User:Caliburn/s/fa/Space of Compact Linear Transformations is Linear Subspace of Space of Bounded Linear Transformations

Theorem
Let $\mathbb F$ be a subfield of $\C$.

Let $\struct {X, \norm \cdot_X}$ be a normed vector space over $\mathbb F$.

Let $\struct {Y, \norm \cdot_Y}$ be a Banach space over $\mathbb F$.

Let $\map {B_0} {X, Y}$ be the space of compact linear transformations $X \to Y$.

Let $\map B {X,Y}$ be the space of bounded linear transformations $X \to Y$.

Then:


 * $\map {B_0} {X, Y}$ is a linear subspace of $\map B {X, Y}$.

Proof
From Compact Linear Transformation is Bounded, we have:


 * $\map {B_0} {X, Y} \subseteq \map B {X, Y}$

From Zero Linear Transformation is Compact, we have:


 * $\map {B_0} {X, Y} \ne \O$

So, by One-Step Vector Subspace Test, we aim to show that:


 * for each $A_1, A_2 \in \map {B_0} {X, Y}$ and $\alpha \in \mathbb F$ we have $A_1 + \alpha A_2 \in \map {B_0} {X, Y}$.

Let $A_1, A_2 \in \map {B_0} {X, Y}$.

Let $\alpha \in \mathbb F$.

We want to show that:


 * each bounded sequence $\sequence {x_n}$ in $X$ has a subsequence $\sequence {x_{n_k} }$ such that $\sequence {\paren {A_1 + \alpha A_2} x_{n_k} }$ converges.

Let $\sequence {x_n}$ be a bounded sequence in $X$.

Since $A_1$ is compact:


 * there exists a subsequence $\sequence {x_{m_j} }$ such that $\sequence {A_1 x_{m_j} }$ converges.

Clearly:


 * $\sequence {x_{m_j} }$ is bounded.

So, since $A_2$ is compact:


 * there exists a subsequence $\sequence {x_{n_k} }$ of $\sequence {x_{m_j} }$ such that $\sequence {A_2 x_{n_k} }$ converges.

From Limit of Subsequence equals Limit of Sequence: Normed Vector Space, we have:


 * $\sequence {A_1 x_{n_k} }$ converges.

So, we have:


 * $\sequence {\paren {A_1 + \alpha A_2} x_{n_k} }$ converges

with $\sequence {x_{n_k} }$ a subsequence of $\sequence {x_n}$ as required.

So:


 * $A_1 + \alpha A_2 \in \map {B_0} {X, Y}$