Lecture Notes on:
Math 6710: Applied Linear Operators
and Spectral Theory (Applied
Mathematics)
Chris Miles
miles@math.utah.edu
Last Updated: June 18, 2015
Contents
1
2
3
4
Metric spaces
1.1 Preliminaries . . . . . . . . . . .
1.2 Open sets, closed sets, topology
1.3 Completeness . . . . . . . . . .
1.4 Banach Fixed-point Theorem . .
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Normed Vector Spaces, Banach Spaces
2.1 Preliminaries . . . . . . . . . . . . . . . .
2.2 Norms . . . . . . . . . . . . . . . . . . . .
2.3 Finite dimensional normed vector spaces
2.4 Compactness and finite dimension . . . .
Linear operators
3.1 Preliminaries . . . . . . .
3.2 Bounded linear operators
3.3 Linear functionals . . . .
3.4 Algebraic dual spaces . .
3.5 Dual spaces . . . . . . .
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1
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12
12
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14
14
15
Inner product spaces
4.1 Preliminaries . . . . . . . . .
4.2 Completions, Sobolev spaces
4.3 Orthogonal projections . . . .
4.4 Total orthonormal sets . . . .
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17
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20
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i
ii
CONTENTS
4.5
4.6
5
6
7
Representation of functions on Hilbert spaces . . . . . . . . . . . . . . . . 21
Lax-Milgram theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
Linear operators II
5.1 Hilbert adjoint . . . . . . . . . . . . . . . . .
5.2 Boundedness . . . . . . . . . . . . . . . . .
5.3 Reflexive spaces . . . . . . . . . . . . . . .
5.4 Weak convergence and weak* convergence
5.5 Compact operators . . . . . . . . . . . . . .
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25
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28
Spectral Theory
6.1 Preliminaries . . . . . . . . . . . . . . . . . . .
6.2 Spectral properties of bounded linear operators
6.3 Spectral properties of compact operators . . .
6.4 Bounded self-adjoint operators . . . . . . . . .
6.5 Return to Fredholm theory . . . . . . . . . . . .
6.6 Sturm-Liouville problem . . . . . . . . . . . . .
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30
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34
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39
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40
41
Distribution Theory
7.1 Multi-index notation . . . .
7.2 Definitions . . . . . . . . . .
7.3 Distributions . . . . . . . .
7.4 Operations on distributions
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Index
42
Disclaimer: These notes are for my personal use. They come with no claim of correctness or
usefulness.
Errors or typos can be reported to me via email and will be corrected as quickly as possible.
Acknowledgments: The content of these notes is based on lectures delivered (excellently) by
David C. Dobson for Math 6710 in Fall 2013.
The notes themselves are derived from notes originally taken by Gregory Handy.
Many proofs have been omitted for brevity but can be found in either:
• E. Kreyszig, Introductory Functional Analysis with Applications, Wiley (1989).
• F. G. Friedlander and M. Joshi, Introduction to the Theory of Distributions, Cambridge
(1999).
1
Metric spaces
1.1
Preliminaries
Definition 1.1. A metric space is a set X , together with a mapping d : X → R, with properties
for all x, y, z ∈ X :
(D1) d(x, y) ≥ 0 with d(x, y) = 0 iff x = y
(D2) d(x, y) = d(y, x)
(D3) d(x, y) ≤ d(x, z) + d(z, y)
Example 1.1. R is a metric space with d(x, y) := |x − y|.
Example 1.2. R3 is also a metric space with the Euclidean metric.
Example 1.3. Let `∞ := {x = (ξ1 , ξ2 , . . .) : ξj ≤ cx }, where cx does not depend on j , just x.
The distance defined on this metric space is: d(x, y) := supj |ξi − ηi |.
Example 1.4. Let C[a, b] be the set of all continuous functions on [a, b], then d(x, y) :=
maxt∈[a,b] |x(t) − y(t) is a metric space. We know the maximum is attained due to the continuity
of x, y .
Example 1.5. The following is known as the discrete metric space. X can be anything and the
metric is:
(
1 if x = y
d(x, y) :=
0
otherwise
.
This is typically used as a counter-example.
Example 1.6. Let B(A) be the set of all bounded functions on A, i.e., f : A → R. Then
∃cf : supt∈A |f (t)| ≤ cf . We define this metric to be d(x, y) := supt∈A |x(t) − y(t)|.
1
2
CHAPTER 1. METRIC SPACES
o
P
p
x = (ξ1 , ξ2 , . . .) : ∞
|ξ
|
<
∞
. The distance defined on this
j=1 j
P
1/p
∞
metric space is: d(x, y) :=
. Note that p = 2 is very similar to the Euclidean
j=1 |ξj − ηj
Example 1.7. Let `p :=
n
norm and will be a noteworthy case. It is worth definition two noteworthy inequalities:
Theorem 1.1 (Hölder’s inequality).
∞
X
|ξj ηj | ≤
j=1
p
X
!1/p
|ξj |p
·
j=1
p
X
!1/q
|ηj |p
,
j=1
where p, q are conjugate, i.e. 1/p + 1/q = 1. Note, Cauchy-Schwarz inequality is a special case
when p = q = 2.
Theorem 1.2 (Minkowski’s inequality).
p
X
!1/p
|ξj + ηj |p
j=1
1.2
≤
p
X
!1/p
|ξj |p
j=1
+
p
X
!1/p
|ηj |p
j=1
Open sets, closed sets, topology
For x0 ∈ X , define the following:
B(x0 , r) := {x ∈ X : d(x0 , x) < r} (open ball)
B̃(x0 , r) := {x ∈ X : d(x0 , x) ≤ r} (closed ball)
s(x0 , r) := {x ∈ X : d(x0 , x) < r} (sphere)
Definition 1.2. A subset M ⊂ X is called open if each point x0 of M has a ball B(x0 , ) ⊂ M
for some > 0. Also, K ⊂ X is closed if K C := X − K is open. A neighborhood N of x0 is
any set which contains an open ball around x0 .
Definition 1.3. x0 is an interior point of M if M is in a neighborhood of x0 . i.e. ∃ > 0 :
B(x0 , ) ⊂ M .
Definition 1.4. The collection of all interior points of M is called M 0 or Int(M ).
The collection J of all open sets on (X, d) is called a topology for (X, d) if it satisfies:
(T1) ∅ ∈ J
(T2) The union of any collection (infinite or finite) of elements of J is in J .
(T3) Any finite intersection is in J .
Note, no metric is needed for a topology but a metric always induces a topology.
1.3. COMPLETENESS
3
Definition 1.5. Let ρ : X → Y where (X, d) and (Y, D̄) are metric spaces. T is called
¯ x, T x0 ) < whenever d(x, x0 ) < δ .
continuous at x0 ∈ X if: ∀ > 0, ∃δ > 0 : d(T
Alternatively: T (B(x0 , δ)) ⊂ B(T x0 , ).
Definition 1.6. Let M ⊂ X . A point x0 ∈ X is called an accumulation point (or a limit
point), if every neighborhood of x0 contains at least one point M , distinct from x0 .
Definition 1.7. X plus all of its accumulation points is called the closure, denoted X . Interestingly, in a metric space B̃(x0 , ) is not necessarily equal to B(x0 , ).
Definition 1.8. A subset M ⊂ X is called dense if M = X .
Definition 1.9. A metric space is called separable if it has a countable dense subset.
Example 1.8. X = [0, 1], d(x, y) := |x − 1|. M = (0, 1) is dense in X .
Example 1.9. Q ⊂ R is dense, thus R is separable.
Example 1.10. C is separable for the same reason.
Example 1.11. `∞ is not separable.
Proof. Choose x ∈ [0, 1] and write as binary x = ξ1 +ξ2 /2+ξ3 /22 +· · · and let y = (ξi ) ∈ `∞ .
There are uncountably many of these but the distance between them is always 1. Thus, consider a
ball of radius 1/3 around each point. No countable dense subset exists. Thus, not separable.
Example 1.12. `p for 1 ≤ p < ∞ is separable.
Proof. Idea: (η1 , η2 , . . . , ηn , 0, . . .) is close enough.
1.3
Completeness
Definition 1.10. A sequence (xn ) in a metric space (X, d) is convergent if there is an
x ∈ X : limn→∞ d(xn , x) → 0.
Definition 1.11. A sequence (xn ) is Cauchy if ∀ > 0, ∃N : d(xn , xm ) < whenever
n, m > N .
Fact: every convergent sequence is a Cauchy Sequence is convergent, but the reverse is not
necessarily true.
Definition 1.12. A space is said to be complete if all Cauchy sequences convergence.
Definition 1.13. A sequence is bounded if the set of its elements is bounded.
4
CHAPTER 1. METRIC SPACES
Theorem 1.3. Convergent sequences are bounded. Furthermore, convergent sequences have
unique limits and if xn → x, yn → y , then d(xn , yn ) → d(x, y).
Theorem 1.4. Let M ⊂ X , then x ∈ M iff ∃(xn ) ⊂ M : xn → x. Also, M is closed iff
every sequence in M which converges, converges to a point in M .
Theorem 1.5. A subspace M of X is complete iff M is closed as a subspace of X .
Example 1.13. `∞ is complete.
(n)
(j)
Proof. Idea: Consider (xn = {ξ1 , . . .}), a Cauchy sequence, then ξ1 forms a Cauchy sequence
of real numbers, thus converges and defines our limit.
Example 1.14. Consider c, the subspace of `∞ consisting of the convergent sequences is
complete by noting that it is closed.
Example 1.15. `p for 1 ≤ p < ∞ is complete.
Example 1.16. C[a, b] is complete.
Proof. Let (xn ) be a Cauchy sequence in C[a, b]. For a fixed t0 , xn (t0 ) is a Cauchy sequence
of real numbers and therefore converges. Now, note that x(t) is continuous by noting that
|x(t) − x(t + h)| < |x(t) − xn (t)| + |xn (t) − xn (t + h)| + |xn (t + h) − x(t + h)| < , thus
x(t) is continuous.
Rb
Example 1.17. Q, polynomials on [a, b], and C[a, b] with the metric d(x, y) = a |x(t) −
y(t)| dt are all examples of incomplete spaces. Note the last example comes from approximating
a step function.
ˆ exists and is defined
Definition 1.14. The completion of a metric space (X, d), denoted (X̂, d)
by an isometry (distance preserving) map to a dense subset of X̂ . The completion is unique up to
isometry.
1.4
Banach Fixed-point Theorem
This theorem is also known as the contraction mapping principle. In general if we have a mapping
T : X → X and we want to solve T x = x, the solution is called a fixed point.
Definition 1.15. T is a contraction on a set S if ∃a < 1 : d(T x, T y) < a d(x, y) for all
x, y ∈ S .
Theorem 1.6 (Banach-fixed point theorem). If (X, d) is a complete metric space and T is a
contraction, then T has a unique fixed point x ∈ X .
We can define the iteration xn+1 = T xn for n = 0, 1, . . .. Using a geometric series argument,
an
we see that d(xn , x) ≤ 1−a
d(x0 , x), which gives us an estimate of the error.
