Subsequence Characterisation of Compact Linear Transformations

Theorem
Let $\mathbb F \in \set {\R, \C}$.

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

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

Then $A$ is compact :


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

Proof
Let $\operatorname {ball} X$ be the closed unit ball of $X$.

Necessary Condition
Suppose that:


 * $A$ is compact.

That is:


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

Let $\sequence {x_n}$ 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 $\sequence {y_n}$ by:


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

for each $n \in \N$.

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

So:


 * $\sequence {y_n}$ is a sequence in $\operatorname {ball} X$.

Now define a sequence $\sequence {z_n}$ by:


 * $\ds z_n = A y_n$

for each $n \in \N$.

Then:


 * $\sequence {z_n}$ is a sequence in $\map A {\operatorname {ball} X}$

Since:


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

we have:


 * $\sequence {z_n}$ 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 $\sequence {z_n}$ 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 $\sequence {x_n}$ was an arbitrary bounded sequences, we have:


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

Sufficient Condition
Suppose that:


 * for all bounded sequences $\sequence {x_n}$ 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 $\sequence {y_n}$ 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 Subset of Normed Vector Space contains Limits of Sequences iff Closed:


 * 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 $\sequence {y_n}$ be a sequence in $\overline {\map A {\operatorname {ball} X} }$.

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

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:


 * $\sequence {x_n}$ is bounded.

So, by assumption:


 * there exists a subsequence $\sequence {x_{n_j} }$ of $\sequence {x_n}$ 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:

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 $\sequence {y_n}$.

Since $\sequence {y_n}$ was arbitrary, we have that:


 * any sequence $\sequence {y_n}$ 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.