Proof. (⇒) $\( T \)$ compact, $\( \mathopen{}\left({x}_{n}\right)\mathclose{} \)$ bounded implies $\( \overline{ \mathopen{}\left(T\mathopen{}\left( {x}_{n}\right)\mathclose{}\right)\mathclose{} } \)$ compact.

(⇐) $\( B \)$ bounded, $\( \mathopen{}\left({y}_{n}\right)\mathclose{} \)$ a sequence in $\( \overline{T\mathopen{}\left( B\right)\mathclose{}} \)$ then get $\( \mathopen{}\left({x}_{n}\right)\mathclose{} \)$ in $\( B \)$ such that $\( \mathopen{}\left\lVert{}T\mathopen{}\left( {x}_{n}\right)\mathclose{}-{y}_{n}\right\rVert\mathclose{}\lt \frac{1}{n} \)$. Since then $\( \mathopen{}\left(T\mathopen{}\left( {x}_{{n}_{k}}\right)\mathclose{}\right)\mathclose{} \)$ converges for subsequence $\( \mathopen{}\left({x}_{{n}_{k}}\right)\mathclose{} \)$, get $\( \mathopen{}\left({y}_{{n}_{k}}\right)\mathclose{} \)$ to converge (to the same limit).

Proof. If $\( S \)$ and $\( T \)$ are in $\( \mathcal{K}\mathopen{}\left( X, Y\right)\mathclose{} \)$ and $\( \mathopen{}\left({x}_{n}\right)\mathclose{} \)$ is bounded in $\( X \)$, then there exists $\( \mathopen{}\left( {x}_{n}'\right)\mathclose{} \)$ such that $\( \mathopen{}\left(T\mathopen{}\left( {x}_{n}'\right)\mathclose{}\right)\mathclose{} \)$ converges, and there is also a further subsequence $\( \mathopen{}\left( {x}_{n}''\right)\mathclose{} \)$ such that $\( \mathopen{}\left(S\mathopen{}\left( {x}_{n}''\right)\mathclose{}\right)\mathclose{} \)$ converges. Then we have that $\( \mathopen{}\left( \mathopen{}\left(S+T\right)\mathclose{}\mathopen{}\left( \mathopen{}\left( {x}_{n}'' \right)\mathclose{}\right)\mathclose{} \right)\mathclose{} \)$ is convergent, which shows $\( S+T\in \mathcal{K}\mathopen{}\left( X, Y\right)\mathclose{} \)$.

Scalar multiplication is clearly ok. For closed, suppose $\( \mathopen{}\left({T}_{n}\right)\mathclose{} \)$ is a sequence of compact operators converging in norm to $\( T\in \mathcal{L}\mathopen{}\left( X, Y\right)\mathclose{} \)$ ($\( \mathopen{}\left\lVert{}{T}_{n}-T\right\rVert\mathclose{} \to 0 \)$). Take $\( \mathopen{}\left({x}_{n}\right)\mathclose{} \)$ in $\( X \)$ with $\( \mathopen{}\left\lVert{}{x}_{n}\right\rVert\mathclose{}\leq M \)$ for all $\( n \)$. Using compactness of $\( {T}_{1} \)$, $\( {T}_{2} \)$, …, get successive subsequences $\( \mathopen{}\left( {x}_{n}'\right)\mathclose{} \)$, $\( \mathopen{}\left( {x}_{n}''\right)\mathclose{} \)$, … such that $\( \mathopen{}\left( {T}_{1}\mathopen{}\left( {x}_{n}'\right)\mathclose{} \right)\mathclose{} \)$ converges, $\( \mathopen{}\left( {T}_{j}\mathopen{}\left( {x}_{n}''\right)\mathclose{} \right)\mathclose{} \)$ converges for $\( j \)$ in $\( \mathopen{}\left\{\, 1, 2\,\right\}\mathclose{} \)$, $\( \mathopen{}\left( {T}_{j}\mathopen{}\left( {x}_{n}'''\right)\mathclose{} \right)\mathclose{} \)$ converges for $\( j \)$ in $\( \mathopen{}\left\{\, 1, 2, 3\,\right\}\mathclose{} \)$, etc. Let $\( \mathopen{}\left({x}_{{m}_{k}}\right)\mathclose{} \)$ be the $\( k \)$th item