1.4. BANACH FIXED-POINT THEOREM
5
Corollary 1.1. Let T be a mapping of a complete metric space (X, d) onto itself. Suppose T
is a contraction on some closed ball Y with radius r around a point x0 and also assume that
d(x0 , T x0 ) < (1 − a)r, then fixed point iteration converges to a unique fixed point in Y .
Thus, if we have a contraction on a smaller space, as long as it maps onto itself, we still have our
contraction mapping.
Example 1.18. We can apply this to linear algebra. Consider X = Rn with d(x, y) =
maxj |ξj − ηj |. Consider trying to solve x = Cx + B , where C = (cij ), B = (b1 , b2 , . . . bn ).
We hope to solve this by (I − C)x = B , the form of our fixed point iteration. Thus,
define T : X → X as T x := Cx + B , then we again hope to solve T x = x. Consider
|T x − T y| = |C(x − y)|. Note, in particular:
n
X
|T xj − T yj | = cjk (ξk − ηk )
k=1
≤
=
n
X
k=1
n
X
|cjk | max |ξk − ηk |
k
|cjk |d(x, y)
k=1
and take the max over j on both sides, yielding: K = maxj
parameter.
Pn
k=1
|cjk | as our contraction
A similar result holds for Jacobi iteration, where A = (A − D) + D, where D are the
diagonal entries of A. We now are solving the problem (A − D)x + Dx =
C, meaning
x = −D−1 (A − D)x + D−1 C . Using the same criteria we end up with
Pn
k=1
ajk ajj < 1 for a
contraction. Note this is equivalent to row diagonal dominance.
Gauss-Seidel iteration is also similar, where A = −L + D − U , where L, U are the lower and
upper triangular parts of A respectively. We now are solving the problem (D − L)x − U x = C ,
meaning x = (D − L)−1 U x + (D − L)−1 C . The condition above is sufficient but a looser
exists. Note this method converges in fewer iterations but is more difficult to parallelize than the
Jacobi iteration.
We can also use the contraction mapping to prove the following theorem from ordinary differential
equation theory:
Theorem 1.7 (Picard). Given the initial value problem:
(
x0 = f (x, t)
x(t0 ) = x0 ,
(1.1)
where f is continuous on R = {(t, x0 ) : |t − t0 | ≤ α, |x0 | ≤ b}. Since f is continuous, it
is bounded by |f (x, t)| ≤ c in this interval. Also, suppose f satisfies the Lipschitz condition:
6
CHAPTER 1. METRIC SPACES
∃K : |f (x, t) − f (y, t)| ≤ k|x − y|, then the IVP defined by 1.1 has a unique solution on the
interval J = [t0 − β, t0 + β], where β < min {a, b/c, 1/k}.
Proof. Integrating the IVP yields:
t
Z
x(t) = x0 +
f (τ, x(τ )) dτ
t0
Rt
Thus, define the integral operator: (T x)(t) := x0 + t0 f (τ, x(τ )) dτ , meaning we are solving
T x = x. Consider X = C[J], where J is defined above and define d(x, y) = maxt∈J |x(t) −
y(t)|. Note this is max not sup because we have continuous functions on an interval. Let
J ⊂ C[J] be the subspace consisting of functions that satisfy |x(t) − x0 | ≤ cβ , which is closed
and therefore complete. Now consider T : J → J , which yields:
Z t
|T x(t) − x0 | = f (τ, x(τ )) dτ t0
≤ c|t − t0 |
≤ cβ.
since x ∈ J , τ ∈ J
We must now show that T is a contraction:
Z t
|T x(t) − T y(t)| ≤ f (τ, x(τ )) − f (τ, y(τ )) dτ t0
≤ |t − t0 |K max |x(τ ) − y(τ )|
t∈J
= K|t − t0 | d(x, y)
≤ Kβ d(x, y).
This suggests that d(T x, T y) ≤ Kβ d(x, y), thus if Kβ < 1, we have a contraction, which is
true by the definition of β . Thus, we have the Picard iteration:
Z
t
xn+1 (t) = x0 +
f (τ, xn (τ )) dτ.
t0
Example 1.19. We can also consider integral equations. Consider those of the first kind to be
Rb
of the form: a k(t, τ )x(τ ) dτ , which is the continuous extension of multiplying by a matrix.
We assume k is continuous on this square and therefore is bounded.
We can also define integral equations of the second kind which take the form: x(t) −
Rb
µ a k(t, τ )x(τ ) dτ + v(t). Here, µ is a parameter and k is again assumed to be continuous. We typically hope to solve for x, which can be done by rearranging: (I − µk)x = v .
Rb
Thus, define the integral operator: T x(t) := µ a k(t, τ )x(τ ) dτ + v(t), meaning we hope to
1.4. BANACH FIXED-POINT THEOREM
7
T x(t) = x(t). We must show that T is a contraction:
Z b
|T x(t) − T y(t)| = µ
k(t, τ ) [x(τ )y(τ )] dτ aZ b
k(t, τ ) [x(τ )y(τ )] dτ ≤ |µ| a
≤ |µ|c(b − a) max |x(τ ) − y(τ )|
τ
= |µ|c(b − a) d(x, y)
Thus, if c|µ|(b − a) < 1, we have a contraction. Also note that if the integral upper bound is
changed to t, we can redefine our kernel to be 0 unless τ ≤ t.
2
Normed Vector Spaces, Banach
Spaces
2.1
Preliminaries
Definition 2.1. A vector space over a field K is a non-empty set of elements (vectors) together
with two algebraic operations (+, ×) that satisfy the usual vector space axioms not listed here.
Example 2.1. Rn , Cn are both vector spaces.
Example 2.2. C[a, b] is a vector space.
Example 2.3. `p are vector spaces for all p.
Definition 2.2. A subspace of a vector space X is a non-empty subset that is closed under
addition and multiplication with scalars and the subset is a vector space itself.
As usual, we can write elements in a vector space as (finite) linear combinations of the other
elements. A linearly independent set is one in which: α1 x1 +· · ·+αn xn = 0 =⇒ α1 = αn = 0,
otherwise they are linearly dependent.
Definition 2.3. A vector space is called finite-dimensional if there is a non-negative integer n
such that the space contains n linearly independent vectors but any subset of the space with n + 1
vectors is linearly dependent. In this case, we consider dim X = n.
Example 2.4. X = {0}, dim X = 0.
Example 2.5. dim Rn = dim Cn = n.
Example 2.6. dim `p = ∞.
If B is any linearly independent subset of X such that span B = X then B is called a (Hamel)
basis of X . Every x has a unique representation as a linearly combination of elements in B .
8
2.2. NORMS
9
Note, every vector space has a Hamel basis but it can’t always be found. Every Hamel basis has
the same cardinality. All proper subspaces have lesser dimension.
2.2
Norms
Definition 2.4. A norm k · k on a vector space X is a (continuous by Lipschitz, see reverse
triangle inequality) function X → R that satisfies the usual norm properties, including the
triangle inequality.
The norm provides us with the notion of the size of an element. Equipping a space with a norm
has a profound impact and provides complex structure.
Definition 2.5. A normed vector space (NVS) is a vector space with a norm defined. A norm
on X automatically defines a metric: d(x, y) = kx − yk and this is called the induced norm.
|x−y|
Note, not every metric is induced by a norm (for instance, d(x, y) = 1+|x−y| is not).
Definition 2.6. A Banach space is a normed vector space which is also complete. Note, every
normed vector space has a unique completion.
1/2
Example 2.7. Rn , Cn where kxk = (|x1 |2 + · · · |xn |2 )
Example 2.8. `p with kxk =
P
∞
j=0
|ξj kp
1/p
.
.
Example 2.9. C[a, b] with kxk = maxt |x(t)| is a Banach space, but inducing this same space
with the norm kxk =
R
b
a
1/2
|x(t)| dt
provides a normed vector space, but not a complete
2
one. The completion of this space is L2 [a, b], a very important space.
Theorem 2.1. A subspace Y ⊂ X is complete iff Y is closed in X . Note, needs to be a
subspace.
Definition 2.7. A sequence {xn } is said to be convergent if there exists an x ∈ X such that
limn→∞ kxn − xk → 0.
P∞
Definition 2.8. If j=1 kxj k ≤ ∞, the sum is considered absolutely convergent . Note:
{absolute convergence =⇒ convergence} ⇔ {X is complete}.
Definition 2.9. If a NVS X contains
P∞a sequence (e1 , e2 , . . .) with the property that every vector
can be represented uniquely x = j=0 αj ej , then (en ) is said to be a Schauder basis of X .
Example 2.10. In `p , the natural choice of basis vectors (1, 0, 0, . . .), (0, 1, 0, . . .), . . . form a
Schauder basis.
10
CHAPTER 2. NORMED VECTOR SPACES, BANACH SPACES
Theorem 2.2. If X has a Schauder basis, it is separable.
2.3
Finite dimensional normed vector spaces
Lemma 2.1. Let {x1 , . . . , xn } constitute a linearly independent set in a NVS X . Then there
exists a number c > 0 such that for every selection of scalars {α1 , . . . , αn }:
kα1 x1 + · · · αn xn k ≥ c(|α1 | + · · · |αn |).
Proof. Follows from Bolzano-Weierstrauss.
Theorem 2.3. Every finite dimensional subspace of a normed vector space is complete (regardless
of if the original vector space is complete).
Theorem 2.4. Every finite dimensional subspace of a normed vector space is closed in the
original space.
Definition 2.10. A norm k · k on a NVS X is said to be equivalent to another norm k · k0 if
there are constants a, b > 0 such that akxk0 ≤ kxk ≤ bkxk0 for all x ∈ X .
Theorem 2.5. On any finite dimensional NVS, all norms are equivalent.
2.4
Compactness and finite dimension
Definition 2.11. A metric space is (sequentially) compact if every sequence has a convergent
subsequence. Note: a subset is compact if every sequence in the subset converges to an element
of the subset.
Theorem 2.6. Every compact subset M of a metric space is closed and bounded. Note: the
converse is not always true.
Example 2.11. Consider (en ) ∈ `2 . The set is bounded and closed but every element is distance
1 apart and therefore has no convergent subsequence.
Theorem 2.7. If X is finite dimensional, then {closed and boundedness}⇔ compactness.
Proof. Uses Bolzano-Weierstrauss.
Lemma 2.2 (F. Riesz). Let Y and Z be a subspace of a NVS X , where Y is closed and a proper
subset of Z . Then for every number θ ∈ (0, 1) : ∃z ∈ Z : kzk = 1 and kz − yk > 0 for all
y ∈Y.
2.4. COMPACTNESS AND FINITE DIMENSION
11
Theorem 2.8. If the closed unit ball B = {x : kxk ≤ 1} is compact then X is finite dimensional.
Proof. Uses Lemma 2.2.
Theorem 2.9. Let T : X → Y be continuous, where X, Y are metric spaces, then the image
T (M ), where M is compact, is compact.
