorem says the product
Q of compact spaces is compact.
Proof We consider the product x∈BX [−1, 1] with the topology from R.
We can embed BX ? into this by φ 7→ (φ(x)) so φ : BX → [−1, 1]. This is a 1-1
embedding. If φ 6= ψ there is an x ∈ BX where φ(x) 6= ψ(x). The topology inherited by
BX ? from the product topology is exactly the weak-? topology. Neighbourhoods restrict
only finitely many values φ(x1 ), ..., φ(xn ). The product is compact by Tychonov’s theorem.
It suffices to prove that the image of BX ? is closed. Suppose f is in the closure of the
image. We want that if x, y ∈ BX , and λ, µ ∈ R with |λ| + |µ| ≤ 1 then
f (λx + µy) = λf (x) + µf (y)
as then this will imply that f is the restriction to BX of some linear functional.
Given ε > 0, the functions g satisfying
1. |g(x) − f (x)| < ε
2. |g(y) − f (y)| < ε
3. |g(λx + µy) − f (λx + µy)| < ε
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MA3G8 Functional Analysis II Lecture Notes Spring 2013
form an open set in the product, since we are restricting three values of g. So this set
meet sthe image of BX ? , say φ is a linear functional whose image satisfies the inequalities,
hence
|f (λx + µy) − λf (x) − µf (y)| = |g(x) − f (x) + g(y) − f (y) + g(λx + µy) − f (λx + µy)|
< (1 + |µ| + |λ|)ε
< 2ε
and letting ε → 0 we have f (λx + µy) = λf (x) + µf (y)
3
Q.E.D.
The Category Theorems
If C1 ⊃ C2 ⊃ ... is a decreasing sequence of non empty closed subsets of a compact space
∞
then ∩∞
1 Ci 6= ∅. This is not the case in R, since [0, ∞) ⊃ [1, ∞) ⊃ .. and ∩k=1 [k, ∞) = ∅.
Theorem 3.1 (Cantor) If C1 ⊃ C2 ⊃ ... is a decreasing sequence of non empty closed
subsets of a complete metric space X and diam(Cn ) → ∞ as n → 0 then ∩∞
1 Cn 6= ∅
Proof For each n, choose xn ∈ Cn . Since xn ∈ Cm if n ≥ m then d(xn , xm ) ≤
diam(Cm ) → 0 and so the sequence (xn ) is Cauchy, and let x be the limit. Since xn ∈ Cm
for all n ≥ m and Cm is closed then x ∈ Cm for all m and so x ∈ ∩Cm .
Q.E.D.
Definition 3.2 A subset E of a metric space X is called nowhere dense if its closure
contains no non-empty open subset of X.
Theorem 3.3 (Baire) A complete metric space X is not the countable union of nowhere
dense subsets of itself.
Corollary 3.4 R is not countable.
This is since it is the union of all singletons {x}.
Proof Let E1 , E2 , ... be nowhere dense in X. We want to find x 6∈ ∪Ei . We can assume,
by taking closures, that the Ei s are closed and each contains no non empty open set.
Choose x1 6∈ E1 . There is a ball centred at x1 which does not meet E1 , and hence a
closed ball B(x1 , r1 ) ∩ E1 = ∅. We may assume that r1 < 1.
The open ball B(x1 , r1 ) is not contained in E2 since E2 is nowhere dense. So choose
x2 ∈ B(x1 , r1 ) but x2 6∈ E2 . Since E2 is closed there is a ball B(x2 , r2 ) which is contained
in B(x1 , r1 ) with B(X2 , r2 ) ∩ E2 = ∅. We may assume that r2 < 21 .
Continuing in this way we build B(x1 , r1 ) ⊃ B(x1 , r1 ) ⊃ B(x2 , r2 ) ⊃ B(x2 , r2 ) ⊃ ...
with ri < 1i and B(xi , ri ) ∩ Ei = ∅.
By Cantor’s theorem, there is a point in all the B(xi , ri ) and hence in none of the Ei s.
Q.E.D.
We shall use this to prove the “analogue” of the 1-1 implies onto theorem for a finite
dimension vector space. This is the Open mapping theorem.
Baire’s theorem allows you to construct nasty things. There are too few nice things
so there must be some nasty ones. At the end of the section we shall show that there is a
continuous 2π periodic function on [−π, π] whose Fourier series diverges at 0. This uses
the Uniform Boundedness Principle.
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MA3G8 Functional Analysis II Lecture Notes Spring 2013
Theorem 3.5 (UBP) Suppose φ1 , φ2 , ... are bounded linear functionals on a Banach
space X. If for each x ∈ X the values {φn (x)} form a bounded set, then there is some M
with ||φn || ≤ M for all n.
This at first seems strange, if not simply wrong. Take for example the following.
Suppose that X = l1 . Then take φ1 big in direction e1 , and zero in the rest, φ2 huge
in e2 and zero in the rest, etc. Then at e7 φ7 is gigantic, but everything else is zero.
However the functionals are not uniformly bounded. This is not a contradiction as we
can find some clever point x which picks up some bad behaviour from lots of the φ s.
Suppose
that φi : l1 → R is the coordinate functional φi (x) = xi . Then if x ∈ l1 , we
P
have
|φi (x)| = ||x||1 < ∞ which is essentially the opposite to the above. Thus clever
means very clever.
This example is as bad as it gets.
Theorem 3.6 (Sharp UBP/Plank)
If (φi ) are unit functionals (operator norm 1) and
P
(wi ) are positive numbers with
wi < 1 then there is a unit vector x ∈ X with |φn (x)| >
wn for all n.