in the $\( m \)$th sequence. This makes $\( \mathopen{}\left({x}_{{m}_{k}}\right)\mathclose{}_{k=1}^{\infty} \)$ eventually a subsequence of $\( \mathopen{}\left( {x}_{n}'\right)\mathclose{} \)$, $\( \mathopen{}\left( {x}_{n}''\right)\mathclose{} \)$, …. It follows that $\( \mathopen{}\left( {T}_{n}\mathopen{}\left( {x}_{{m}_{k}}\right)\mathclose{} \right)\mathclose{}_{k=1}^{\infty} \)$ converges for every $\( n \)$. Now $$\[ \mathopen{}\left\lVert{}T\mathopen{}\left( {x}_{{m}_{i}}\right)\mathclose{}-T\mathopen{}\left( {x}_{{m}_{k}}\right)\mathclose{}\right\rVert\mathclose{}\leq \mathopen{}\left\lVert{}\mathopen{}\left(T-{T}_{n}\right)\mathclose{}\mathopen{}\left( {x}_{{m}_{i}}\right)\mathclose{}\right\rVert\mathclose{}+\mathopen{}\left\lVert{}{T}_{n}\mathopen{}\left( {x}_{{m}_{i}}-{x}_{{m}_{k}}\right)\mathclose{}\right\rVert\mathclose{}+\mathopen{}\left\lVert{}\mathopen{}\left({T}_{n}-T\right)\mathclose{}\mathopen{}\left( {x}_{{m}_{k}}\right)\mathclose{}\right\rVert\mathclose{}\leq 2M\mathopen{}\left\lVert{}T-{T}_{n}\right\rVert\mathclose{}+\mathopen{}\left\lVert{}{T}_{n}\mathopen{}\left( {x}_{{m}_{i}}-{x}_{{m}_{k}}\right)\mathclose{}\right\rVert\mathclose{}\text{.} \]$$ Given $\( ε\gt 0 \)$, there exists $\( n \)$ such that $\( \mathopen{}\left\lVert{}T\mathopen{}\left( {x}_{{m}_{i}}\right)\mathclose{}-T\mathopen{}\left( {x}_{{m}_{k}}\right)\mathclose{}\right\rVert\mathclose{}\lt \frac{ε}{2}+\mathopen{}\left\lVert{}{T}_{n}\mathopen{}\left( {x}_{{m}_{i}}-{x}_{{m}_{k}}\right)\mathclose{}\right\rVert\mathclose{} \)$ and this is less than $\( \frac{ε}{2}+\frac{ε}{2}= ε \)$ for sufficiently large $\( i \)$ and $\( k \)$. So the sequence is Cauchy, and we're done since $\( Y \)$ is complete.

1. We may assume $\( λ= 1 \)$. Suppose $\( {x}_{n}-T\mathopen{}\left( {x}_{n}\right)\mathclose{} \to w \)$. Let $\( {d}_{n}= \operatorname{d}\mathopen{}\left( {x}_{n}, \operatorname{Ker}\mathopen{}\left( 1-T\right)\mathclose{}\right)\mathclose{}= \inf_{T\mathopen{}\left( y\right)\mathclose{}= y}{}\mathopen{}\left\lVert{}{x}_{n}-y\right\rVert\mathclose{} \)$. Harmlessly perturb each $\( {x}_{n} \)$ by a vector in $\( \operatorname{Ker}\mathopen{}\left( 1-T\right)\mathclose{} \)$ to make $\( \mathopen{}\left\lVert{}{x}_{n}\right\rVert\mathclose{}\lt {d}_{n}+1 \)$. We claim that $\( \sup_{n}{}{d}_{n}\lt \infty \)$. Otherwise, pass to a subsequence and assume $\( {d}_{n} \to \infty \)$. Let $\( {z}_{n}= \frac{{x}_{n}}{{d}_{n}+1} \)$, so $\( \mathopen{}\left\lVert{}{z}_{n}\right\rVert\mathclose{}\lt 1 \)$. Again passing to a subsequence, we may assume $\( T\mathopen{}\left( {z}_{n}\right)\mathclose{} \to y \)$, because $\( T \)$ is compact. Also $\( {z}_{n}-T\mathopen{}\left( {z}_{n}\right)\mathclose{}= \frac{{x}_{n}-T\mathopen{}\left( {x}_{n}\right)\mathclose{}}{{d}_{n}+1} \to 0 \)$. It follows that $\( {z}_{n} \to y \)$ and that $\( T\mathopen{}\left( y\right)\mathclose{}= y \)$. Thus $$\[ \frac{{d}_{n}}{{d}_{n}+1}\leq \frac{\mathopen{}\left\lVert{}{x}_{n}-\mathopen{}\left({d}_{n}+1\right)\mathclose{}y\right\rVert\mathclose{}}{{d}_{n}+1}= \mathopen{}\left\lVert{}{z}_{n}-y\right\rVert\mathclose{} \to 0 \text{,} \]$$ a contradiction. With the claim established, we know that $\( \mathopen{}\left({x}_{n}\right)\mathclose{} \)$ is bounded, so after passing to a subsequence we get that $\( T\mathopen{}\left( {x}_{n}\right)\mathclose{} \to y \)$, so $\( {x}_{n} \to w+y \)$, and thus $\( w= \lim_{n}{} \mathopen{}\left(1-T\right)\mathclose{}{x}_{n} = \mathopen{}\left(1-T\right)\mathclose{}\mathopen{}\left(w+y\right)\mathclose{} \)$.

2. Suppose contrariwise that $\( λ\in \mathop{\sigma}\mathopen{}\left( T\right)\mathclose{}\setminus \mathopen{}\left\{\, 0\,\right\}\mathclose{} \)$ but $\( \operatorname{Ker}\mathopen{}\left( λ-T\right)\mathclose{}= \mathopen{}\left\{\, 0\,\right\}\mathclose{} \)$. As usual, we may assume $\( λ= 1 \)$. The range of $\( 1-T \)$ cannot be all of $\( X \)$, since if it were, the injective operator $\( 1-T \)$ would be invertible by the open mapping theorem. Thus $\( \mathopen{}\left(1-T\right)\mathclose{}\mathopen{}\left( X\right)\mathclose{} \)$ is a closed proper subspace of $\( X \)$, and $\( 1-T \)$ is a linear homeomorphism of $\( X \)$ with $\( \mathopen{}\left(1-T\right)\mathclose{}\mathopen{}\left( X\right)\mathclose{} \)$. It follows that the subspaces $\( {\mathopen{}\left(1-T\right)\mathclose{}}^{n}\mathopen{}\left( X\right)\mathclose{} \)$ ($\( n\in \mathopen{}\left\{\, 0, 1, 2, \dotsc\,\right\}\mathclose{} \)$) are all closed, with $\( {\mathopen{}\left(1-T\right)\mathclose{}}^{n+1}\mathopen{}\left( X\right)\mathclose{}\subsetneq {\mathopen{}\left(1-T\right)\mathclose{}}^{n}\mathopen{}\left( X\right)\mathclose{} \)$ for each $\( n \)$, Lemma I.65 yields a sequence $\( \mathopen{}\left({x}_{n}\right)\mathclose{} \)$ with $$\[ \mathopen{}\left\lVert{}{x}_{n}\right\rVert\mathclose{}= 1 \text{,} \]$$ $$\[ {x}_{n}\in {\mathopen{}\left(1-T\right)\mathclose{}}^{n}\mathopen{}\left( X\right)\mathclose{} \text{,} \]$$ $$\[ \operatorname{d}\mathopen{}\left( {x}_{n}, {\mathopen{}\left(1-T\right)\mathclose{}}^{n+1}\mathopen{}\left( X\right)\mathclose{}\right)\mathclose{}\gt \frac{1}{2} \]$$ for every $\( n \)$. Take $\( m\gt n \)$ and write $\( y= \mathopen{}\left(T-1\right)\mathclose{}\mathopen{}\left( {x}_{n}\right)\mathclose{}+\mathopen{}\left(1-T\right)\mathclose{}\mathopen{}\left( {x}_{m}\right)\mathclose{}-{x}_{m} \)$. Then $\( y\in {\mathopen{}\left(T-1\right)\mathclose{}}^{n+1}\mathopen{}\left( X\right)\mathclose{} \)$. Since $\( T\mathopen{}\left( {x}_{n}\right)\mathclose{}-T\mathopen{}\left( {x}_{m}\right)\mathclose{}= {x}_{n}+y \)$, we conclude that $\( \mathopen{}\left\lVert{}T\mathopen{}\left( {x}_{n}\right)\mathclose{}-T\mathopen{}\left( {x}_{m}\right)\mathclose{}\right\rVert\mathclose{}\gt \frac{1}{2} \)$ for all $\( m\neq n \)$, which contradicts $\( T\in \mathcal{K}\mathopen{}\left( X\right)\mathclose{} \)$.

