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.
=== Definition 1 ===
 
=== Definition 2 ===
== 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: 
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.
=== 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:
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.
Category:Compact Linear Transformations