Proof (Plank =⇒ UBP) If (||φn ||) is unbounded we can choose a subsequence φnk
with ||φnk || ≥ 4k . Let ψk = φnk /||φnk || which is a unit functional. Choose x with
1
|ψk (x)| ≥ 2k+1
and ||x|| ≥ 1. Then |φnk (x)| ≥ 2k−1 → ∞.
Q.E.D.
Proof (UBP) Suppose (φn (x)) is bounded for each x ∈ X. For each k ∈ N, let Ck =
{x ∈ X : |φn (x)| ≤ k, ∀n}. These sets are closed and they cover X. Thus by Baire’s
theorem at least one is not nowhere dense, and hence includes a ball of positive radius.
In fact Ck = kC1 , so C1 includes such a ball, say the ball of radius r > 0 around y.
C1 is convex and symmetric so if u ∈ C1 then −u ∈ C1 . If ||x|| ≤ r we have y + x and
−y + x in C1 and so the average x = 21 (y + x) + 12 (−y + x) ∈ C1 by convexity. Thus if
||x|| ≤ r then |φn (x)| ≤ 1 for all n. Hence ||φn || ≤ 1r for all n.
Q.E.D.
3.1
Open Mapping Theorem
Recall that if T : Rn → Rn is linear then it is one to one if and only if it is onto.
Theorem 3.7 (Open mapping) If T : X → Y is a linear map from a Banach space X
to a Banach space Y which is onto then T is an open map, i.e. T (BX ) ⊃ rBY for some
r > 0.
Corollary 3.8 (Inverse mapping theorem) If X and Y are Banach spaces and the
linear map T : X → Y is one to one and onto and bounded then T −1 is bounded.
This looks like a cheat since since we assume one to one and onto, but it isnt because
we conclude that T is invertible in the sense that it has a bounded inverse (in the sense
of analysis).
The proof of the OMT has two steps. The first uses Baire and completeness of Y and
the second uses an iterative argument and completeness of X. They are linked by a clever
remark.
Proof (OMT theorem 3.7) We shall start by showing that for some positive r, the
closure T (Bα ) includes rBY .
For each k let Ck = T (kBX ) for k = 1, 2, ... These closed sets cover Y since T is onto
and hence by Baire’s theorem at least one has non empty interior. Since Ck = kC1 in
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MA3G8 Functional Analysis II Lecture Notes Spring 2013
fact C1 has non empty interior. Suppose C1 includes the ball of radius r > 0 around y.
Since C1 is convex and symmetric, it includes the ball of radius r around 0, rBY . Thus
T (Bα ) ⊃ rBY .
In particular, if ||y|| ≤ r there is an x ∈ X with ||x|| ≤ 1 so that
r
||y − T x|| ≤
(3.1)
2
We shall now prove that T (2BX ) ⊃ rBY , and so T (BX ) ⊃ 2r BY .
Suppose that y ∈ Y and ||y|| ≤ r. Choose x1 ∈ BX with ||y − T x1 || ≤ 2r by (3.1).
Now by applying (3.1) to the vector y − T x1 we can find x2 with norm ||x2 || ≤ 21 with
||y − T x1 − T x2 || ≤
r
4
1
Continuing we obtain (xj ) with ||xi || ≤ 2i−1
and ||y − T (x1 + ... + xi )|| ≤ 2ri . Since X
P∞
Pi
is complete, the sum
P∞ 1 xj converges
P∞ 1 to x and ||y − T x|| = lim ||y − T ( 1 xj )|| = 0 so
y = T x and ||x|| ≤ 1 ||xj || ≤ 1 2i−1 = 2
Q.E.D.
We could try to find ui such that ||y − T ui || → 0. If the ui have a convergent
subsequence then its limit is in BX and has image Y . This though requires compactness.
Proof (Inverse mapping theorem 3.8) By the OMT, we know that if ||y|| ≤ r there
is an x with ||x|| ≤ 1 and y = T x for some r > 0. Thus ||T −1 y|| = ||x|| ≤ 1 and so
||T −1 || ≤ 1r .
Q.E.D.
3.2
Closed Graph Theorem
If T : X → Y is a linear map, its graph is the set {(x, T x) : x ∈ X} ⊂ X × Y .
If X and Y are Banach spaces then X × Y is a Banach space with
||(x, y)|| = ||x|| + ||y||
but note that this isn’t the only norm we can use. We can thus ask whether the graph is
closed in X × Y . The answer is yes if whenever xn → x and T xn → y then y = T x.
If T is continuous, the graph is closed since in this case xn → x =⇒ T xn → T x, even
if we do not know that T xn converges.
Theorem 3.9 (Closed Graph) If T is a linear map between Banach spaces X and Y
then T is bounded if and only if its graph is closed in X × Y .
Proof Let G = {(x, T x) : x ∈ X} be the graph of T . It is a closed subspace of X × Y
so is a Banach space in the inherited norm. Define
π1 : G → X
by π1 (x, T x) = x
π2 : X × Y → Y
by π2 (x, y) = y
These are obviously linear. π2 is continuous because
||y|| ≤ ||x|| + ||y|| = ||(x, y)||
π1 is continuous because
||x|| ≤ ||x|| + ||T x|| = ||(x, T x)||
π1 is one to one and onto, since if u = x we have (u, T u) = (x, T x). By the IMT (or
OMT), π1 has a bounded inverse. Since T = π2 ◦ π1−1 , it is bounded.
Q.E.D.