3. Suppose contrariwise that for some $\( δ\gt 0 \)$ there are distinct $\( {λ}_{1} \)$, $\( {λ}_{2} \)$, … and non-zero vectors $\( {x}_{1} \)$, $\( {x}_{2} \)$, … with $\( T\mathopen{}\left( {x}_{j}\right)\mathclose{}= {λ}_{j}{x}_{j} \)$ and $\( \mathopen{}\left\lvert{}{λ}_{j}\right\rvert\mathclose{}\geq δ \)$ for all $\( j \)$. Let $\( {Y}_{n} \)$ be the linear span of $\( {x}_{1} \)$, $\( {x}_{2} \)$, …, $\( {x}_{n} \)$. As in elementary linear algebra, the $\( {x}_{j} \)$'s are linearly independent, so each $\( {Y}_{n} \)$ is properly contained in $\( {Y}_{n+1} \)$. By Lemma I.65, we can find unit vectors $\( {y}_{n} \)$ with $\( {y}_{n}\in {Y}_{n} \)$ and $\( \operatorname{d}\mathopen{}\left( {y}_{n+1}, {Y}_{n}\right)\mathclose{}\gt \frac{1}{2} \)$. Notice that $\( T\mathopen{}\left( {Y}_{n}\right)\mathclose{}\subseteq {Y}_{n} \)$. Take $\( m\gt n \)$. Write $\( {y}_{m}= α{x}_{m}+ y' \)$, where $\( y'\in {Y}_{m-1} \)$. We have $$\[ T\mathopen{}\left( {y}_{m}\right)\mathclose{}-T\mathopen{}\left( {y}_{n}\right)\mathclose{}= {λ}_{m}α{x}_{m}+T\mathopen{}\left( y'\right)\mathclose{}-T\mathopen{}\left( {y}_{n}\right)\mathclose{}-{λ}_{m}{y}_{m}+{λ}_{m}{y}_{m}= {λ}_{m}{y}_{m}-{λ}_{m} y'+T\mathopen{}\left( y'\right)\mathclose{}-T\mathopen{}\left( {y}_{n}\right)\mathclose{}\in {λ}_{m}{y}_{m}+{Y}_{m-1} \text{.} \]$$ We conclude that $\( \mathopen{}\left\lVert{}T\mathopen{}\left( {y}_{n}\right)\mathclose{}, T\mathopen{}\left( {y}_{m}\right)\mathclose{}\right\rVert\mathclose{}\gt \frac{δ}{2} \)$ for all $\( m\neq n \)$, which contradicts $\( T\in \mathcal{K}\mathopen{}\left( X\right)\mathclose{} \)$.

Proof.

1. Since $\( \overline{ T\mathopen{}\left( \mathrm{B}_{H}\mathopen{}\left( 1\right)\mathclose{}\right)\mathclose{} } \)$ is compact, it is separable and there exists a countable subset $\( S\subseteq \mathrm{B}_{H}\mathopen{}\left( 1\right)\mathclose{} \)$ such that $\( \overline{T\mathopen{}\left( S\right)\mathclose{}}= \overline{ T\mathopen{}\left( \mathrm{B}_{H}\mathopen{}\left( 1\right)\mathclose{}\right)\mathclose{} } \)$. So, $\( T\mathopen{}\left( \bigcup_{n=1}^{\infty}{}nS\right)\mathclose{} \)$ is a countable dense subset of $\( \overline{T\mathopen{}\left( H\right)\mathclose{}} \)$.
