Equivalence of Definitions of Compact Linear Transformation

Theorem

Let $\struct {X, \norm \cdot_X}$ and $\struct {Y, \norm \cdot_Y}$ be normed vector spaces.

Let $T : X \to Y$ be a linear transformation.

The following definitions of the concept of compact linear transformations are equivalent:

Definition 1

Let $\operatorname {ball} X$ be the closed unit ball in $\struct {X, \norm \cdot_X}$.

We say that $T$ is a compact linear transformation if and only if:

$\map \cl {T \sqbrk {\operatorname {ball} X} }$ is compact in $\struct {Y, \norm \cdot_Y}$

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where $\cl$ denotes topological closure.

Definition 2

We say that $T$ is a compact linear transformation if and only if:

for each bounded sequence $\sequence {x_n}_{n \mathop \in \N}$ in $X$:

the sequence $\sequence {T x_n}_{n \mathop \in \N}$ has a subsequence convergent in $\struct {Y, \norm \cdot_Y}$.

Proof

Necessary Condition

Suppose that:

$A$ is compact.

That is:

$\overline {\map A {\operatorname {ball} X} }$ is compact.

Let $\sequence {x_n}_{n \mathop \in \N}$ be a bounded sequence in $X$.

Then there exists a positive real number $M > 0$ such that:

$\norm {x_n}_X \le M$

for all $n \in \N$.

Now define a sequence $\sequence {y_n}_{n \mathop \in \N}$ by:

$\ds y_n = \frac {x_n} M$

for each $n \in \N$.

Then for each $n \in \N$ we have:

\(\ds \norm {y_n}_X\)

\(=\)

\(\ds \norm {\frac {x_n} M}_X\)

\(\ds \)

\(=\)

\(\ds \frac {\norm {x_n}_X} M\)

using positive homogeneity of the norm

\(\ds \)

\(\le\)

\(\ds 1\)

since $\norm {x_n}_X \le M$

So:

$\sequence {y_n}_{n \mathop \in \N}$ is a sequence in $\operatorname {ball} X$.

Now define a sequence $\sequence {z_n}_{n \mathop \in \N}$ by:

$\ds z_n = A y_n$

for each $n \in \N$.

Then:

$\sequence {z_n}_{n \mathop \in \N}$ is a sequence in $\map A {\operatorname {ball} X}$

Since:

$\map A {\operatorname {ball} X} \subseteq \overline {\map A {\operatorname {ball} X} }$

we have:

$\sequence {z_n}_{n \mathop \in \N}$ is a sequence in $\overline {\map A {\operatorname {ball} X} }$

Since $\overline {\map A {\operatorname {ball} X} }$ is compact, we have that:

there exists a convergent subsequence $\sequence {z_{n_j} }$ of $\sequence {z_n}_{n \mathop \in \N}$ with limit in $\overline {\map A {\operatorname {ball} X} }$.

So:

the sequence $\sequence {A y_{n_j} }$ converges.

Let:

$\ds y = \lim_{j \mathop \to \infty} A y_{n_j}$

Then:

$\ds y = \lim_{j \mathop \to \infty} \map A {\frac {x_{n_j} } M}$

so, since $A$ is linear:

$A x_{n_j} \to M y$

as $j \to \infty$.

So:

the sequence $\sequence {x_{n_j} }$ is such that $\sequence {A x_{n_j} }$ converges.

Since $\sequence {x_n}_{n \mathop \in \N}$ was an arbitrary bounded sequences, we have:

all bounded sequences $\sequence {x_n}_{n \mathop \in \N}$ in $X$ have a subsequence $\sequence {x_{n_j} }$ such that the sequence $\sequence {A x_{n_j} }$ converges.

$\Box$

Sufficient Condition

Suppose that:

for all bounded sequences $\sequence {x_n}_{n \mathop \in \N}$ in $X$, there exists a subsequence $\sequence {x_{n_j} }$ such that the sequence $\sequence {A x_{n_j} }$ converges.

From the definition of a compact operator, we aim to show that:

$\overline {\map A {\operatorname {ball} X} }$ is compact.