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3.3
Application to Fourier Analysis
There is a continuous 2πP
periodic function on [−π, π] whose Fourier series diverges at 0.
int where
The Fourier series is ∞
−∞ an e
Z π
1
f (x)e−inx dx
an =
2π −π
The N th partial sum is
N
X
an e
int
−N
Z
Z π
N
N
X
1
1 X int π
−inx
f (x)e
dx =
f (x)
=
e
ein(t−x) dx
2π
2π −π
−π
−N
−N
The sum inside is
N
X
ein(t−x) =
−N
e−N (t−x) − ei(N +1)(t−x)
1 − ei(t−x)
1
=
1
e−(N + 2 )(t−x) − ei(N + 2 )(t−x)
(t−x)
(t−x)
e−i 2 − ei 2
sin((N + 21 )(t − x))
=
sin( 21 (t − x))
and thus
N
X
an e
−N
int
1
=
2π
f (x)
sin((N + 12 )(t − x))
dx
sin( 12 (t − x))
f (x)
sin((N + 12 )(x))
dx
sin( 21 (x))
π
Z
−π
In particular, the value at t = 0 is
N
X
−N
an =
1
2π
Z
π
−π
The bit on the left hand side is supposed to converge to f (0), and so you would expect
the graph of the kernel DN to spike at 0. We would like this Fourier series to approximate
f , meaning that
Z π
1
f (x)DN (x)dx ≈ f (0)
2π −π
We would like the kernel to reproduce f . Thus the kernel should have integral 1 with its
mass concentrated near
R πt, it is an “approximation to the identity”
1
The map f 7→ 2π −π f (x)DN (x)dx = SN (f ) is a bounded linear functional on the
space of continuous 2π periodic functions, with norm ||f || = maxx∈[−π,π] |f (x)|. By UBP,
we can deduce that there is an f where (SN (f )) is unbounded if (||SN ||) is unbounded.
We claim that
Z π 1
1
sin(N + 2 )x ||SN || =
dx = ||DN ||1
2π −π sin 12 x SN acts on L∞ [−π, π] by the same integral formula and its norm is at most ||DN ||1 .
On L∞ the norm clearly is ||DN ||1 because
R π we can apply it to sgn(DN ) in L∞ and
1
||sgn(DN )||∞ = 1 but SN (sgn(DN )) = 2π
−π (sgn(DN ))DN = ||DN ||1 . However sgn(DN )
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MA3G8 Functional Analysis II Lecture Notes Spring 2013
is not continuous but we can approximate by continuous functions of L∞ norm 1 and which
are 2π periodic.
, ..., ± NN+π1 and
Now DN vanishes at ± N π+ 1 , ± N2π
+1
2
2
2
kπ
1
||DN ||1 =
π
Z
0
π
N+ 1
Z 2
N
sin(N + 1 )x X
1
2
dx ≥
sin 21 x π
k=1
(k−1)π
N+ 1
2
1
≥
π
N
X
kπ
N+ 1
2
Z
k=1 (k−1)π
N+ 1
2
sin(N + 1 )x 2
dx
sin 12 x sin(N + 1 )x
2
dx
sin 2Nkπ+1 because x 7→ sin x is increasing on [0, π]. Now if x > 0 we have sin x ≤ x so for each k,
1
≤ 2(Nkπ+1) and so we get that
sin kπ
2(N +1)
2(N + 1)
||DN ||1 ≥
π2
N
X
k=1
Z
1
k
N
2(N + 1) X 1
=
π2
k
k=1
=
kπ
N+ 1
2
1
sin(N + )x dx
2 (k−1)π
N+ 1
2
Z
0
π
1
N+ 2
1
sin(N + )xdx
2
N
4 X1
π2
k
k=1
4
> 2 log N → ∞
π
and so we have the result that we want
4
Unbounded Operators
We motivate the following with a classical example. We have a taut string fixed at 0 and
1 and at time t = 0 it is plucked into shape x 7→ f (x, 0). Thereafter its shape is f (x, t).
Then f obeys the wave equation
∂2f
∂2f
c2 2 = 2
∂x
∂t
with the boundary conditions f (0, t) = 0 = f (1, t). We now expand f (x, t) as a Fourier
series to get it as
∞
X
an (t) sin(nπx)
f (x, t) =
1
Note that sin nπx vanishes at 0 and 1 so we immediately satisfy the boundary conditions,
and this is also the reason why we do not consider the cosine terms. By extending to an
odd function on [−1, 1] we have an expression in L2 . If we can differentiate term by term
then
∞
∞
X
X
∂2f
∂2f
00
=
an (t) sin(nπx)
=−
an (t)n2 π 2 sin(nπx)
∂t2
∂x2
1
1
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MA3G8 Functional Analysis II Lecture Notes Spring 2013
By the wave equation we have
a00n (t) = −c2 n2 π 2 an (t)
P
and so an (t) = An cos(cnπt) + Bn sin(cnπt). At t = 0 we have f (x, 0) = ∞
1 Cn sin(nπx)
and if we furthermore assume that ∂f
|
=
0,
i.e.
we
have
a
slope
but
no
speed,
then we
∂t t=0
get that Bn = 0 for all n and thus
an (t) = An cos(cnπt)
and An = Cn and so
f (x, t) =
∞
X
An cos(cnπt) sin(nπx)
1
The method above worked because the functions sin nπx are eigenfunctions of the
map f 7→ f 00 satisfying the boundary conditions f (0) = 0 = f (1) and are orthogonal in
L2 [0, 1]. This is what we expect from a self adjoint operator. The Laplacian f 7→ f 00 is
not a bounded operator on any useful space, so we need a theory of unbounded operators.
