Compact Operators

Compact Operator

Let

ℋ

be a hilbert space and let

T : ℋ \to ℋ

be a bounded linear operator then we say that it is compact if for every bounded subset

B \subseteq ℋ

, the image

T (B)

is relatively compact, meaning that the closure of

T (B)

is compact in the norm topology of

ℋ

The set of compact transformations in $L (X, Y)$ will be denoted by $K (X, Y)$ .

Bounded Operator in Terms of Sequences

Let

X

and

Y

be normed spaces. A linear transformation

T \in L (X, Y)

is compact if, for any bounded sequence

(x_{n})

X

, the sequence

(T x_{n})

Y

contains a convergent subsequence.

$⟹$ If $T$ is assumed compact and $(x_{n})$ is bounded ie) the set $im (x_{n}) = {x_{n} : n \in ℕ_{1}}$ is bounded, and therefore $T (im (x_{n}))$ is compact in $Y$ and therefore contains a convergent subsequence

$⟸$ Now suppose that every bounded sequence $(x_{n})$ contains a subsequence $x_{n_{k}}$ such that $(T x_{n_{k}})$ converges in $Y$ . Our goal is to show that given any bounded subset $B \subseteq X$ we must show that $T (B)$ is compact, but recall that every compact set is closed as we are working in a metric space induced by the norm, therefore if we prove that $T (B)$ is compact, then it is closed so that $T (B) = T (B)$ showing that $T (B)$ is compact.

So moving towards this given any bounded subset $B \subseteq X$ , let $(y_{n})$ be any sequence in $T (B)$ . Then $y_{n} = T x_{n}$ for some sequence $(x_{n})$ in $B$ . Since $B$ was bounded, then so is $(x_{n})$ , therefore by our assumption we know $(T x_{n})$ contains a convergent subsequence, that is $(y_{n})$ contains a convergent subsequence therefore $T (B)$ is compact, and as discussed this shows that $T (B)$ is compact.

A Compact Operator Is Bounded

Let

X

and

Y

be normed spaces an let

T \in K (X, Y)

. Then

T

is bounded.

Suppose that

T

is not bounded. Then for each integer

n \geq 1

there exists a unit vector

x_{n}

such that

‖ T x_{n} ‖ \geq n

. Since the sequence

{x_{n}}

is bounded, by the compactness of

T

there exists a subsequence

{T x_{n (r)}}

which converges. This contradicts

‖ T x_{n (r)} ‖ \geq n (r)

. (ie. convergence implies boundedness)

Note that the above shows that: $K (X, Y) \subset B (X, Y)$ .

The Compact Operators Are a Linear Subspace of the Bounded Operators

Suppose that

S, T \in K (X, Y)

and that

α, β \in ℂ

then

α S + β T \in K (X, Y)

Let

{x_{n}}

be a bounded sequence in

X

. Since

S

is compact, there is a subsequence

{x_{n (r)}}

such that

{S x_{n (r)}}

converges. Then, since

{x_{n (r)}}

is bounded and

T

is compact, there is a subsequence

{x_{n (r (s))}}

of the sequence

{x_{n (r)}}

such that

{T x_{n (r (s))}}

converges. Since the sum of convergent sequences converges, it follows that the sequence

{α S x_{n (r (s))} +

β T x_{n (r (s))}}

converges. Thus

α S + β T

is compact.

The Composition of Linear Operators Is Compact If at Least One Is

Suppose that

S \in B (X, Y), T \in B (Y, Z)

and at least one of the operators

S, T

is compact, then

T \circ S \in B (X, Z)

is compact.

Let

{x_{n}}

be a bounded sequence in

X

. If

S

is compact then there is a subsequence

{x_{n (r)}}

such that

{S x_{n (r)}}

converges. Since

T

is bounded (and so is continuous), the sequence

{T S x_{n (r)}}

converges. Thus

T S

is compact. If

S

is bounded but not compact the the sequence

{S x_{n}}

is bounded. Then since

T

must be compact, there is a subsequence

{S x_{n (r)}}

such that

{T S x_{n (r)}}

converges, and again

T S

is compact.

Riesz'

Suppose that

X

is a normed vector space, and that

Y

is a closed subspace of

X

where

Y \neq X

and

α \in (0, 1)

then there exists a

‖ x_{α} ‖ = 1

and

‖ x_{α} - y ‖ > α

for all

y \in Y

TODO: Add the proof here.

The Unit Disk nor the Unit Circle Is Compact in an Infinite Dimensional Space

TODO: Add the content for the theorem here.

Suppose that

x_{1} \in K

then since

X

is not finite dimensional it must be the case that

span (x_{1}) \neq X

, now we know that

span (x_{1})

is closed and therefore we obtain some

x_{2} \in K

such that

‖ x_{2} - α x_{1} ‖ \geq \frac{3}{4}

The Identity Operator on an Infinite Dimensional Space Is Not Compact

X

is an infinite dimensional normed vector space then

{id}_{X}

is not compact

Recall that since there exists a sequence of unit vectors

(x_{n}) \in X

that does not have a convergent subsequence. This is a problem because

(x_{n})

is a bounded sequence, but

({id}_{X} (x_{n})) = (x_{n})

doesn't have a convergent subsequence so

{id}_{X}

is not compact.

week 4

A Compact Operator on a Infinite Dimensional Space Is Not Invertible

X

is an infinite-dimensional normed space and

T \in K (X)

, then

T

is not invertible

Suppose that

T

were invertible, so that

T^{- 1}

exists, and that

{id}_{X} = T^{- 1} \circ T

, so that

{id}_{X}

is compact, but then

X

was infinite dimensional, so

{id}_{X}

is not compact so that

T

is not invertible.

