Theorem: Compact Operator on Hilbert Space Direct Sum

Let $\sequence {\HH_n}_{n \mathop \in \N}$ be a sequence of Hilbert spaces.
Denote by $\HH = \bigoplus_{n \mathop \in \N} \HH_n$ their Hilbert space direct sum.
For each $n \in \N$, let $T_n \in \map B {\HH_n}$ be a bounded linear operator.
Suppose that:
:$\ds \sup_{n \mathop \in \N} \norm {T_n} < \infty$
where $\norm {\, \cdot \, }$ signifies the norm on bounded linear operators.
Define $T \in \map B \HH$ by: 
:$\forall h = \sequence {h_n}_{n \mathop \in \N}: T h = \sequence {T_n h_n}_{n \mathop \in \N} \in \HH$
(That $T$ is indeed bounded follows from Bounded Linear Operator on Hilbert Space Direct Sum.)
Then $T$ is compact {{iff}} the following conditions hold:
:For each $n \in \N$, $T_n$ is compact
:$\ds \lim_{n \mathop \to \infty} \norm {T_n} = 0$

Proof:
Assume:
:$(1):$ For each $n \in \N$, $T_n$ is compact
:$(2): \ds \lim_{n \mathop \to \infty} \norm {T_n} = 0$
We show $\map \cl {T \sqbrk {\operatorname {ball} \HH } }$ is compact in $\HH$.
By $(1)$, $K_k := \map \cl {T_k \sqbrk {\operatorname {ball} \HH_k } }$ is compact in $\HH_k$ for each $k\in\N$.
Further, each $K_k \subseteq \HH_k$ is a sequentially compact subset since:
:Compact Metric Space is Totally Bounded
:Complete and Totally Bounded Metric Space is Sequentially Compact
By Countable Product of Sequentially Compact Spaces is Sequentially Compact:
:$\prod_{k \in \N} K_k \subseteq \prod_{k \in \N} \HH_k$
is a sequentially compact subset with respect to the product topology.
Recall Subspace of Complete Metric Space is Relatively Compact iff Every Sequence has Cauchy Subsequence.
Let $\sequence {T h^n}_n$ be a sequence in $T \sqbrk {\operatorname {ball} \HH }$.
As:
:$T \sqbrk {\operatorname {ball} \HH } \subseteq \prod_{k \in \N} K_k$
there is a convent subsequence $\sequence {T h^{n_j} }_j$ in $\prod_{k \in \N} \HH_k$.
That is, for each $k\in\N$:
:$\ds \lim_{j\mathop\to\infty}T _k h^{n_j}_k$ converges in $\HH_k$.
As Convergent Sequence is Cauchy Sequence:
:$(3): \sequence {T _k h^{n_j}_k}_j$ is a Cauchy sequence in $\HH_k$ for each $k$.
Let $\epsilon > 0$ be arbitrary.
By $(2)$, there is an $N \in \N$ be so large that $\sup_{k \ge N} \norm {T_k} < \epsilon$.