Theorem 2.10. A continuous function T : M → R over a compact M attains both a minimum
and maximum over M .
3
Linear operators
3.1
Preliminaries
Definition 3.1. A linear operator T is a mapping with the following properties:
(T1) The domain D(T ) is a vector space and the range, R(T ) lies in a vector space over the
same field.
(T2) T (ax + by) = aT x + bT y for all x, y ∈ D(T ) and scalars a, b.
Definition 3.2. Define the null space of an operator to be N(T ) = {x ∈ D(T ) : T x = 0}.
Example 3.1. I : X → X where Ix := x is the identity operator.
Example 3.2. If A ∈ Rm×n then Ax is defined by matrix operation over a finite dimensional
space.
Example 3.3. Let X = {p[0, 1] → R : p is polynomial} and define Dp(t) :=
unbounded.
Example 3.4. Q : L1 [0, 1] → L1 [0, 1] defined by Qf (t) :==
R1
1
Z
kQf k1 =
kQf (t)k dt
Z 1 Z t
=
k
f (s) dsk dtk
0
0
Z 1Z t
≤
|f (s)| ds dt
0
0
Z 1
≤
|f (s)| ds
0
0
= kf k1 .
12
0
dp
.
dt
Note this is
f (s) ds. Note in this case:
3.2. BOUNDED LINEAR OPERATORS
13
Thus, from this, we can conclude that Q maps [0, 1] onto itself.
Example 3.5. Let k be a continuous function on [0, 1]2 , then define K : C[0, 1] → C[0, 1] to
R1
be (Kf )(t) := 0 k(s, t)f (s) ds for t ∈ [0, 1]. This is again the continuous matrix extension.
Theorem 3.1. The operator T is injective (or one-to-one) if T x1 = T x2 =⇒ x1 = x2 . If
T : D(T ) → R(T ) is injective, then ∃T −1 : R(T ) → D(T ) defined by T −1 y = x where
T x = y . For linear operators, note that T −1 exists if N(T ) = {0}. Furthermore, if T −1 exists
and T is a linear operator, then T −1 is also a linear operator.
3.2
Bounded linear operators
Definition 3.3. Let X, Y be normed vector spaces and T : D(T ) ⊂ X → Y be a linear
operator. T is considered a bounded linear operator (BLO) if ∃c > 0 : kT xkY ≤ ckxkX for
all x ∈ D(T ).
For a bounded linear operator, the smallest constant c defines the norm on the operator:
kT xk
sup kT xk.
x∈D(T ) kxk x∈D(T )
kT k := sup
kxk=1
x6=0
Example 3.6. The identity operator is clearly bounded with kIk = 1.
Example 3.7. For a matrix A ∈ Rn×m then kAk = supkxk=1 kAxk is the spectral radius and
is therefore a bounded linear operator.
Example 3.8. Note that the derivative operator is not bounded because Dxn (t) = n → ∞,
where xn (t) := ntn−1 .
Example 3.9. Our standard integral operator (Kf )(t) :=
noting:
R1
0
k(s, t)f (s) ds is bounded by
Z 1
kKf k = max k(s, t)f (s) ds
t∈[0,1]
0
Z 1
≤ max
|k(s, t)| |f (s)|ds
t
0
Z 1
Z 1
≤ max
|f (s)| ds
|k(s, t)| ds
t
0
0
{z
}
|
=c
≤ ckf k.
Thus, we can conclude kKk ≤ kKf k/kf k ≤ c and therefore K is a BLO.
14
CHAPTER 3. LINEAR OPERATORS
Theorem 3.2. If a normed space X is finite dimensional, then every linear operator on X is
bounded.
Theorem 3.3. Let T : D(T ) ⊂ X → Y be a linear operator, then T is bounded iff T is
continuous (even at a single point).
Note that if T is bounded then N(T ) is closed. Also, if T1 , T2 satisfy the norm property:
kT1 T2 k ≤ kT1 k kT2 k.
3.3
Linear functionals
Consider the special case of a linear operator where f : D(F ) ⊂ X → K , where K is the field
associated with X , typically R or C, then this operator is called a linear functional. As with
linear operators, we can define the notion of boundedness:
Definition 3.4. A bounded linear functional (BLF) is a linear functional if there ∃c such that
|f (x)| ≤ ckf k. From this, we can define the norm of the functional to be:
kf xk
.
x∈D(f ) kxk
kf k := sup
x6=0
Example 3.10. Let y0 ∈ Rn be fixed, then the dot product with y0 , that is f x = x · y0 defines
a linear functional where |f (x)| = |x · y| ≤ kxkky0 k by Cauchy-Schwartz which suggests
kf k ≤ ky0 k.
R
Example 3.11. Consider y0 ∈ Lq and define Rf : Lp → R by f (x) := x(t)y0 (t) dt. It’s
clear f is linear and we can also note |f (x)| ≤ |x(t)||y0 (t)| dt ≤ kx(t)kp ky0 kq by Hölder’s
inequality. Thus, we have a BLF. Note we can obtain equality in the integral by choosing
x(t) := sgn y(t)|y(t)|q−1 .
Example 3.12. Consider x ∈ C[0, 1] and t0 ∈ [0, 1], then f (x) := x(t0 ) is a BLF. This can be
seen by noting |f (x)| = |x(t0 )| ≤ maxt |x(t)| = kxk, thus kf k ≤ 1.
Example 3.13. Consider again the space of polynomials on [0, 1] and let f (x) :=
that if we consider xn (t) = tn this is again unbounded.
3.4
df
(1).
dx
Note
Algebraic dual spaces
Definition 3.5. The set of all linear functionals over a vector space X forms a vector space,
which we will denote X ∗ and is called the algebraic dual.
3.5. DUAL SPACES
15
Note, since X ∗ is itself a vector space, it also has an algebraic dual, denoted X ∗∗ . We can
identify elements of this vector space: for x ∈ X , define g ∈ X ∗∗ which maps g : X ∗ → K by
g(f ) = f (x).
From this, we can define a mapping c : X → X ∗∗ by c(x) = g , where g is defined above. Note
that c is linear. This mapping is called the canonical embedding of X → X ∗∗ .
Definition 3.6. If c is bijective, X and X ∗∗ are isomorphic, in which case X is considered to
be algebraically reflexive.
Theorem 3.4. Finite dimensional vector spaces are always algebraically reflexive.
3.5
Dual spaces
Definition 3.7. The set of all bounded linear functionals on a normed vector space X is said
to be the dual space X 0 , where the definition of bounded linear functional remains the same as
defined in Definition 3.4.
Denote B(X, Y ) to be the space of all BLO from X to Y . This space is itself a vector space.
In fact, it is also a normed vector space by taking kBk = supkxk=1 kBxk. From this, we can
rewrite the dual space as X 0 = B(X, K).
Theorem 3.5. Let X be a NVS and Y be a Banach space, then B(X, Y ) is a Banach space.
Proof. We already know that we have a NVS. We need only show completeness, but this follows
from linearity of the operator.
Note we can think of this from a linear algebra perspective. Each x ∈ X has a unique
representation x = ξ1 e1 + · · · + ξn en , but conversely every n-tuple of coefficients {ξ1 , . . . , ξn }
defines a unique x under the same mapping. Thus, these spaces are isomorphic.
Denote T x := ξ1 T e1 + · · · ξm T em and let B = (b1 , . . . , bn ) be a basis for Y . For each k
there is a unique
τ1k b1 + · · · τnk bP
T ek = y , which then suggests
n sinceP
Pm representation:PTmek = P
n
n
m
that T x = k=1 ξk T (ek ) = k=1 ξk j=1 τjk bk = j=1 ( k=1 τjk ξk ) bj . Thus, n × m
matrices are just mappings from dim X = m → dim Y = n. The elements of X can be thought
of as m dimensional column vectors (i.e. a m × 1 matrix) and the elements of X ∗ are effectively
1 × m matrices, or row vectors.
From this, we can conclude that X = X ∗ for finite dimensional X and since every BLO on a
finite dimensional space is bounded, we know X ∗ = X 0 .
16
CHAPTER 3. LINEAR OPERATORS
Example 3.14. (Rn )0 = Rn . This follows from the above analysis. |f (x)| = | ξk γk | ≤
P
P
P
P
|ξk γk | ≤ ( |ξk |)1/2 ( |γk |)1/2 = kxk ( |γk |)1/2 . Note we can obtain equality by
choosing ξk = γk , thus kf k = kgk, where g = (γ1 , . . . , γn ).
P
Example 3.15. (`1 )0 =P`∞ . Note that `1 has a standard Schauder
basis
P
Pand can be written as
n
1 0
x ∈ `1 =⇒ lim
ξ
e
,
so
f
∈
(`
)
=⇒
f
(
ξ
e
)
=
ξk f (ek ). Note nowe
k k
Pn→∞ k=1 k k
P
that |f (x)| ≤
|ξk γk | ≤ supk |γk | |ξk | = supk |γk |kxk1 =⇒ kf k ≤ supk kγk k =
kgk∞ . We can again obtain equality in this case telling us that kf k = kgk∞ .
4
Inner product spaces
4.1
Preliminaries
Definition 4.1. An inner product space is a vector space X together with an inner product.
Definition 4.2. An inner product is a mapping h·, ·i : X × X → K , where K is the field.
Note the inner product satisfies the typical set of axioms.
p
Also note the inner product defines a norm: kxk = hx, xi. This is the norm induced by an
inner product, thus every inner product space is a normed vector space. One key inequality for
inner product spaces is the following:
Theorem 4.1 (Schwarz inequality). For every x, y ∈ X :
|hx, yi| ≤ kxkkyk
or equivalently:
| hx, yi |2 ≤ hx, xi hy, yi .
We can also provide two other properties about inner product spaces that are occasionally useful.
Theorem 4.2 (Parallelogram Law).
kx + yk2 + kx − yk2 = 2 kxk2 + kyk2
Theorem 4.3 (Polarization Identities). For a real inner product space:
hx, yi =
1
kx + yk2 − kx − yk2
4
and for a complex inner product space:
1
kx + yk2 − kx − yk2
4
1
= hx, yi =
kx + iyk2 − kx − iyk2 .
4
< hx, yi =
17
18
CHAPTER 4. INNER PRODUCT SPACES
Definition 4.3. A complete inner product space is called a Hilbert space.
Example 4.1. Rn is a Hilbert space with the inner product defined to be the dot product.
Example 4.2. Cn is also a Hilbert space with hx, yi := x · ȳ .
Example 4.3. `2 is a Hilbert space with hx, yi :=
P
ξj η̄j .
Example 4.4. `p for p 6= 2 is not a Hilbert space.
Example 4.5. L2 [a, b] is a Hilbert space where hx, yi :=
Rb
a
x(t)ȳ(t) dt.
Example 4.6. Again,Lp [a, b] is not a Hilbert space for p 6= 2.
Example 4.7. C[a, b] is not a Hilbert space with the ∞ norm.
Just as the norm is continuous, as is the inner product. Thus xn → x, yn → y =⇒ hxn , yn i →
hx, yi.