Compact Symmetric Operators That Commute Can Be Simultaneously Diagonalized

Suppose that

ℋ

is a hilbert space, if

T, S : ℋ \to ℋ

are compact symmetric operators which commute, i.e.

(T S =

S T

), show that they can be diagonalized simultaneously. In other words, there exists an orthonormal basis for

ℋ

which consists of eigenvectors for both

T

and

S

Suppose

λ_{i}

are the collection of eigenvalues of

T

and

E_{λ_{i}} (T)

are the corresponding eigenspaces. Then we will show that an orthogonal collection in

E_{λ_{i}}

also are eigenvectors of

S

. Suppose that

f_{1}, f_{2}, \dots f_{n}

forms a basis for

E_{λ}

, then

T f_{j} = λ_{1} f_{j} .

So that,

T S f_{j} = S T f_{j} = λ S f_{j},

i.e.,

S f_{j} \in E_{λ} (T)

, i.e.

S f_{j} = \sum_{i = 1}^{n} α_{i, j} f_{i}

Thus,

S : E_{λ} \to E_{λ}

can be represented as an

n \times n

matrix with entries

α_{i, j}

. From the symmetry of

S

, it follows that

α_{i, j} = α_{j, i}

and thus, the orthogonal matrix has a collection of orthogonal eigenvectors of the mapping

S

. This, shows that every eigenvector of

T

with eigenvalue not equal to 0 is also an eigenvector of

S

. For

λ = 0

, a similar proof shows that

S : null (T) \to null (T)

and it follows from the spectral theorem, that there exists an orthogonal basis of

null (T)

which are the eigenvectors of

S

too.

Complex Form of a Linear Operator

Let

ℋ

be a separable Hilbert space and assume that

T : ℋ \to ℋ

is a compact normal bounded operator, then there exist a pair of commuting compact self-adjoint operators

A, B : ℋ \to ℋ

such that

T = A + B i

Define the real part $A$ and imaginary part $B$ of $T$ as follows: $A : = \frac{T + T^{*}}{2}, B : = \frac{T - T^{*}}{2 i},$ where $T^{*}$ denotes the adjoint of $T$ .

We now show that $A$ and $B$ are self-adjoint, we verify that $A^{*} = A$ and $B^{*} = B$ . $A^{*} = {(\frac{T + T^{*}}{2})}^{*} = \frac{T^{*} + {(T^{*})}^{*}}{2} = \frac{T^{*} + T}{2} = A .$ so that, $A$ is self-adjoint, onto $B$ $B^{*} = {(\frac{T - T^{*}}{2 i})}^{*} = \frac{T^{*} - T}{- 2 i} = \frac{T - T^{*}}{2 i} = B .$ Thus, $B$ is also self-adjoint.

$A, B$ are also compact as the compact linear operators form a linear subspace of the bounded operators.

hw5

If a Bounded Linear Operator Has Finite Rank Then It Is Compact

Let

X, Y

be normed spaces and

T \in B (X, Y)

then if

T

has finite rank then

T

is compact

Since

T

has finite rank then the space

Z = im (T)

is a finite dimensional normed space. Now given any bounded sequence

(x_{n})

X

then the sequence

(T x_{n})

is bounded in

Z

and thus by the Bolzano-Weierstrass theorem this sequence must have a convergence subsequence, meaning that

T

is compact.

hw8

Every Compact Operator Is Bounded

Let

X, Y

be normed vector spaces, then for any compact linear operator

T : X \to Y

is bounded

It's clear that the unit sphere

U = {x \in X : ‖ x ‖ = 1}

is bounded by construction, since

T

is compact, then we know that

T (U)

is compact therefore we know that

T (U) \subseteq T (U)

is bounded, and thus we have:

\sup_{‖ x ‖ = 1} | T x | < \inf

therefore

T

is bounded

Every Compact Operator Is Continuous

Let

X, Y

be normed vector spaces, then for any compact linear operator

T : X \to Y

is continuous

Since we just realized that it must be bounded, then it also must be continuous, as needed.

Finite Rank Operator

Given a linear operator

T

we say that it has finite rank whenever:

\dim (im (T)) < \infty

A Bounded Finite Rank Operator Is Compact

Suppose that

T

is a bounded finite rank linear operator, then

T

is compact

Let

(x_{n})

be any bounded sequence in

X

, then since we have

{‖ T x_{n} ‖}_{Y} \leq {‖ T ‖}_{op} ‖ x_{n} ‖

then we deduce that

(T x_{n})

is a bounded, thus

{T x_{n} : n \in : ℕ 1}

is a subspace of a finite dimensional normed vector space and thus closed, therefore it is compact, and therefore we know that it has a convergent subsequence. Therefore

T

is compact, as needed.

If a Linear Operator Has a Finite Dimensional Domain Then It Is Compact

Suppose that

T

is a linear operator such that

\dim (dom (T)) < \infty

then

T

is compact.

Since the domain is finite dimensional then we know

T

is a finite rank operator, therefore by the above proposition we know that

T

is compact.