The Laplacian is “ like ” a self adjoint operator though, since using integration by parts
twice we get:
Z 1
00
hf , gi =
f 00 (x)g(x)dx
0
Z 1
0
1
= f (x)g(x)|0 −
f 0 (x)g 0 (x)dx
0
Z 1
= f 0 (x)g(x)|10 − f (x)g 0 (x)|10 +
f (x)g 00 (x)dx
0
so if f and g are
and f (0) = 0 = f (1) and g(0) = 0 = g(1) then we get hf 00 , gi = hf, gi.
So f 00 looks self adjoint on smooth functions satisfying the Dirichlet boundary conditions.
The boundary conditions are thus part of the definition of the operator. Dirichlet and
Neumann boundary conditions make it “like” a self adjoint operator.
The apparent self adjointness is not enough though for spectral theory.
C2
Definition 4.1 A linear operator T is called densely defined in the Hilbert space H if
it is defined and linear on a dense subspace T : D(T ) → H.
Note that f 00 is defined and linear on C 2 which is dense in L2
Definition 4.2 We say that S extends T if D(S) ⊃ D(T ) and S(x) = T (x) for x ∈
D(T ).
Definition 4.3 We say that T is symmetric if
hT x, yi = hx, T yi
if x, y ∈ D(T ).
Definition 4.4 We define the adjoint of a densely defined T as follows. We define
D(T ? ) to be all y for which x 7→ hT x, yi is a bounded linear functional on D(T ). In this
case the functional extends uniquely to H. It can be represented uniquely by an inner
product x 7→ hx, zi for some z ∈ H. We call this z by T ? y, so
hT x, yi = hx, T ? yi
if x ∈ D(T ) and y ∈ D(T ? ).
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MA3G8 Functional Analysis II Lecture Notes Spring 2013
Definition 4.5 T is called self adjoint if D(T ) = D(T ? ) and T = T ?
Note that T is symmetric is the same as saying that T ? extends T .
Warning though that D(T ? ) might not even be dense.
The map f 7→ f 00 on C 2 [0, 1] with f (0)R = 0 = Rf (1) is not self adjoint, but it is
1
1
symmetric. There are functions g for which 0 f 00 g = 0 f g 00 but with g not in the given
domain, because g 00 is not continuous.
The integration by parts repeated works fine if f (0) = 0 = f (1) = g(1) = g(0) and f 0
and g 0 are absolutely continuous, meaning
Z x
Z x
0
0
v(t)dt
u(t)dt
g (x) =
f (x) =
0
0
and then f 0 g 0 makes sense by Hölder and f 00 = u and g 00 = v. To find the correct domain
on which to study
f 00 we set H 1 [0, 1] to be the space of absolutely continuous functions
Rx
f (x) = c + 0 u(t)dt with u ∈ L2 [0, 1] and H 2 [0, 1] to be the space of differentiable
functions f on [0, 1] with f 0 (x) ∈ H 1 .
RxRt
f 00 may not exist everywhere but f (x) = a + bx + 0 0 u(s)dsdt and u ∈ L2 . Let D
be the space of functions
R
D = {f ∈ H 2 [0, 1] : f (0) = 0 = f (1)}
Theorem 4.6 The map f 7→ f 00 is self adjoint on D.
R
Proof We need to show that if the map f 7→ f 00 g is a bounded linear functional on D
R
R
RxRt
then g ∈ D and g(x) = a + bx + 0 0 u(s)dsdt and f 00 g = f u, as then we would have
?
D(T ) = D(T ? ) and hT
R x,00yi = hx, T yi as required.
R
If the map f 7→ f g is a bounded linear functional then we can write it as f h
RxRt
for some h ∈ L2 by Riesz-Fréchet. Then let G(x) = 0 0 h(s)dsdt. Then G ∈ H 2 and
G00 (x) = h.
We want to check that G is almost g. Then
Z 1
Z 1
00
0
1
f (x)G(x)dx = f (x)G(x)|0 −
f 0 (x)G0 (x)dx
0
0
Z 1
0
1
0
1
= f (x)G(x)|0 − f (x)G (x)|0 +
f (x)G00 (x)dx
0
Z 1
= f 0 (x)G(x)|10 +
f (x)G00 (x)dx
0
Recall that
Z 1
R
f 00 g
=
R
f h but also
G00
= h so we have
Z 1
Z
00
0
1
0
1
f (x)G(x)dx = f (x)G(x)|0 +
f (x)h(x)dx = f (x)G(x)|0 +
0
0
1
f 00 g
0
We apply to f which not only satisfies f (0) = 0 = f (1) but to those which satisfy as well
f 0 (0) = 0 = f 0 (1).
Thus G − g is orthogonal to all f 00 = u ∈ L2 for functions f ∈ D satisfying also
f 0 (0) = 0 = f 0 (1).R
x
Since f 0 (x) = 0 u(t)dt the condition f 0 (1) = 0 is equivalent to the statement
Z 1
u(t)dt = 0
0
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MA3G8 Functional Analysis II Lecture Notes Spring 2013
and since f (x) =
Z
0=
1
Rx
0
f 0 (t)dt so we have
f 0 (t)dt = xf 0 |10 −
Z
1
xf 00 (x)dx = −
Z
xf 00 (x)dx = −
0
0
0
1
Z
1
xu(x)dx
0
and so G − g is orthogonal to all elements of L2 satisfying
Z 1
Z 1
u=0
xu(x)dx = 0
0
0
thus G − g is orthogonal to all u ∈ L2 which are themselves orthogonal to linear functions.
Thus G − g must be linear. This means
g(x) = a + bx + G(x) ∈ H 2
R1
R1
R1
and g 00 = G00 = h. This means 0 f 00 g = 0 f h = 0 f g 00 but also, integrating by parts
twice gives
Z
Z
1
1
f 00 g = f 0 (x)g(x)|10 +
f (x)g 00 (x)dx
0
0
and so in fact f 0 (x)g(x)|10 = 0 for all f ∈ D. We can find f ∈ D with arbitrary values of
f 0 (0), f 0 (1) and so we must have g(0) = 0 = g(1).