4.2
Completions, Sobolev spaces
Theorem 4.4. Let X be an inner product space, then there is a Hilbert space H and an
isomorphism T : X → W , where W ⊂ H and is dense. H is unique up to isomorphism (that
is, preserves inner product). In other words, every inner product space has a completion.
Now, we can see that C[a, b], along with C 1 [a, b], the subspace of C[a, b] where derivatives are
continuous and C01 [a, b], the subspace of C 1 [a, b] where the endpoints are both 0 are all Banach
spaces but not Hilbert spaces.
We know that L2 [a, b] is the completion of C[a, b], but what are the completion of the other
spaces? Define H1 [a, b] to be the completion of C 1 [a, b]. It is called a Sobolev space. The inner
product in this case is defined to be:
Z
hx, yiH 1 [a,b]
:=
b
Z
x(t)ȳ(t) dt +
a
b
x0 (t) + ȳ 0 (t) dt.
a
Similarly, the Sobolev space H01 [a, b] can be defined, corresponding to elements such that
limt→a+ = limt→a− = 0.
Example 4.8. Using this notion of Sobolev spaces, we can now begin to form alternate
formulations of problems, which we will revisit later. Consider the following ODE:
x00 − cx = f,
t ∈ (a, b),
x(a) = x(b) = 0.
4.3. ORTHOGONAL PROJECTIONS
19
Note, we can now multiply both sides of this equation by some function y(t) and integrate from
a to b to obtain:
Z
b
Z
00
b
x (t)y(t) dt − c
a
Z
b
x(t)y(t) dt =
a
f (t)y(t) dt
a
Integration by parts yields:
Z
−
b
0
0
Z
x (t)y (t) dt − c
a
b
Z
b
x(t)y(t) dt =
a
f (t)y(t) dt
a
and if we assume c = 1 then we have the following form:
− hx, yiH 1 = hf, yiL2 .
that in general
we can define H k (0, 1) with the norm: kxkH k :=
RNote
R
|x0 (t)|2 dt + · · · |xk (t)|2 dt. We know that H 0 = L2 .
4.3
R
|x(t)|2 dt +
Orthogonal projections
We know x, y are said to be orthogonal if hx, yi = 0 and is denoted x ⊥ y . Also, x can be
orthogonal to a set M if hx, yi = 0 for all y ∈ M . The set of all x that are orthogonal to a set Y
is denoted the orthogonal complement, denoted Y ⊥ .
Definition 4.4. A set M ⊂ X is convex if for every x, y ∈ M , the element λx + (1 − λ)y ∈ M
for λ ∈ [0, 1].
Theorem 4.5. Let X be an inner product space and M be a non-empty, convex, complete subset
of X . Then for x ∈ X there is a unique y ∈ M such that:
kx − yk = inf kx − ỹk.
ỹ∈m
Furthermore, if M is a subspace, then x − y is orthogonal to M .
If X is a vector space with subspaces Y, Z , we can write X as a decomposition, that is:
X = Y ⊕ Z , meaning every element x ∈ X can be written uniquely as x = y + z , where
y ∈ Y and z ∈ Z . In this case, X is called the direct sum of Y and Z . If Y, Z are orthogonal,
this is called an orthogonal decomposition.
Example 4.9. Consider Rn , then Y = {y = (ξ1 , . . . , ξn−1 , 0)} and Z = {y = (0, . . . , ηn )},
then Rn = Y ⊕ Z .
Theorem 4.6. Let H be a Hilbert space and Y be a closed, complete, convex subspace, then H
can always be written as an orthogonal decomposition:
H = Y ⊕ Y ⊥.
(4.1)
20
CHAPTER 4. INNER PRODUCT SPACES
In the above case, we can define P : X → Y to be the orthogonal projection such that P x = y ,
where x = y + z . Note that P 2 = P , N(P ) = Y ⊥ .
Definition 4.5. A set M is said to be orthogonal if hx, yi = 0 for all x, y ∈ M . It is called
orthonormal if kxk = 1 for all x ∈ M .
Example 4.10. In Rn , the standard basis elements are orthonormal.
Example 4.11. In L2 (0, 1), the set M = {sin(2πnx)} {cos(2πnx)} forms an orthogonal
set. It can be rescaled to be orthonormal. Note, this is the Fourier basis.
S
Note, for orthogonal elements, the triangle
the Pythagorean theorem, that is:
P inequality
P becomes
2
2
2
2
2
kx + yk = kxk + kyk and also k ξi k = kξi k .
If M = {e1 , . . . , en } is an orthogonal set and let Y = span M , then any y ∈ Y can be written
as y = α1 e1 + · · · + αn en =⇒ hyi ej = αj .
Pn
Note, now the sum i=1 hx, ei i ei is well defined for any x ∈ X since we know x = y + z for
y ∈ Y, z ∈ Y ⊥ . We then can define the projection of x onto y as:
y=
n
X
hx, ej i ej =
j=1
n
X
hy + z, ej i ej =
j=1
n
X
hy, ej i ej
j=1
Note that we know kxk2 = kyk2 + kzk2 which suggests that kyk2 ≤ kxk2 , which leads to the
following inequality:
Theorem 4.7 (Bessel’s inequality). Let (ek ) be an orthogonal sequence in X , then for every
x ∈ X:
∞
X
|hx, ej i| ≤ kxk2
j=1
Note that hx, ek i are called Fourier coefficients.
Also notice that if M is any orthonormal set in X then the number of non-zero Fourier coefficients
must be countable, which follows from Bessel’s inequality by arguing: how many coefficients
can we have with | hx, ek i | > 1/n? If it was infinite, then Bessel’s inequality would not hold,
thus it must be finitely many.
4.4
Total orthonormal sets
Definition 4.6. A set M is considered total in a normed vector space X if the span M is dense
in X .
4.5. REPRESENTATION OF FUNCTIONS ON HILBERT SPACES
21
A few facts worth noting are that for a Hilbert space H , H must contain a total orthonormal set.
All total orthonormal sets for H have the same cardinality, denoted the Hilbert dimension. A
given subset M of H contains a countable total orthonormal set. Also, we have the following
variation of Bessel’s Inequality:
Theorem 4.8 (Parseval’s equality). An orthonormal set M is total if and only if for every
x ∈ H:
∞
X
| hx, ej i | = kxk.
j=1
Note the sum is to ∞ but there are only countably non-zero coefficients.
Note that if two Hilbert spaces have the same Hilbert dimension, they are isomorphic, therefore
“once you’ve seen one real Hilbert space of dimension n, you’ve seen them all”. Besicovich
spaces are “almost periodic” functions which form a non-separable Hilbert space but it rarely
occurs in applications.
P∞
Note, this analysis provides the justification for the Fourier series expansion: f (t) = k=1 ak sin(2πkt)+
P
∞
2
j=1 bj cos(2πjt) + c0 . As mentioned before, these form a total orthonormal set in L .
What about the Fourier transform? f (t) = fˆ(ω)eiωt dω ? Do the exponential functions
correspond to a uncountable orthonormal set? No. They aren’t even in L2 .
R
We can also consider the Gram-Schmidt Process which produces a total orthonormal set
{y1 , . . . , yn } from a linearly independent set {x1 , . . . , xn }.
The construction of the first element is easy: y1 = kxx11 k . Now, the projection theorem says that
x2 = y + z , where y ∈ span{y1 } and z ⊥ y . Thus, x2 = hx2 , y1 i y1 + v2 , which suggests that
v2 = x2 − hx2 , y1 i y1 6= 0 since x1 , x2 are linearly independent. Finally, set y2 = kvv22 k to get
normality.
Similarly, for x3 we know that x3 = hx3 , y1 i y1 + hx3 , y2 i y2 + v3 and we can repeat the process.
Note, this is guaranteed to work if the xj are linearly independent, but this method is numerically
unstable due to dividing by a potentially arbitrarily small number.
4.5
Representation of functions on Hilbert spaces
Recall that X 0 was the set of all bounded linear functionals on X . We know if X is reflexive
then X X̃ 0 . To check the reflexivity of a Hilbert space we have the following theorem.
Theorem 4.9 (Riesz representation theorem). Let H be a Hilbert space. For every bound
linear functional f ∈ H 0 , there is a unique element z ∈ H such that for all x ∈ H :
f (x) = hx, zi .
22
CHAPTER 4. INNER PRODUCT SPACES
Furhtermore, kf k = kzk.
Proof. First note that N(f ) is closed since f is bounded. If N(f ) = H , then z = 0 and f = 0
and we are done, so assume that H = N(f ) ⊕ N f ⊥ and chose z0 ∈ N(f )⊥ \ {0}.
We know immediately that f (z0 ) 6= 0, so note that



f (x)z0 
f (x)


f x −
f (z0 ) = 0.
 = f (x) −
f (z0 ) 
f (z0 )

| {z }
∈N(f )
From this, we can conclude:
0=
f (x)
x−
z0 , z0
f (z0 )
= hx, z0 i −
f (x)
hz0 , z0 i
f (z0 )
and therefore:
f (z0 )
=
f (x) = hx, z0 i ·
kz0 k2
*
f (z0 )
x,
z0
kz0 k
+
.
f (z )
So, if we choose z = kz00k z0 , we then have f (x) = hx, zi as desired. Uniqueness is obvious. To
show the norm relation:
kzk2 = | hz, zi | = |f (z)| ≤ kf kkzk =⇒ kzk ≤ kf k,
and similarly:
|f (x)| = | hxi z| ≤ kxkkzk =⇒
|f (x)|
≤ z =⇒ kf k ≤ kzk.
kxk
Thus, we can conclude kf k = kzk.
Thus, we know that there is a bijective mapping between X and X 0 , meaning every Hilbert space
is reflexive.
4.6
Lax-Milgram theorem
Definition 4.7. Let X and Y be real vector spaces, then a function X × Y → R is considered
a bilinear form if it satisfies the typical linearity properties.
4.6. LAX-MILGRAM THEOREM
23
Example 4.12. a(x, y) := hx, yi is a bilinear form.
Note, if the field is the complex numbers and the second term becomes conjugated in linearity,
the form is considered sesquilinear.
We say the form is bounded if there exists a constant c such that: |a(x, y)| ≤ ckxkX kykY . As
usual, the norm of this operator is defined to be the smallest c that satisfies this property.
Theorem 4.10. Let a be a bounded bilinear form over real Hilbert spaces a : H1 × H2 → R,
then there exists a uniquely determined linear operator S : H1 → H2 such that:
a(x, y) = hSx, yi .
Furthermore, kak = kSk.
Proof. For a fixed x, the bilinear form becomes a BLF and this follows immediately from the
Riesz Representation Theorem.
Definition 4.8. A bilinear form b(x, y) is called coercive (or uniformly elliptic) if there exists a
constant c such that:
b(x, x) ≥ ckxk2 .
