Compact Operator on Hilbert Space Direct Sum

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Theorem

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 if and only if 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 $i,j \ge M$:

\(\ds \norm {T h^{n_i} - T h^{n_j} }^2\) \(=\) \(\ds \sum _{k \mathop \in \N} \norm {T_k h^{n_i}_k - T_k h^{n_j}_k }_{\HH_k}^2\)
\(\ds \) \(=\) \(\ds \sum _{k \mathop = 0}^{N-1} \norm {T_k h^{n_i}_k - T_k h^{n_j}_k }_{\HH_k}^2 + \sum _{k \mathop = N}^\infty \norm {T_k h^{n_i}_k - T_k h^{n_j}_k }_{\HH_k}^2\)
\(\ds \) \(\le\) \(\ds \epsilon^2 + \sum _{k \mathop = N}^\infty \norm {T_k h^{n_i}_k - T_k h^{n_j}_k }_{\HH_k}^2\)
\(\ds \) \(\le\) \(\ds \epsilon^2 + \sum _{k \mathop = N}^\infty \norm {T_k}^2 \norm {h^{n_i}_k - h^{n_j}_k }_{\HH_k}^2\)
\(\ds \) \(\le\) \(\ds \epsilon^2 + \epsilon^2 \sum _{k \mathop = N}^\infty \norm {h^{n_i}_k - h^{n_j}_k }_{\HH_k}^2\)
\(\ds \) \(\le\) \(\ds \epsilon^2 + \epsilon^2 \sum _{k \mathop = 0}^\infty \norm {h^{n_i}_k - h^{n_j}_k }_{\HH_k}^2\)
\(\ds \) \(=\) \(\ds \epsilon^2 + \epsilon^2 \norm {h^{n_i} - h^{n_j} }^2\)
\(\ds \) \(\le\) \(\ds \epsilon^2 + \epsilon^2 \paren {\norm {h^{n_i} } + \norm {h^{n_j} } }^2\)
\(\ds \) \(\le\) \(\ds 5 \epsilon^2\) as $h^{n_i}, h^{n_j} \in \operatorname {ball} \HH$

This means, $\sequence {T h^{n_j} }_j$ is a Cauchy sequence in $\HH$.

$\Box$


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.

$\Box$


Finally, we show $\lim_{n \mathop \to \infty} \norm {T_n} = 0$.

Aiming for a contradiction, suppose 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.

$\blacksquare$


Sources

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