By $(3)$, we can choose an $M \in \N$ so large that for all $i,j \ge M$:
:$\ds \max _{k \mathop\in \set {0,\ldots,N-1} }\norm {T h^{n_i}_k - T h^{n_j}_k }_{\HH_k} < \frac \epsilon {\sqrt N}$
Then for all $i,j \ge M$:
{{begin-eqn}}
{{eqn | l = \norm {T h^{n_i} - T h^{n_j} }^2
      | r = \sum _{k \mathop \in \N} \norm {T_k h^{n_i}_k - T_k h^{n_j}_k }_{\HH_k}^2
}}
{{eqn | r = \sum _{k \mathop = 0}^{N-1} \norm {T_k h^{n_i}_k - T_k h^{n_j}_k }_{\HH_k}^2 + \sum _{k \mathop = N}^\infty \norm {T_k h^{n_i}_k - T_k h^{n_j}_k }_{\HH_k}^2
}}
{{eqn | o = \le
      | r = \epsilon^2 + \sum _{k \mathop = N}^\infty \norm {T_k h^{n_i}_k - T_k h^{n_j}_k }_{\HH_k}^2
}}
{{eqn | o = \le
      | r = \epsilon^2 + \sum _{k \mathop = N}^\infty \norm {T_k}^2 \norm {h^{n_i}_k - h^{n_j}_k }_{\HH_k}^2
}}
{{eqn | o = \le
      | r = \epsilon^2 + \epsilon^2 \sum _{k \mathop = N}^\infty \norm {h^{n_i}_k - h^{n_j}_k }_{\HH_k}^2
}}
{{eqn | o = \le
      | r = \epsilon^2 + \epsilon^2 \sum _{k \mathop = 0}^\infty \norm {h^{n_i}_k - h^{n_j}_k }_{\HH_k}^2
}}
{{eqn | r = \epsilon^2 + \epsilon^2 \norm {h^{n_i} - h^{n_j} }^2
}}
{{eqn | o = \le
      | r = \epsilon^2 + \epsilon^2 \paren {\norm {h^{n_i} } + \norm {h^{n_j} } }^2
}}
{{eqn | o = \le
      | r = 5 \epsilon^2
      | c = as $h^{n_i}, h^{n_j} \in \operatorname {ball} \HH$
}}
{{end-eqn}}
This means, $\sequence {T h^{n_j} }_j$ is a Cauchy sequence in $\HH$.
{{qed|lemma}}
Conversely, assume $T$ is compact.
For each $n \in \N$, the projection $\pi_n : \HH \to \HH_n$ is continuous, as:
:$\ds \forall h \in \HH : \norm {\map {\pi_n} h}_{\HH_n} \le \norm h$
Furthermore:
:$T_n \sqbrk {\operatorname {ball} \HH_n } = \pi_n \sqbrk {T \sqbrk {\operatorname {ball} \HH } } \subseteq \pi_n \sqbrk {\map \cl {T \sqbrk {\operatorname {ball} \HH } } }$
Now $\pi_n \sqbrk {\map \cl {T \sqbrk {\operatorname {ball} \HH } } }$ is compact, as Continuous Image of Compact Space is Compact.
By Compact Subset of Normed Vector Space is Closed and Bounded:
:$\map \cl {T_n \sqbrk {\operatorname {ball} \HH_n } } \subseteq \pi_n \sqbrk {\map \cl {T \sqbrk {\operatorname {ball} \HH } } }$
Therefore $\map \cl {T_n \sqbrk {\operatorname {ball} \HH_n } }$ is compact, as Closed Subspace of Compact Space is Compact.
That is, $T_n$ is compact.
{{qed|lemma}}
Finally, we show $\lim_{n \mathop \to \infty} \norm {T_n} = 0$.
{{AimForCont}} there is an $\epsilon > 0 $ and sequence $\sequence {n_j}_j$ such that:
:$\ds \norm {T_{n_j} } \ge 2 \epsilon$
For each $j$, we can choose $a_j \in \operatorname {ball} \HH_{n_j}$ such that:
:$ \norm {T_{n_j} a_j}_{\HH_{n_j} } \ge \epsilon$
Let:
:$h^j := \tuple { \underbrace {0,\ldots,0}_{n_j}, a_j,0,\ldots}$
Then $h^j \in \operatorname {ball} \HH$, and:
:$\norm {T h^j} = \norm {\tuple { \underbrace {0,\ldots,0}_{n_j}, T_{n_j} a_j,0,\ldots} } = \norm {T_{n_j} a_j}_{\HH_{n_j} } \ge \epsilon$
Moreover, for all $\forall j \ne k$
:$\norm {T h^j - T h^k}^2 = \norm {T h^j}^2 + \norm {T h^k}^2 \ge 2 \epsilon^2$
as $\innerprod {T h^j} {T h^k} = 0$.
Therefore, $\sequence {T h^j}_j$ does not have a subsequence.
That is, $T \sqbrk {\operatorname {ball} \HH }$ is not relatively compact.
This is a contradiction.
{{qed}}