Theorem 4.11 (Lax-Milgram). Let H be a real Hilbert space and let b : H × H → R be a
bounded, coercive, bilinear form. Let B : H → H 0 be defined by (Bx)(y) = b(x, y), then
B −1 : H 0 → H exists and is bounded.
Proof. Let S : H → H be the operator such that b(x, y) = hSx, yi, then we know from
coercivity:
ckxk2 ≤ |b(x, x)| ≤ kSxkkxk =⇒ ckxk ≤ kSxk
Therefore, we know S is injective since if Sx = Sy =⇒ ckx − yk ≤ kSx − Syk = 0 =⇒
x = y . We now claim R(S) is closed. Let y ∈ R(S) and therefore ∃(yn ) ∈ S such that
yn → y and we know yn = Sxn . Since yn is convergent, it must be Cauchy, which suggests xn is
Cauchy and therefore convergent. Thus, Sx = y so y ∈ R(S), meaning that the range is closed.
Suppose that R(S) 6= H then ∃z ∈ R(S)⊥ \ {0} such that hSx, zi = 0 and choose x = z
which suggests that hSz, zi = 0 but this violates the coercivity estimate, meaning R(S) = H .
Thus, S is injective and the range is the whole space meaning it is onto and therefore bijective,
meaning S −1 exists and is bounded.
Next consider ckxk ≤ kSxk by the injectivity of S , which suggests that Sx = y =⇒ x = S −1 y
and S is bounded. We can then define B −1 : H 0 → H by the following analysis: let g ∈ H 0 =⇒
∃f ∈ H : g(y) = hf, yi and Bx = g means (Bx)(y) = g(y) = hf, yi =⇒ B −1 g = S −1 y
which uniquely determines B −1 .
24
CHAPTER 4. INNER PRODUCT SPACES
In general, we can now solve problems of the form b(x, y) = g(y) for all y and turn them into
the form Bx = g in our functional space.
Example 4.13. Consider the following boundary value problem:
(
−u00 + au = f
u(0) = u(1) = 0.
(4.2)
Assume a(t) ∈ C[0, 1] and a(t) ≥ a0 > 0 in this interval. We can also assume f ∈ L2 (0, 1) to
proceed. Consider a test function v ∈ C01 [0, 1] and multiply both sides and integrate:
−u00 v + auv = f v
Z 1
Z 1
Z 1
00
−
u v dt +
auv dt =
f v dt
0
0
0
Z 1
Z 1
Z 1
0 0
u v dt +
auv dt =
f v dt.
0
0
(4.3)
0
Thus, we now have the weak version of our problem. That
solving
(4.2) and (4.3) is equivalent.
R 0 is,
R
0
We can now define our bilinear operator b(u, v) := u v + auv 0 for u, v ∈ C01 [0, 1]. Then
note:
Z
Z
0 0
|b(u, v)| ≤ | u v | + | auv|
Z
Z
0 0
≤ |u v | + max a(t) |uv|
t
0
0
≤ ku kL2 kv kL2 + max a(t)kukL2 kvkL2
t
0
≤ max{1, kak∞ } [ku kL2 kv 0 kL2 + kukL2 kvkL2 ]
≤ 2 max{1, kak∞ } kukH 1 kvkH 1
|
{z
}
=c
Thus, our bilinear form is bounded. We need now show coercivity as well:
Z
0 2
Z
a(u)2
Z
Z
0 2
≥ (u ) + a0 u2
Z
Z
0 2
2
≥ min{1, a0 }
(u ) + u
b(u, u) =
(u ) +
= min{1, a0 }kuk2H1
R
Now, we have b(u, v) = g(v) ⇔ Bu = g . We know g(v) = f v and also |g(v)| ≤
kf kL2 kgkL2 ≤ kf kL2 kvkH 1 , thus g is bounded (and therefore continuous), meaning our weak
problem has a unique solution u ∈ H01 [0, 1] which depends continuously on f .
5
Linear operators II
5.1
Hilbert adjoint
Definition 5.1. Let H1 , H2 be Hilbert spaces and T : H1 → H2 be a BLO. The adjoint of T ,
denoted T ∗ is defined to be: hT X, yiH2 := hx, T ∗ yiH1 .
Note, “Hilbert adjoint” and “adjoint” are equivalent when working in a Hilbert space. The adjoint
always exists and is unique with kT ∗ k = kT k. If we have a sesquilinear form h(x, y) then this
can be written as h(y, x) = hy, T xi = hT ∗ x, yi.
Example 5.1. Consider a matrix A : Rn → Rm as a linear operator. Then A = (aij ) ∈ Rm×n .
In this case:
hAx, yi = Ax · y = (Ax)T y = xT AT y = x, AT y =⇒ AT = A∗ .
Example 5.2. Consider K : L2 [0, 1] → L2 [0, 1] defined by the usual integral operator
R1
(Ku)(t) = 0 k(t, s)u(s) ds. We can interchange the integrals as we did the sums above and
R1
the resulting adjoint becomes K ∗ v(s) = 0 k(t, s)v(t) ds.
A few other notable facts include: kT ∗ T k = kT T ∗ k = kT k2 and (S + T )∗ = S ∗ + T ∗ . Also
note that if a BLO satisfies T = T ∗ , it is considered self-adjoint . If it is not bounded, it is
considered symmetric.
5.2
Boundedness
Theorem 5.1 (Hahn-Banach). Let X be a vector space, Z be a subspace of X and consider
ρ : X → R satisfy ρ(x + y) ≤ ρ(x) + ρ(y) and ρ(ax) ≤ |a|ρ(x), that is, we have a sub-linear
functional. Let f be a functional on Z such that |f (x)| ≤ ρ(x), then there exists a linear
extension f˜ of f onto X satisfying |f˜| ≤ ρ.
25
26
CHAPTER 5. LINEAR OPERATORS II
We can now present some consequences of the Hanh-Banach theorem.
Theorem 5.2. Let f be a BLF on a subspace Z of the NVS X , then there is a BLF f˜ which
extends to X and has the same norm.
Corollary 5.1. Let X be a normed vector space. Choose x0 6= 0 ∈ X . Then there is a BLF
f˜ ∈ X 0 such that kf˜k = 1 and f˜(x0 ) = kx0 k.
Theorem 5.3. Let X be a NVS. For every x ∈ X , kxk = supf ∈X 0
property that f (x0 ) = 0∀f ∈ X 0 =⇒ x0 = 0.
f 6=0
|f (x)|
.
kf k
Also, if x0 has the
Theorem 5.4 (Uniform Boundedness Theorem). Let X be a Banach space and let {τn } be a
sequence of BLO such that τn : X → Y , where Y is a normed vector space. If for each fixed
x ∈ X : {τn (x)} is bounded, then there is a c such that kτn k ≤ c.
5.3
Reflexive spaces
Recall the canonical embedding map c : X → X ∗∗ , where given x ∈ X , g ∈ X ∗∗ , we can
define g(f ) = f (x) =⇒ cx = g . Also note c is an isometry. X is then isomorphic with
R(c) ⊂ X 00 and recall a space is reflexive if R(c) = X 00 , that is X X̃ 00 .
Example 5.3. `p with 1 < p < ∞ is reflexive by recalling (`p )0 = `q .
Example 5.4. Lp with 1 < p < ∞ follows from an identical argument.
Example 5.5. Note that p = 1, ∞ is not reflexive for either case.
Theorem 5.5. Let X be a normed vector space, then:
(1) If X is reflexive, then X is a Banach space. (since X 0 must be Banach)
(2) If X is finite-dimensional, then X is reflexive.
(3) If X is a Hilbert space, then X is reflexive. (by Riesz representation theorem)
Lemma 5.1. Let X be a normed vector space and Y ⊂ X be a closed, proper subspace. Choose
x0 ∈ X − Y and define: δ = inf y∈Y kx0 − yk, then there exists f ∈ X 0 such that f = 0 on Y
and that kf k = 1 and f (x0 ) = δ .
Theorem 5.6. Let X be a normed vector space. If X 0 is separable, then X 0 is also separable.
Proof. Follows from previous lemma.
5.4. WEAK CONVERGENCE AND WEAK* CONVERGENCE
27
Thus, we can conclude that if X is reflexive, then if either X or X 00 is separable, then the other
must be separable as well. Also note that if X is separable and X 0 is not, X cannot be reflexive.
For instance, note that `1 is separable but `∞ is not, thus `1 is not reflexive.
5.4
Weak convergence and weak* convergence
Definition 5.2. The sequence {xn } in a normed vector space X is said to strongly converge if
kxn − xk → 0 as n to∞.
Definition 5.3. The sequence {xn } in a normed vector space X is said to weakly converge if
kf (xn ) − f (x)k → 0 as n to∞ for all f ∈ X 0 .
In this case, we denote this limit xn * x and say x is the weak limit of {xn }. It’s clear that
strong convergence =⇒ weak convergence, but the converse is not necessarily true. Note, if we
have a finite dimensional space, the converse is true.
Example 5.6. Consider
X = `p and define (xn ) := (en ). Since (`p )0 = `q , we can represent
P
∞
f ∈ `q as f (x) := j=1 ξj ηj . Note, |ηj | → 0. Therefore f (xn ) → 0 as well, but xn is not
Cauchy and doesn’t converge, but xn * 0.
Note that the weak limit is always unique and that if a sequence is weakly convergent, then every
subsequence is as well. Also, {kxn k} must also be bounded, but the proof is tricky.
Let {fn } be a sequence of bound linear functionals in X 0 . Then recall that we say fn → f , that
is, converges strongly if kfn − f k → 0 and fn *, that is, converges weakly if g(fn − f ) → 0
for all g ∈ X 00 . We can introduce yet another notion of convergence.
Definition 5.4. If the above sequence has the property that fn (x) − f (x) → 0 for all x ∈ X ,
∗
then we say fn * f , or fn converges to f in the sense of weak* in X ∗ .
Weak convergence is stronger than weak* convergence. That is, weak implies weak* . To see
this, note that if fn * then g(fn ) − g(f ) → 0 for all g ∈ X 00 , thus every x ∈ X maps to a
g ∈ X 00 by the canonical map g(f ) = f (x). Thus, g(fn ) = fn (x) and g(f ) = f (x), thus
fn (x) − f (x) → 0.
Also note that if X is reflexive, meaning X = X 00 , then weak and weak* are equivalent.
q
p
Suppose {un } is a sequence in Lq , where 1 < q <
R ∞. L is
R reflexive and its pdual space is L ,
q
where 1/p + 1/q = 1. un * u in L means that un v → uv for all v ∈ L and in this case,
weak* is the same.
Next consider {un } as a sequence in L∞ . Here, un * means that for all f ∈ (L∞ )0 : f (un ) → f ,
∗
∞ 0
but we don’t know what
L∞ = (L1 )0 , which now means that un * u
R (L ) Ris. We can note that
which suggests that un v → uv for all v ∈ L1 .