That is:

all sequences $\sequence {y_n}_{n \mathop \in \N}$ in $\overline {\map A {\operatorname {ball} X} }$ have a convergent subsequence $\sequence {y_{n_k} }$ with limit in $\overline {\map A {\operatorname {ball} X} }$.

Note that from Topological Closure is Closed, we have:

$\overline {\map A {\operatorname {ball} X} }$ is closed in $Y$.

So, from the definition of a closed set in a normed vector space:

if a sequence in $\overline {\map A {\operatorname {ball} X} }$ converges, it has limit in $\overline {\map A {\operatorname {ball} X} }$.

So it suffices to find a subsequence with limit in $Y$.

Let $\sequence {y_n}_{n \mathop \in \N}$ be a sequence in $\overline {\map A {\operatorname {ball} X} }$.

We now construct a sequence $\sequence {x_n}_{n \mathop \in \N}$ in $\map A {\operatorname {ball} X}$ that approximates $\sequence {y_n}_{n \mathop \in \N}$.

For each $n \in \N$, we have $y_n \in \overline {\map A {\operatorname {ball} X} }$, so there exists some $h \in \map A {\operatorname {ball} X}$ such that:

$\ds \norm {y_n - h}_Y < \frac 1 n$

by the definition of closure.

For each $n \in \N$, let $z_n$ be any such $h$.

Since $z_n \in \map A {\operatorname {ball} X}$, there exists $x_n \in \operatorname {ball} X$ such that:

$z_n = A x_n$

Since $x_n \in \operatorname {ball} X$, we also have:

$\norm {x_n}_X \le 1$

for each $n \in \N$.

So:

$\sequence {x_n}_{n \mathop \in \N}$ is bounded.

So, by assumption:

there exists a subsequence $\sequence {x_{n_j} }$ of $\sequence {x_n}_{n \mathop \in \N}$ such that the sequence $\sequence {A x_{n_j} }$ converges.

Let:

$\ds y = \lim_{j \mathop \to \infty} A x_{n_j}$

We show that:

$y_{n_j} \to y$

We have:

\(\ds \norm {y_{n_j} - y}_Y\)

\(=\)

\(\ds \norm {y_{n_j} - y + A x_{n_j} - A x_{n_j} }_Y\)

\(\ds \)

\(\le\)

\(\ds \norm {y_{n_j} - A x_{n_j} }_Y + \norm {A x_{n_j} - y}_Y\)

since the norm satisfies the triangle inequality

\(\ds \)

\(<\)

\(\ds \frac 1 {n_j} + \norm {A x_{n_j} - y}_Y\)

since $\norm {A x_n - y}_Y < n^{-1}$ for each $n \in \N$

Let $\epsilon$ be a positive real number.

Since $n_j \to \infty$ by the definition of subsequence, we have:

$\ds \frac 1 {n_j} \to 0$

So there exists $N_1 \in \N$ such that:

$\ds \frac 1 {n_j} < \frac \epsilon 2$

for $n > N_1$.

By the definition of convergence, since:

$\ds y = \lim_{j \mathop \to \infty} A x_{n_j}$

there exists $N_2 \in \N$ such that:

$\ds \norm {A x_{n_j} - y}_Y < \frac \epsilon 2$

for $n > N_2$.

Let:

$N = \max \set {N_1, N_2}$

Then, for $n > N$, we have:

$\norm {y_{n_j} - y}_Y < \epsilon$

So, since $\epsilon > 0$ was arbitrary we have:

$y_{n_j} \to y$

So $\sequence {y_{n_j} }$ is a convergent subsequence of $\sequence {y_n}_{n \mathop \in \N}$.

Since $\sequence {y_n}_{n \mathop \in \N}$ was arbitrary, we have that:

any sequence $\sequence {y_n}_{n \mathop \in \N}$ in $\overline {\map A {\operatorname {ball} X} }$ has a convergent subsequence $\sequence {y_{n_j} }$ with limit in $\overline {\map A {\operatorname {ball} X} }$.

So:

$\overline {\map A {\operatorname {ball} X} }$ is compact.

Hence:

$A$ is compact.

$\blacksquare$