2. Notice $\( \mathopen{}\left\lVert{}{P}_{n}\mathopen{}\left( y\right)\mathclose{}-y\right\rVert\mathclose{} \to 0 \)$ for all $\( y\in \overline{T\mathopen{}\left( H\right)\mathclose{}} \)$ (since we have $\( {\mathopen{}\left\lVert{}{P}_{n}\mathopen{}\left( y\right)\mathclose{}-y\right\rVert\mathclose{}}^{2}= \sum_{j=n+1}^{\infty}{}{\mathopen{}\left\lvert{}\mathopen{}\left\langle{}y, {e}_{j}\right\rangle\mathclose{}\right\rvert\mathclose{}}^{2} \)$). $\( \mathopen{}\left\lVert{}{P}_{n+1}\mathopen{}\left( T\mathopen{}\left( x\right)\mathclose{}\right)\mathclose{}-T\mathopen{}\left( x\right)\mathclose{}\right\rVert\mathclose{}\leq \mathopen{}\left\lVert{}{P}_{n}\mathopen{}\left( T\mathopen{}\left( x\right)\mathclose{}\right)\mathclose{}-T\mathopen{}\left( x\right)\mathclose{}\right\rVert\mathclose{} \)$ for all $\( x \)$. So, $\( \mathopen{}\left\lVert{}{P}_{n+1}\mathopen{}\left( T\right)\mathclose{}-T\right\rVert\mathclose{}\leq \mathopen{}\left\lVert{}{P}_{n}\mathopen{}\left( T\right)\mathclose{}-T\right\rVert\mathclose{} \)$. Let $\( δ= \inf_{n}{}\mathopen{}\left\lVert{}{P}_{n}\mathopen{}\left( T\right)\mathclose{}-T\right\rVert\mathclose{} \)$. If $\( δ\gt 0 \)$, then we get $\( {x}_{n}\in H \)$ with $\( \mathopen{}\left\lVert{}{x}_{n}\right\rVert\mathclose{}= 1 \)$ such that $\( \mathopen{}\left\lVert{}{P}_{n}\mathopen{}\left( T\mathopen{}\left( {x}_{n}\right)\mathclose{}\right)\mathclose{}-T\mathopen{}\left( {x}_{n}\right)\mathclose{}\right\rVert\mathclose{}\geq \frac{δ}{2} \)$. Compactness of $\( T \)$ gives $\( T\mathopen{}\left( {x}_{{n}_{k}}\right)\mathclose{} \to y \)$ (in $\( \overline{T\mathopen{}\left( H\right)\mathclose{}} \)$). So $$\[ \frac{δ}{2}\leq \mathopen{}\left\lVert{}{P}_{{n}_{k}}T\mathopen{}\left( {x}_{{n}_{k}}\right)\mathclose{}-T\mathopen{}\left( {x}_{{n}_{k}}\right)\mathclose{}\right\rVert\mathclose{}\leq \mathopen{}\left\lVert{}{P}_{{n}_{k}}T\mathopen{}\left( {x}_{{n}_{k}}\right)\mathclose{}-y\right\rVert\mathclose{}+\mathopen{}\left\lVert{}{P}_{{n}_{k}}\mathopen{}\left( y\right)\mathclose{}-y\right\rVert\mathclose{}+\mathopen{}\left\lVert{}y-T\mathopen{}\left( {x}_{{n}_{k}}\right)\mathclose{}\right\rVert\mathclose{}\leq 2\mathopen{}\left\lVert{}T\mathopen{}\left( {x}_{{n}_{k}}\right)\mathclose{}-y\right\rVert\mathclose{}+\mathopen{}\left\lVert{}{P}_{{n}_{k}}\mathopen{}\left( y\right)\mathclose{}-y\right\rVert\mathclose{} \to 0 \text{,} \]$$ and $\( \frac{δ}{2}= 0 \)$.