28
CHAPTER 5. LINEAR OPERATORS II
Example 5.7. Consider the following boundary value problem:
(
u00 − a(x)u = f
u(0) = u(1) = 0,
where a(x) corresponds to some periodic material coefficient, say a periodic square wave from
1 to 2 such that an (x) = a(x/n). We can now construct the weak form by multiplying by a
v ∈ H01 (0, 1):
Z
0 0
uv +
Z
Z
an uv = −
fv
Note that {an }R is a sequenceR in L∞ (0, 1) and that |uv| ≤ kukL2 kvkL2 so uv ∈ L1 . This then
means that if an (uv) → a(uv) for some a, then the solution converges.
R
R
Another possible an choice is an (x) = 2 + sin(2πnx). In this case, an (x)1 dx = 2, and also,
R1
R
R
(1
−
θ(x))
dx
=
1
a
,
and
finally
v
→
2v dx by the Riemann-Lebesgue lemma, thus
n
0
∗
∞
sin nx * in L .
5.5
Compact operators
Recall that a set K is compact if every sequence {xn } in K has a convergent subsequence {xnk }
and converges to an element in K .
Definition 5.5. Let X, Y be normed vector spaces. An operator T : X → Y is said to be
compact if T is linear and if for every bounded subset M of X , then T (M ) is compact.
Lemma 5.2. Every compact operator T : X → Y is bounded. If dim X = ∞, then
I : X → X is not compact.
To see this, note that u = {x ∈ X : kxk = 1} is bounded and T (u) is compact and hence
bounded, meaning supkxk=1 kT xk ≤ ∞ =⇒ T is bounded. To see the second statement, note
that I(u) · u is not compact when dim X = ∞.
Theorem 5.7 (Compactness criterion I). An equivalent criteria for compactness can be
summarized as: T is compact if and only if it maps every bounded sequence {xn } in X onto a
sequence {T xn } in Y which has a convergent subsequence.
C(X, Y ) is the space of all compact operators T : X → Y and is a vector space since
C(X, Y ) ⊂ B(X, Y ), which is Banach if Y is Banach.
Theorem 5.8 (Compactness criterion II). Yet another equivalent criteria is: If T is bounded
and linear and either:
(1) dim(X) < ∞
5.5. COMPACT OPERATORS
29
(2) dim(R(T )) < ∞
(3) dim(Y ) < ∞,
then T is compact. In other words, if we have a finite dimensional bounded operator, it is indeed
compact.
Theorem 5.9. If (Tn ) is a sequence of compact operators and kTn − T k → 0, then T is also
compact.
Proof. Let (xn ) be a bounded sequence in X . We will construct a subsequence yn such that T yn
converges.
We know T1 is compact, so ∃ a subsequence x1,m of xm such that T1 x1,m is Cauchy. We can
repeat this process for T1 , . . . , Tn . Thus we have a sequence (Tk xj,k ) that is Cauchy whenever
k ≤ j . We can then define the “diagonal sequence” ym = xm,m . For any fixed k , (Tk ym ) is
Cauchy because eventually k ≤ m. Since {xm } is bounded, we have that {ym } is also bounded.
We want to show that (T ym ) is Cauchy. Given > 0, choose p, N large enough such that
kTp − T k < /3c and kTp yk − Tp yj k ≤ /3
kT yk − T yj k ≤ kT yk − Tp yk k + kTp yk − Tp yj k + kTp yj − T yj k
≤ kT − Tp kkyk k + /3 + kT − Tp kkyj k < .
Example 5.8. Consider T : `2 → `2 defined by T x := (ξ1 /1, ξ2 /2, ξ3 /3, . . .), which can be
thought of as multiplying by an infinite diagonal matrix. We hope to show that T is compact.
Define Tn to be T truncated to n terms. Note that since dim Tn = n, it is compact. We also
2 P∞
2
P∞
1
1
2
2
kxk2 =⇒
note that kTn x − T xk2 =
j=n+1 |ξj | ≤ n+1
j=n+1 |ξj /j| ≤ n+1
1
kTn x − T xk ≤ n+1
kxk =⇒ kTn − T k → 0. Therefore, by our previous theorem, T is
compact.
Theorem 5.10. Let X, Y be NVS and let T be compact, then if xn * x then T xn → T x.
6
Spectral Theory
6.1
Preliminaries
Suppose T : X → X where dim(X) < ∞ and X is a NVS, then we can represent T as a
matrix A = (aij ). Consider:
Ax = λx.
(6.1)
Recall that λ ∈ C is an eigenvalue if (6.1) has a solution, in which case x 6= 0 is the corresponding
eigenvector. The nullspace of (A − λI) is the eigenspace corresponding to the eigenvalue λ.
The collection of eigenvalues of A is called the spectrum of A, denoted σ(A).
Note that eigenvalues are the same for similar matrices, that is A2 = C −1 A1 C , where C is
non-singular, thus σ(A) is unambiguously defined. Also, det(A − λI) is a polynomial of degree
n in λ. The Fundamental Theorem of Algebra says that there must be at least one root and no
more than n including multiplicities.
This works for finite dimensional operator, but defining the spectrum of an infinite dimensional
operator becomes less straightforward. Thus, consider X 6= {0}be a complex NVS and let
T : D(T ) ⊂ X → X be a linear operator.
For λ ∈ C, define Tλ := T − λI and Rλ (T ) := Tλ−1 = (T − λI)−1 when it exists. Rλ (T ) is
called the resolvent operator because it is solving something.
Definition 6.1. The resolvent set ρ(T ) is the set of all λ ∈ C such that:
(R1) Rλ (T ) exists and therefore Tλ−1 exists.
(R2) Rλ (T ) is bounded and therefore Tλ−1 is bounded.
(R3) The domain of Rλ (T ) is dense in X , that is D(Rλ (T )) = X . Note D(Rλ (T )) =
R(T − λI).
30
6.2. SPECTRAL PROPERTIES OF BOUNDED LINEAR OPERATORS
31
We can then define the spectrum σ(T ) to be C − ρ(T ). Thus, one or more of the conditions
must fail for λ ∈ σ(T ). We can then split σ(T ) into three disjoint sets depending on which
property is violated:
(1) σp (T ), the point spectrum, consists of all λ such that Rλ (T ) = (T − λI)−1 does not
exists. Note that in this case (T − λI)x = 0 has a non-zero solution, meaning that we can
think of this as T x = λx, thus denoting λ as an eigenvalue. This x is also considered an
eigenvector.
(2) If Rλ (T ) = (T − λI)−1 exists but is not bounded, then λ exists in the continuous spectrum,
denoted σc (T ).
(3) Again, if Rλ (T ) = (T − λI)−1 exists but the domain is not dense in X , then λ ∈ σr (T ),
the residual spectrum.
By definition, we know σ(T ) = σp (T ) ∪ σc (T ) ∪ σr (T ), thus we can characterize an operator
by its spectrum.
−1
Example 6.1. Consider the identity operator I . Then Rλ (I) = (I − λI)−1 = [(1 − λ)I] =
1
I . Note that Rλ (I) is not defined for λ = 1 so 1 ∈ σp (I). If λ 6= 1, then Rλ (I) exists and
1−λ
is bounded and is defined for all X , thus λ ∈ ρ(I).
Example 6.2. Consider T : `2 → `2 defined by T x := (ξ1 , ξ2 /2, ξ3 /3, . . .). If λ = 1/n,
then (T − λI) = (T − n1 I) =⇒ (T − n1 I)en = 0, thus Rλ (T ) does not exist, meaning
1
is an eigenvalue with eigenvector en . Note that when λ = 0, (T − λI) = T and T −1 =
n
(ξ1 , 2ξ2 , 3ξ3 , . . .), which exists and is defined on a dense subset of `2 , which are all sequences
with finitely many non-zero empties. Note that kT −1 en k = n, thus this is unbounded and
therefore 0 ∈ σc (T ).
Example 6.3. Consider T : `2 → `2 defined by T x := (0, ξ1 , ξ2 , . . .), that is, we have the right
shift operator. First consider λ = 0. Then (T − λI) = T and therefore T −1 x = (ξ2 , ξ3 , . . .).
Note, Rλ exists and is bounded, but the range of T are only sequences with a 0 in the first term,
which is not dense in `2 , meaning λ ∈ σr (T ).
6.2
Spectral properties of bounded linear operators
Theorem 6.1. Let T : X → X be a BLo where X P
is a Banach space, then if kT k < 1 then
k
(I − T )−1 exists, is bounded, and (I − T )−1 = I + ∞
k=1 T .
The above theorem is known as a Neumann series for a bounded linear operator. The proof
is clear by considering a geometric series and that we know that in a Banach space, absolute
convergence implies convergence.
32
CHAPTER 6. SPECTRAL THEORY
Theorem 6.2. Let X be a complex Banach space and T : X P
→ X be a BLO. The resolvent set,
k k+1
ρ(T ) is an open set in C and for every λ0 ∈ ρ(T ), Rλ (T ) = ∞
k=0 (λ − λ0 ) Rλ0 (T ) and the
series is absolutely convergent for |λ − λ0 | < 1/kRλ0 k.
Theorem 6.3. Let X be a complex Banach space and T : X → X be a BLO, then σ(T ) is
compact and σ(T ) ⊂ {λ ∈ C : |λ| ≤ kT k}.
Theorem 6.4. If X 6= {0} is a Banach space then σ(T ) 6= ∅.
Definition 6.2. The spectral radius r(T ) of an operator T is defined to be r(T ) = supλ∈σ(T ) |λ|.
Note if σ(T ) is compact, this sup is reached.
Although it is immediately clear r(T ) ≤ kT k, it is more difficult to show that r(T ) =
limn→∞ kT n k1/n .
Theorem 6.5. Let T be a LO on a complex NVS X . Eigenvectors corresponding to distinct
eigenvalues are linearly independent.
6.3
Spectral properties of compact operators
Theorem 6.6. Let X be a NVS and T : X → X be compact. The set of all eigenvalues σp (T )
is countable with the only accumulation point possible for σp (T ) is 0.
From this, we can know that the set of all eigenvalues of a compact linear operator can be arranged
into a sequence whose only possible point of accumulation is 0.
Theorem 6.7. Let X be a NVS. Let T : X → X be compact. Then the eigenspace
corresponding to every non-zero eigenvalue λ ∈ σp (T ) is finite dimensional.
Theorem 6.8 (Fredholm alternative). Let X be a NVS and let T : X toX be a compact linear
operator. Then either:
(1) The problem (I − T )x = 0 has a nontrivial solution x ∈ X .
(2) For each y ∈ X , (I − T )x = y has a unique solution.
Note, the two cases are mutually exclusive. In case (2), the inverse operator (I − T )−1 whose
existence is asserted, is bounded: kxk ≤ k(I − T )−1 kkyk.
Lemma 6.1. Let X be a NVS and Y be a proper closed subspace of X , then given θ < 1, there
is a vector y ∈ X − Y with kyk = 1 and inf x∈Y kx − yk ≥ θ.
Lemma 6.2. Let X be a NVS, and T : X → X be compact, S : X → X be bounded, then
T S and ST are both compact.
6.4. BOUNDED SELF-ADJOINT OPERATORS
33
Both these lemmas combined can be used to prove the statement of the Fredholm alternative.