We have proved that the domain of the adjoint is not too big. Conversely, if f, g ∈ D
then integration by parts shows
Z 1
Z 1
f 00 g =
f g 00
0
0
Q.E.D.
4.1
Closed Operators
Definition 4.7 A densely defined operator T in a Hilbert space H with domain D(T ) is
called closed if its graph is closed, i.e. if xn ∈ D(T ) and xn → x and T (xn ) → y in H
then x ∈ D(T ) and T (x) = y.
The closed graph theorem can thus be reworded as if T is closed and D(T ) = H then
T is bounded. In other words, closed is almost bounded, if you are not defined on the
whole space.
Example 4.1 Suppose H = l2 and define T : C00 → l2 by
T (x1 , x2 , ..., xn , 0, 0, ...) = (x1 , 2x2 , ..., nxn , 0, ...)
where C00 is the set of all finitely non zero sequences. This operator is not closed. Let
u1 = (1, 0, ...),
u2 = (1, 1/4, 0, ...), ...
un = (1, 1/4, 1/9, ..., 1/n2 , 0, ...), ...
and then T u1 = u1 , T un = (1, 1/2, ..., 1/n, 0, ...) and un → (1, 1/4, 1/9, ...) = x and
T un → (1, 1/2, 1/3, ...) = y but x 6∈ C00 = D(T ).
However, we can extend this operator to be closed. Choose
X
D = {(xi )∞
n2 x2n < ∞}
1 :
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MA3G8 Functional Analysis II Lecture Notes Spring 2013
P 2 2
and observe that
n xn < ∞ enforces T x ∈ l2 . You might call this the natural domain.
Then T is defined on D and we claim that it is closed.
Suppose ui → x and T ui → y = (y1 , y2 , ...) in l2 . The map
(θ1 , θ2 , ...) 7→ (θ1 , θ2 /2, θ3 /3, ...)
is bounded on l2 , and we denote it by S. Then if u ∈ D we have S(T u) = u. So
ui = S(T ui ) → S(y)
since S is bounded and so x = Sy and also
X
n2 x2n =
X
n2
y 2
n
n
=
X
yn2 < ∞
and so x ∈ D and so T x = y since xi = 1i yi .
In other words what we have done in the above is to define the domain of definition
so that the “inverse” S works as it should.
Consider the map f 7→ f 0 in L2 [0, 1], initially defined on C 1 [0, 1] the space of continuously differentiable functions. The operator is not symmetric on this space, because of
the change of sign when you perform integration by parts.
However, f 7→ if 0 is symmetric with the right boundary conditions.
The map is not closed on C 1 ; we find fn → f in L2 with fn0 → g in L2 but f 6∈ C 1 .
Choose f to be the function 0 at 0 and 1, to be 1/2 at 1/2 and piecewise linear. Then
choose g to be the function 1 up to 1/2 and -1 from 1/2 to 1, as in the pictures (YET
TO DRAW).
Then choose continuous approximations gn converging to g in L2 , for example


0 ≤ x ≤ 1/2 − 1/n
1
gn (x) = linear 1/2 − 1/n ≤ x ≤ 1/2 + 1/n


−1
1/2 + 1/n ≤ x ≤ 1
Rx
and so gn are indeed continuous and gn → g in L2 . Set fn (x) = 0 gn (t)dt and so fn ∈ C 1
and fn0 = gn by construction. Note that f 6∈ C 1 . It
R x is easy to see that fn → f in L2
because the map which takes u to the function
x
→
7
0 u isR a bounded linear operator on
R
L2 . Thus since fn0 → g in L2 we have fn → g in L2 and g = f .
We now extend the domain so that the function is closed.
Theorem 4.8 The map f 7→ f 0 is closed on the domain
{f ∈ H 1 [0, 1] : f (0) = f (1)}
which is called periodic H 1
Note that this is the natural space for f 0 .
Proof
Suppose fn is in periodic H 1 and fn → f in L2 and fn0 → g in L2 . Let G(x) =
Rx
1
0
0 g(t)dt and note that G ∈ H and
R x G = g. We now want to show that G is almost f .
The map that takes u to x 7→ 0 u(t)dt is bounded on L2 and so
Z x
Z x
0
fn (t)dt →
g(t)dt = G(x)
0
0
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MA3G8 Functional Analysis II Lecture Notes Spring 2013
in L2 and also
Z
fn (x) = fn (0) +
x
fn0 (t)dt
0
and these converge to f . Thus fn (0) → f − G in L2 and so f − G is constant. We can
thus write
f (x) = c + G(x)
R1
and so f ∈ H 1 and f 0 = G0 = g. Also note that 0 = fn (1) − fn (0) = 0 fn0 for all n and
R1
R1
R1
since fn0 → g in L2 then 0 g = 0 and so f (1) − f (0) = 0 f 0 = 0 g = 0 and so f is in
periodic H 1
Q.E.D.
For if 0 , we have
1
Z
0
if ḡ =
if ḡ|10
Z
−
0
1
Z
0
if ḡ =
0
1
¯0
f ig
0
We now show that the integral as an operator is bounded. We are showing
Z x
u(t)dt
u 7→ x 7→
0
is bounded on L2 , i.e.