Theorem 6.9. Let T : X → X be compact and X be a NVS. Then every non-zero λ ∈ σ(T )
is an eigenvalue.
It can also be shown that 0 ∈ σ(T ) for all compact linear operators with infinite dimension. This
follows from the fact that if 0 ∈ ρ(T ), then T −1 would eist and be bounded, meaning T −1 T = I
would be compact, but note that I can only be a compact operator under finite dimensions, thus
we have a contradiction.
6.4
Bounded self-adjoint operators
Recall that if T : H → H is bounded on the Hilbert space H then the adjoint T ∗ : H → H
defined by hT x, yi := hx, T ∗ yi exists and is bounded by kT ∗ k = kT k. Note that self-adjointness
implies T ∗ = T .
Theorem 6.10. Let T : H → H be a bounded self-adjoint linear operator on the Hilbert
space H . If T has eigenvalues, they must be real and the eigenvectors corresponding to distinct
eigenvalues are orthogonal. Note, this is the same result that holds for symmetric matrices.
Theorem 6.11. Let T : H → H be a bounded self-adjoint linear operator, then σr (T ) = ∅.
Theorem 6.12. Let T : H → H be a BSALO on the Hilbert space H . Then λ ∈ ρ(T ) iff
∃c > 0 such that k(T − λI)xk ≥ ckxk for all x ∈ H .
Example 6.4. Define (T x)(t) = tx(t) for t ∈ [0, 1] with T : C[0, 1] → C[0, 1], which
immediately implies that σ(t) = [0, 1]. We argued previously that σ = σr in this case as the
domain of Rλ or the range of T − λI is not dense in C[0, 1]. If we consider this same function
over L2 the spectral properties change. It stays bounded, but we can first ask if there are any
eigenvalues by considering: (T − λ)x = 0 = (t − λ)x = 0 for all λ ∈ [0, 1] =⇒ x = 0, thus
there are no eigenvalues. It can also be shown that T is self adjoint, meaning that σr (T ) = ∅.
y(t)
Now, consider (t − λ)x(t) = y(t) =⇒ x(t) = t−λ =⇒ kxk = k(T − λ)−1 yk. For
instance, consider λ = 0, then kxk → ∞, thus σ = σc .
Theorem 6.13. Let T : H → H be a BSALO on the Hilbert space H , then σ(T ) is real.
Theorem 6.14. For T : H toH , a BSALO, then: kT k = max{|m|, |M |} = supx6=0 |q(x)|
and furthermore m, M are both spectral values.
Theorem 6.15 (Spectral theorem). Let T : H → H be a compact, self adjoint, linear operator
on a Hilbert space H , then there exists a complete orthonormal basis for H consisting of
eigenvectors for T .
We know some basic facts already:
34
CHAPTER 6. SPECTRAL THEORY
(1) Eigenspaces Eλ corresponding to λ 6= 0 are finite dimensional.
(2) Eλ1 ⊥ Eλ2 if λ 6= λ2 .
(3) There are countably many eigenvalues.
(4) The spectrum consists of only eigenvalues plus possibly 0 ∈ σp (T ) ∪ σc (T ).
We can
P∞also consider some consequences of the Spectral theorem. Particularly we can write
x = j=1 P
hx, uj i uj where x ∈ P
H and {uj } form an orthonormal basis for H , which suggests
∞
∞
that T x = j=1 hx, uj i T uj = j=1 λj hx, uj i uj , which tells us everything about {uj }.
P∞
P∞
So, to solve (I −T )x = y , we need only solve j=1 (1−λj ) hx, uj i uj = j=1 hy, uj i uj =⇒
(1 − λj ) hx, uj i = hy, uj i. In other words, we diagonalize the problem. We can see from
P∞ hy,uj i
this that hx, uj i = hy, uj i /(1 − λj ) and therefore x =
j=1 1−λj uj , which follows from
P∞
x = j=1 hx, uj i uj .
Thus, if we want to solve T x = y , we can simply solve x =
since T is compact, thus T −1 must be unbounded.
P∞
j=1
hy,uj i
uj ,
λj
but we know λj → 0
Overall, the big take-away from this theorem is: compact operators can be approximated with
finite matrices.
6.5
Return to Fredholm theory
We can now formulate a stronger version of the Fredholm alternative for Hilbert spaces.
We first need the lemma:
Lemma 6.3. If T : H → H is a compact linear operator, then T ∗ is also compact.
Theorem 6.16 (Fredholm alternative). Let T : H → H be a BLO on the Hilbert space H ,
then R(T ) = N(T ∗ )⊥ . Furthermore, there are three exclusive alternatives:
(1) (I − T )x = y has a solution x ⇔ y ∈ N(I − T ∗ )⊥ .
(2) (I − T ∗ x) = y has a solution x ⇔ y ∈ N(I − T )⊥ .
(3) If neither N(I − T ∗ ) = 0 nor N(I − T ) = 0, then both equations have solutions x for all
y ∈ H and the inverse operators (I − T )−1 and (I − T ∗ )−1 is bounded.
6.6
Sturm-Liouville problem
We can apply most of the theory covered in this course by considering a single example, the
Sturm-Liouville problem.
6.6. STURM-LIOUVILLE PROBLEM
35
In particular, we want to solve the 1 dimensional equation describing acoustic waves:
1
vx
ρ
− κvtt = 0,
(6.2)
x
where ρ(x) is the density of the medium and κ(x) is the compressibility, which is equivalent to
1/ν(x), where ν is the bulk modulus.
We’d like to separate variables v(x, t) = u(x)g(t), which yields:
g(t)
1 0
u (x) − κu(x)g 00 (t) = 0 =⇒
ρ
x
1 0
u (x)
ρ
x
κu(x)
=
g 00 (t)
= −λ.
g(t)
Note, as with any separation of variables problem, both sides of this equation are constant so we
obtain two ordinary differential equations:
0
1 0
u = −κλu(x)
ρ
g 00 (t) = −λg(t).
We can observe that g(t) = eiωt , λ = ω 2 so λ ≥ 0. Other solutions for g(t) exist but this will
suffice. We can generalize the u equation to the form:
(σu0 )0 − qu + λκu = 0,
q(x) ≥ 0,
σ(x) ≥ σ0 > 0, κ(x) ≥ κ0 > 0.
(6.3)
Note, q(x) corresponds to absorption, and here we will assume we have Dirichlet boundaries,
that is u(0) = u(1) = 0. Let w ∈ C01 [0, 1] be a test function, then we have:
Z
1
0
(σu0 ) w − quw + λκuw dx
Z0 1
Z 1
0
0
=
[(σu )w + quw] dx −
κuw dx.
0=
0
R1
0
R1
Thus, if we let a(u, w) := 0 [σu0 w0 + quw] dx and b(u, w) := 0 κuw dx, we have a problem
of the form a(u, w) = λb(u, w), which is the weak form of our problem and true for any
w ∈ H01 (0, 1). We hope to find λ ∈ C and u ∈ H01 (0, 1).
If a, b are bounded, the Riesz representation theorem implies there are bounded operators:
A : H01 → H01 and B : H01 → H01 such that a(u, w) = hAu, wi and b(u, w) = hBu, wi, in
which case our problem becomes: hAu − λBu, wi = 0 for all w ∈ H01 (Ω) which suggests that
Au − λBu = 0. Thus, we must check if a, b are bounded, that is if |a(u, w)| ≤ ckukkwk.
36
CHAPTER 6. SPECTRAL THEORY
Recall that kuk2H1 =
R
|u0 |2 + |u|2 dx and that hu, wi =
R
u0 w0 + uw dx. Thus:
Z 1
κuw dx
|b(u, w)| = 0
Z 1
κ(x)|u(x)||w(x)| dx
≤
0
Z 1
|u(x)||w(x)| dx
≤ max κ(x)
x∈[0,1]
0
≤ ckukL2 kwkL2
≤ ckukH1 kwkH1 .
Note, a similar argument holds for a:
Z
|a(u, w)| = 0
1
σu w + quw dx
0
0
≤ c1 ku0 kL2 kw0 kL2 + c2 kukL2 kwkL2
≤ ckukH1 kwkH1 .
So we can now consider the form Au−λBu = 0 and we want (T −λI)u = 0. Lax-Milgram says
that if a is coercive then A−1 exists and is bounded. Coercivity, recall, means a(u, u) ≥ c0 kuk2 .
First, we prove the Poincaré’s inequality in 1 dimension by first noting that u(x) =
and therefore:
kuk2L2
Z
Rx
0
u0 (t) dt,
1
u(x)2 dx
0
2
Z 1 Z x
0
=
u (t) dt dx
0
0
Z 1 Z x
0
2
≤
u (t) dt , dx
0
0
Z 1Z 1
≤
u0 (t)2 dtdx
Z0 1 0
=
u0 (t)2 dt
=
by Jensen’s inequality
0
= ku0 k2L2 ,
which suggests that kuk2L2 ≤ ku0 kL2 for u ∈ H01 , which is the statement of Poincaré’s inequality
in one dimension. This then allows us to continue:
6.6. STURM-LIOUVILLE PROBLEM
Z
37
1
σu02 + qu2 dx
a(u, u) =
Z0 1
σu02 dx
0
Z 1
u0 (x)2 dx
≥ inf σ(x)
x
0
Z 1
u0 (x)2 dx
= σ0
Z0 1 0 2
u (x)
u0 (x)2
= σ0
+
dx
2
2
0
Z
σ0 1 0
≥
u (x) + u(x)2 dx Poincaré’s inequality
2 0
σ0
= kuk2H 1
2
≥
By Lax-Milgram A−1 : H01 → H01 exists and is bounded. Thus, (A − λB)u = 0, where A, B
are both BLO now becomes (I − λA−1 B)u = 0 and define T := A−1 B . Let ν = 1/λ then we
now have the form (T − νI)u = 0.
Note, A−1 is bounded,Dso if BEis compact, then T is also compact. We must show that B is
compact. Define B̃ by B̃u, w
H1
:=
R
uw dx since hBu, wi =
R
κuw.
∞
1
Consider {sin(nπx)}
φ := cn sin(nπx) where
n=1 , a complete orthogonal set in H0 . Let
√
R 1 02n
2
cn := √ 2 for normalization. Note that now hφn , φn i = 0 φn + φ2n = 1, meaning we
(πn) +1
have a complete orthonormal set (basis) for H01 . Therefore, B̃ becomes:
D
But note,
1
(nπ)2 +1
B̃φn , φm
E
(R 1
φn φm = 0
= R01 2 2
c sin (nπx) =
0 n
if n 6= m
1
(nπ)2 +1
if n = m.
→ 0. Thus, B̃ is analogous to our prototypical compact operator on `2 and is
compact by the same argument. Note that Bu = B̃(κu).
If (un ) is a sequence
in H01 with kun k ≤ 1, then assuming that κ is smooth, kκun k ≤ c,
R
where kκun k2 = (κun )02 + (κun )2 . Since B̃ is compact, B̃(κun ) = B(un ) has a convergent
subsequence, meaning B is also compact and therefore T is compact.
Thus, we can invoke our properties of compact operators. Specifically, we know that the set of
eigenvalues is at most countable, the only point of accumulation is 0, and the eigenspaces for
ν 6= 0 are finite dimensional.
38
CHAPTER 6. SPECTRAL THEORY
Also note that A, B are self-adjoint. It can be shown that if A−1 is bounded and A is Rself adjoint,
1
then so is A−1 . We can define a new inner product (·, ·) by (u, w) := hBu, wi = 0 κuw dx.
T is then self adjoint with respect to this new inner product. To see this:
(T u, w) = hT u, Bwi = A−1 Bu, Bw = Bu, A−1 Bw = hBu, T wi = (u, T w) .
Thus, T is self adjoint with respect to (·, ·), meaning the eigenvalues are real and the eigenvectors
associated with distinct eigenvalues are orthogonal with respect to (·, ·).
Note, we can also show that λ is always positive by noting that:
hAu, ui = a(u, u) = λ b(u, u) = hBu, ui =⇒ λ ≥ 0
| {z }
| {z }
≥0
≥0
Consider T = A−1 B , then T u = 0 =⇒ A−1 Bu = 0 =⇒ Bu = 0 so N(T ) = N(B).
But note, Bu = 0 =⇒ B̃(κu) = 0 but by construction, we know that (B̃) = {0} and also
κu = 0 =⇒ u = 0, thus N(T ) = {0}, meaning 0 is not an eigenvalue of T so νn → 0 and
therefore λn → ∞.
7
Distribution Theory
7.1
Multi-index notation
A multi-index is an n-tuple of non-negative integers α = (α1 , . . . αn ). We define the order of
|α| = α1 + · · · αn . For partial derivatives, we have the notation:
∂ α u :=
∂ |α|
u
∂ α 1 x1 · · · ∂ α n xn
Example 7.1.
∂ 3u
= ∂ α u =⇒ α = (2, 0, 1).
2
∂ x1 ∂x3
7.2
Definitions
Let Ω ⊂ Rn be open. We can define the following spaces:
(1) C k (Ω) := {u ∈ C(Ω) : ∂ α u ∈ C(Ω), ∀|α| ≤ k}
(2) C k (Ω) := {u ∈ C(Ω) : ∂ α u can be extended continuously to Ω, ∀|α| ≤ k}
(3) C ∞ (Ω) :=
T∞
k=0
C k (Ω)
The support of a function u is the complement of the largest open set on which u = 0. Denote
this set by supp u. Note, for continuous functions u, supp u = {x : u(x) 6= 0}.
Define yet another space: C0∞ (Ω) := {u ∈ C ∞ (Ω) : supp u ⊂ Ω and is compact}. Historically C0∞ (Ω) = D(Ω), the space of distributions.
Finally, the last space we need is Dk = {u ∈ C0∞ (Rn ) : supp u ⊂ K : K is compact}.
39
40
CHAPTER 7. DISTRIBUTION THEORY
Then if K ⊂ Ω, that is, a compact subset of Ω and p ≥ 0 be an integer, then we define:
kφkk,p = sup |∂ α φ(x)| = max k∂ α φkL∞ (K)
x∈k
|α|<p
|α|≤p
Note that kφkk,p = 0 6 =⇒ φ = 0 since we are only evaluating it on the compact subset K .
Definition 7.1. Let {φn } be a sequence in C0∞ (Ω). We say that φn → φ ∈ C0∞ (Ω) if
∃K ⊂ Ω : supp φj ⊂ K for all j and kφn − φkk,p . Note again, K must be compact.
Note, this notion of convergence has very strict requirements.
7.3
Distributions
Definition 7.2. A distribution u ∈ D0 (Ω) is a continuous linear functional on C0∞ (Ω).
Note the action of u ∈ D1 (Ω) on φ ∈ C0∞ (Ω) by u(φ) = hu, φi, but this is not an inner product,
just notationally the same.
The continuity means that φj → φ =⇒ hu, φj i → hu, φi or equivalently φj → 0 =⇒
hu, φj i → 0.
Theorem 7.1. A linear functional u on C0∞ (Ω) is continuous iff for every compact K ⊂ Ω,
∃c, p : | hu, φi | ≤ ckφkk,p for all φ ∈ Dk .
Definition 7.3. If there is an integer p such that for every compact K ⊂ Ω, ∃c : | hu, φi | ≤
ckφkk,p for all φ ∈ Dk , then u is said to have order ≤ p.
Example 7.2. We can consider the δ ‘’function” defined by hδ, φi := φ(0). Note, in this case,
| hδ, φi | =R|φ(0)| ≤ kφkk,0 for every k , so δ is a distribution of order zero. The classical
notation is δ(x)φ(x) dx = hδ, φi.
R
(
Example 7.3.
R Every f ∈ L Ω) defines aRdistribution u by hu, φi := Ω f (x)φ(x) dx. Here,
| hu, φi | = Ω f (x)φ(x) dx ≤ max |φ| |f (x)|dx = ckφkk,0 for any k . We then just need
f to be locally integrable for this to hold.
Example 7.4. If α is any multi-index, u defined by hu, φi := (∂ α φ)(0) is a distribution. Note
here | hu, φi | = |∂ α φ(0)| ≤ kφkk,p where p = |α|.
ExampleR7.5. Take Ω = Rn . Denote φ ∈ C0∞ (Ω) by φ(x, y). Define u ∈ D0 (Ω) by
∞
hu, φi := −∞ φ(0, y) dy . That is, the mass is concentrated at the y -axis.
7.4. OPERATIONS ON DISTRIBUTIONS
7.4
41
Operations on distributions
We can consider multiplication of a distribution by a smooth function. Let a ∈ C ∞ (Ω) and
0
u ∈ D0 (Ω)
R . Define au ∈ D (Ω) by hau, φi := hu, aφi. Note, if u is a function, then
hau, φi = auφ dx = hu, aφi.
E
D
∂φ
:= − u, ∂x
. Note
j
α
that the minus sign comes from integration by parts. For any multi-index α, h∂ u, φi :=
=(−1)|α| hu, ∂ α φi.
We can also consider the derivative of a distribution. Define
D
E
∂u
,φ
∂xj
∞
In general,R if T : C0∞ (Ω)
with T 0 mapping the same spaces
R → C00 (Ω) is linear and continuous
such that Ω (T φ)ψ = Ω φ(T ψ) for all φ, ψ ∈ C0∞ (Ω), then given u ∈ D0 (Ω), define T u by
hT u, φi := hu, T 0 φi. Note that T only has to be defined over smooth functions.
*
z
dual pairing
+
}|
{
u
,
φ
|{z}
|{z}
distribution test function
We force our test functions to be limited but our distributions can be very general in our example,
we pair [H01 (0, 1)]0 with H01 (0, 1).
We can define the linear partial differential operator, defined by L := |α|≤k aα ∂ α wjere aα ∈
R
R
P
C ∞ , k is the order of the operator. Then, note: (Lφ)ψ = |α|≤k φ(∂ α (aα ψ))(−1)α =
R P
φ |α|≤k (−1)α ∂ α (aα ψ), which suggests that hLu, φi = hu, L0 ψi, which is defined for every
P
|
{z
L0 ψ
}
distribution.
We
R can alsoRdefine the translation operator τx defined by (τx φ)(y) := φ(x + y). Note:
(τx φ)ψ = φτ−x ψ by a change of variables, then hτx u, φi = hu, τ−x φi.
Returning back to the δ function, we will try to solve Lu = δ , where L is a partial differential
operator. The solution u fundamental solution of a PDE.
R
For
R test functions φ, ψ , define the convolution to be: (φ ∗ ψ)(x) := φ(x − y)ψ(y) dy =
φ(y)ψ(x−y) dy :=R (ψ ∗φ)(x). Also define the reflection operator to be: (Rφ)(x) := φ(−x).
Note: (φ ∗ ψ)(x) = φ(y)(Rτ−x ψ)(y) dy by noting that Rτ−x = τx R. Thus, we can define
the convolution by (u ∗ φ)(x) := hu, Rτ−x φi.
This provides us with a point-wise definition for u or an alternate way to define distributions
by considering the limit as φ becomes localized. In this case, (δ ∗ φ)(x) = φ(x), that is, δ
becomes the identity operator: u = L−1 δ . Now, if we solve Lu = f , we can take v = u ∗ f so
Lu = δ =⇒ Lu ∗ f = δ ∗ f = f .
Index
B(A), 1
B(X, Y ), 15
C(X, Y ), 28
C[a, b], 1
C 1 [a, b], 18
C k (Ω) , 39
C01 [a, b], 18
C0∞ (Ω), 39
D0 (Ω), 40
Dk , 39
H1 [a, b], 18
L2 [a, b], 9
`∞ , 1
`p , 2
absolutely convergent, 9
accumulation point, 3
adjoint, 25
algebraic dual, 14
algebraically reflexive, 15
Banach fixed-point theorem, 4
Banach space, 9
Bessel’s inequality, 20
bilinear form, 22
bounded linear functional , 14
bounded linear operator (BLO), 13
bounded sequence, 3
canonical embedding, 15
Cauchy sequence, 3
Cauchy-Schwarz inequality, 2
closed set, 2
closure, 3
coercivity, 23
compact operators, 28
compactness, 10
42
complete (space), 3
completion, 4
continuous (map), 3
contraction mapping, 4
convergent sequence, 3
convex set, 19
convolution, 41
dense, 3
discrete metric, 1
distribution, 40
eigenvalue, 30
equivalent norm, 10
finite-dimension, 8
fixed point, 4
Fourier coefficients, 20
Fredholm alternative, 32, 34
fundamental solution, 41
Gauss-Seidel iteration, 5
Gram-Schmidt Process, 21
Hölder’s inequality, 2
Hahn-Banach theorem, 25
Hilbert space, 18
index dual space, 15
induced norm, 9
inner product, 17
inner product space, 17
integral equations, 6
interior, 2
interior point, 2
Jacobi iteration, 5
linear functional, 14
INDEX
linear operator, 12
linear partial differential operator, 41
metric space, 1
Minkowski’s inequality, 2
multi-index, 39
Neumann series, 31
norm, 9
normed vector space (NVS), 9
null space, 12
open set, 2
operator norm, 13
orthogonality, 19
orthonormal set, 20
Picard’s theorem, 5
Poincaré’s inequality, 36
reflection operator, 41
resolvent operator, 30
resolvent set, 30
Riesz representation theorem, 21
Schauder basis, 9
self-adjoint, 25
self-adjoint operator, 33
separable, 3
sesquilinear form, 23
Sobolev space, 18
spectral radius, 32
Spectral theorem, 33
spectrum, 31
Sturm-Liouville problem, 34
subspace, 8
support, 39
test functions, 39
topology, 2
total orthonormal set, 20
translation operator, 41
Uniform boundedness theorem, 26
43
vector space, 8
weak convergence, 27
weak* convergence, 27