2
1 Z x
Z
u(t)dt
0
Z
dx ≤ K
0
1
u(t)2 dt
0
Now
Z
0
1 Z x
0
2
2
Z 1 Z 1
u(t)dt dx =
u(t)χ[0,x] dt dx
0
0
Z 1 Z 1 Z 1
2
≤
u
χ[0,x] dx
0
0
0
Z 1 Z 1
=
u2
xdx
0
0
Z
1 1 2
u
=
2 0
Theorem 4.9 If T is densely defined in H then
1. T ? is closed
2. If D(T ? ) is dense then T has a closed extension, T ?? .
3. If T is symmetric then it has closed extension T ? .
In particular, if T is self adjoint then it is closed.
Proof
1. Suppose xn → x in H and T ? xn → y and xn ∈ D(T ? ). The map u 7→ hT u, xn i is a
bounded linear functional for each n. Then
hT u, xn i → hT u, xi
since xn → x and also
hT u, xn i = hu, T ? xn i → hu, yi
since T ? xn → y. Then hT u, xi = hu, yi and because the map u 7→ hu, yi is a bounded
linear functional we conclude that x ∈ D(T ? ) and T ? x = y.
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MA3G8 Functional Analysis II Lecture Notes Spring 2013
2. Since D(T ? ) is dense, T ?? does exist. By 1 it is closed, but does it extend T .
Suppose x ∈ D(T ). The map u 7→ hT ? u, xi = hu, T xi is defined on D(T ? ) and is a
bounded linear functional since T x ∈ H. So x ∈ D(T ?? ) and hu, T xi = hT ? u, xi =
hu, T ?? xi for all u ∈ D(T ? ). Since this is dense, T ?? x = T x.
3. This is immediate from 1, since T symmetric means T ? extends T , and T ? is closed.
Q.E.D.
4.2
The Spectrum
Definition 4.10 For bounded operator T : H → H on a Hilbert space, we define
1. The resolvent consists of λ ∈ C for which T − λI is invertible.
2. The point spectrum consists of λ ∈ C for which T − λI is not injective
3. The continuous spectrum consists of λ ∈ C for which T − λI is one to one, is
not onto but for which Im(T − λI) is dense.
4. The residual spectrum consists of λ ∈ C with T − λI being one to one but Im(T −
λI) is not dense.
Definition 4.11 For an operator T : H → H (not necessarily bounded) on a Hilbert
space, we define
1. The resolvent consists of λ ∈ C for which T − λI is one to one and Im(T − λI) is
dense in H and (T − λI)−1 is bounded on the image of D(T ).
2. The point spectrum consists of λ ∈ C for which T − λI is not one to one on D(T ).
3. The continuous spectrum consists of λ ∈ C for which T − λI is one to one,
Im(T − λI) is dense but (T − λI)−1 is not bounded on Im(T − λI).
4. The residual spectrum consists of λ ∈ C with T − λI is one to one but Im(T − λI)
is not dense.
Theorem 4.12 If T is a self adjoint operator in a Hilbert space then its residual spectrum
is empty.
Proof Suppose Im(T − λI) is not dense. Lets use S to denote T − λI. There is a
non-zero y ∈ H with hSx, yi = 0 for all x ∈ D(T ) = D(S). So the map x 7→ hSx, yi is a
bounded linear functional, and hence y ∈ D(S ? ) = D(T ? ) = D(T ) as T is self adjoint. So
0 = hx, Syi for all x ∈ D(T ) and so Sy = 0. Thus S = T − λI is not one to one. Q.E.D.
Note that if T is bounded then the two definitions agree. It is enough to check the
resolvent since 2 and 4 are identical. We need to check that if S is bounded and densely
defined then its extension to H is invertible if and only if S is one to one, Im(S) is dense
and S −1 is bounded on Im(S).
If the extension is invertible then clearly S is one to one and S −1 is bounded on Im(S).
Initially S is defined on a dense subspace D(S). If y ∈ H the extension S̃ is onto so there
is an x with S̃x = y. Choose xn ∈ D(S) with xn → x. Then S̃xn → S̃x = y because S̃ is
bounded, but S̃xn = Sxn so y can be approximated from Im(S) so Im(S) is dense.
In the other direction, S −1 has a continuous extension, call it T . We shall show that
S̃ ◦ T and T ◦ S̃ are the identity so S̃ is invertible. Suppose y ∈ H and choose yn ∈ Im(S)
with yn → y. Then S −1 yn → T y so S(S −1 yn ) = yn → S̃(T y) so y = S̃(T y) so 4S̃ ◦ T is
the identity. Similarly T ◦ S̃ is the identity.
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MA3G8 Functional Analysis II Lecture Notes Spring 2013
5
The Laplacian and Quantum Harmonic Oscillator.
We have already seen that the map f 7→ f 00 is self-adjoint on Dirichlet H 2 [0, 1]. Thus it
has no residual spectrum. We also observed that there is a complete orthonormal basis
of L2 [0, 1] consisting of the eigenfunctions x 7→ sin nπx with eigenvalue −n2 π 2 .
We would like an analogue of the spectral theorem
1. The spectrum consists only of the eigenvalues (the operator has purely point spectrum).
2. The corresponding eigenvectors form a complete orthonormal basis.
P
00 on H 2 can be expressed as follows. If f 00 = g =
3. The map
f
→
7
f
θn sin nπx then
P θn
f (x) =
sin n pix.
−n2 π 2
In other words the operator multiplies
the coefficientPby the eigenvalue −n2 π 2 or
P
forPeach f ∈ Dirichlet H 2 , f (x) =
γn sin nπx where
n2 π 2 γn2 < ∞ and f 00 (x) =
2
2
− n π γn sin nπx.
This is also true for the Laplacian on a nice domain Ω ⊂ R2 but we shall only prove
it in one dimension, by directly writing down the resolvent.
We need that if λ 6= −n2 π 2 then (T − λI)−1 is bounded on Im(T ), i.e. everything
that is one to one is in the resolvent, and so not in the spectrum. We know that if
fP(x) = sin nπx P
then f 00 (x) − λf (x) = (−n2 π 2 − λ) sin nπx so the resolvent should be
θn
θn sin nπx →
sin πnx
−n2 π 2 −λ
This is clearly a bounded operator since we divide the sequence of coefficients with respect to an orthonormal basis by non-zero numbers which tend to infinity. The
Poperator is
2 and f 00 − λf = g =
even compact.
We
need
to
check
that
if
f
∈
Dirichlet
H
θn sin nπx
P
θn
−1
then f =
sin nπx, i.e. if g ∈ Im(T −λI) then we have (T −λI) g well defined.
−n2 π 2 −λ
We shall use “boot strapping”. We have
f 00 = λf + g
(5.1)
and g ∈ L2 is nice and f ∈ H 2 is very nice. The fact that f satisfies a differential equation
means that smoothness of f and g passes to f 00 and so that f is much smoother.
To do this we start by observing that f ∈ L2 so we may write
X
f (x) =
γn sin nπx
Our aim is to prove that γn = −n2θπn2 −λ for each n.
P
Note that we can’t state that f 00 (x) = −n2 π 2 γn sin nπx because this might not even
make sense.
P
By the differential equation (5.1) above we have f 00 (x) = (λγn + θn ) sin nπx. Integration is a bounded operator on L2 so we can integrate term by term, so as to get
f (x) = cx + d +
X (λγn + θn )
−n2 π 2
sin nπx
Then if c = d = 0 we have
X (λγn + θn )
−n2 π 2
sin nπx =
X
in L2 and so we get −γn n2 π 2 = λγn + θn as required.
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γn sin πnx
MA3G8 Functional Analysis II Lecture Notes Spring 2013
To eliminate the linear term we use the boundary conditions f (0) = 0 = f (1). On
the face of it this sum at 0 and 1 vanishes since we chose sin nπx to do this, but an L2
convergent series might
P not even have a meaning at a particular point. We need something
stronger. However, (λγn + θn ) < ∞ because f, g ∈ L2 and so
X 1 X λγn + θn X
2
≤
<∞
(λγ
+
θ
)
n
n
−n2 π 2 n4 π 4
P (λγn +θn )
so the series
sin nπx converges uniformly.
−n2 π 2
2
Since f ∈ H , not just L2 it should come as no surprise that its Fourier series converges
much better than in L2 . Thus the sum does vanish at 0 and 1 and f (0) = d and f (1) = c+d
so c = d = 0.
We have that (T − λI)−1 is bounded on Im(T − λI) if λ 6= −n2 π 2 . Setting λ = 0 we
get that the spectral representation for T −1 and hence T .
We now consider higher dimensions. The laplacian is
∆f (x, y) =
∂2f
∂2f
+
∂x2
∂y 2
and we consider the Dirichlet Laplacian. The eigenfunctions are (x,
Py)17→ sin mπx sin nπy
and the eigenvalue is −(n2 + m2 )π 2 . In k dimensions we replace
with
n4 π 4
Z
X
1
1
≈
dx = ∞
2
2
4
2
4
(n1 + ... + nk ) π
Rk |x|
If the dimension k is large enough we don’t get the boundary conditions automatically.
There are no tricks, on Ω a random domain we dont know the eigenfunctions. You need
a big theorem.
On R the map f 7→ f 00 has many more eigenvalues eitx 7→ −t2 eitx so −t2 is an
eigenvalue. On [0, 1] we have discrete eigenvalues, but on R we have the whole of (−∞, 0]
in the spectrum.
A Schrödinger operator is a map L : f 7→ −f 00 +V (x)f where V is a potential. To solve
the Schrödinger equation in the same way as the vibrating (
string we need the spectral
0 x ∈ [0, 1]
properties of L. The Dirichlet Laplacian corresponds to V =
.
∞ x ∈ (−∞, 0) ∪ (1, ∞)
The particle is bound in [0, 1] and this forces the discrete energy levels.
5.1
Quantum Harmonic Oscillator
The simplest example of a Schrödinger operator on R is
T f (x) = −f 00 (x) +
x2
f (x)
4
where we have taken our potential V to be x2 /4. This potential tends to ∞ fairly rapidly.
Particles can wander over all of R but typically they are at distance about 1 from 0. The
2
states are bound. If we put f (x) = p(x)e−x /4 then
−f 00 (x) +
x2
1
2
f (x) − λf (x) = e−x /4(p00 (x) + xp0 (x) − (λ − )p(x))
4
2
and we claim that for n = 0, 1, 2, ... the map p 7→ −p00 (x) + xp0 (x) has eigenvalue n with
eigenfunction a polynomial of degree n. This gives rise to the eigenvalues 12 , 32 , 25 , ... for
the original T .
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MA3G8 Functional Analysis II Lecture Notes Spring 2013
The map L which is p 7→ −p00 (x) + xp0 (x) acts on polynomials of degree n as follows:
L(1) = 0,
L(x2 ) = −2 + 2x,
L(x) = x,
L(x3 ) = −6x + 3x3
and can be represented by

0
0

0
0

0 −2 0 . . .
1 0 −6 . . .

0 2
0 . . .
0 0
3 ...
and this matrix is upper triangular and so its eigenvalues are 0, 1, 2, ... and the eigenfunction corresponding to n is a polynomial of degree at most n. These are related to the
Hermite polynomials. Let
2 /2
F (x, t) = ext−t
=
∞
X
pn (x)
0
n!
tn
We can show that pn is a polynomial of degree n in x. This is left as an exercise. We
want that −p00n (x) + xp0 (x) = npn (x) or equivalently
X (−p00 (x) + xp0 (x))
n
n!
=
X npn (x)
n!
or alternatively
∂2F
∂F
∂F
+x
=t
2
∂x
∂x
∂t
which is true upon differentiating F . We call such an F a generating function.
2
T is supposed to be self adjoint so we would like the eigenfunctions pn (x)e−x /4 to be
orthogonal. We would like to know that if m 6= n then
Z ∞
Z ∞
2
2
e−x /2
e−x /2
pn (x)pm (x) √
dx = 0
p2n (x) √
dx = ||pn ||2
2π
2π
−∞
−∞
−
or in other formulation
Z ∞ X
Z ∞X
2
−x2 /2
pm (x) m X pn (x) n e−x /2
pn (x)
ne
s
t √
(st) √
dx =
dx = W (st)
m!
n!
n!
2π
2π
−∞
−∞
m≥0
n≥0
n≥0
Lets see if this holds:
Z ∞
Z ∞
2
2
e−x /2
e−x /2
2
2
exs−s /2 ext−t /2 √
F (x, s)F (x, t) √
dx =
dx
2π
2π
−∞
−∞
Z ∞
1
dx
2
=
e 2 (x−s−t) est √
2π
−∞
Z ∞
1
dx
2
= est
e− 2 (x−s−t) √
2π
−∞
X (st)n
= est =
n!
and so the pn are orthogonal and ||pn ||2 = n!. In order to show that T has pure point
2
spectrum 21 , 32 , 52 , ... the main thing we need is that the pn (x)e−x /4 form a complete
orthonormal basis in L2 (R). We want that the pn form a complete orthonormal basis in
−x2 /2
L2 (R, e √2π ).
27 of 29
MA3G8 Functional Analysis II Lecture Notes Spring 2013
We know that polynomials are dense in L2 [a, b] because they are dense in C[a, b] with
−x2 /2
the much stronger uniform norm. On R with weight e √2π we can approximate in L2 by
continuous functions with bounded support. We can approximate these on the bounded
support by polynomials but the polynomials will explode elsewhere and not approximate
on the line.
−x2 /2
The usual proof of the density in L2 (R, e √2π ) uses the invertibility of the Fourier
transform. We shall use a different approach.
−x2 /2
If q ∈ L2 (R, e √2π ) then its coefficients with respect to the pn are
2
∞
pn (x) e−x /4
pn
√
an =
q(x)
dx = hq, i
n!
n!
2π
−∞
Z
and so the N th partial sum of the expansion is
y 7→
N
X
0
pn (y)
=
an √
n!
2
N
X
pn (x)pn (y) e−x /2
√
q(x)
dx
n!
2π
−∞
0
{z
}
|
Z
∞
KN (y,x)
and so KN is like the Dirichlet kernel.
It gives the best L2 approximations, but potentially bad with uniform approximations.
We shall use a better kernel built with a smooth cut of. The kernel given above cuts off
sharply at N , and so we take
K̃t (y, x) =
∞
X
pn (y)pn (x)
n!
0
2
e−x /4
tn √
2π
for 0 ≤ t < 1 and we will let t → 1. We try to rewrite this kernel in a better manner. We
know that
∞
X
pn (y) n n
F (y, tu) =
t u
n!
0
but instead of u we have pn (x) and so the idea is to make pn like u. We do this as follows:
Lemma 5.1
n
i pn (x)e
−x2 /2
Z
∞
=
2 /2
un eixu e−u
−∞
du
√
2π
Proof We first multiply by tn /n! and sum. On the left we get
∞
X
in pn (x)e−x
0
On the right we get
Z ∞X
∞
−∞
0
2 /2
1
tn
2
2
2
2
= e−x /2 F (x, it) = e−x /2 exit+t /2 = e− 2 (x−it)
n!
2
tn n ixu e−u /2
√
du =
u e
n!
2π
Z
∞
−∞
∞
Z
2 /2
etu+ixu−u
du
√
2π
1
2 1
2 du
e− 2 (u−ix−t) e 2 (ix+t) √
2π
−∞
Z ∞
1
1
du
2
2
= e 2 (ix+t)
e− 2 (u−ix−t) √
2π
−∞
=
28 of 29
MA3G8 Functional Analysis II Lecture Notes Spring 2013
− 1 (z−t)2
The integral is a contour integral along v 7→ v − ix of e 2√2π . By Cauchy’s integral
theorem, the integral around a rectangle is 0. We thus need the integral on the vertical
2
sides to tend to zero. This does as we are integrating e−z .
Q.E.D.
We thus have
K̃t (y, x) =
∞
X
pn (y)
n!
0
Z
n
Z
∞
t
2 /2
(−iu)n eixu e−u
−∞
du
√
2π
∞
1
2
F (y, −itu)eixu e−u /2 du
2π −∞
Z ∞
1
2 2
2
e−iytu+t u /2 eixu−u /2 du
=
2π −∞
2
Z ∞
(x−yt)2
1
− 12 (1−t2 ) u− ix−iyt
−1
2
1−t
=
e
e 2 1−t2 du
2π −∞
2 Z ∞
2
1 (u−fish)
du
1 − 12 (x−yt)
2
1−t
e− 2 shark √
=√ e
2π
2π
−∞
2
(x−yt)
1
1
1 −
= √ e 2 1−t2 √
2π
1 − t2
=
This kernel as a function of v is the density of a Gaussian with mean yt and variance
1 − t2 and so
Z
q(x)K̃t (y, x)dx ≈ q(yt) ≈ q(y)
29 of 29