# Compact Operator on Hilbert Space Direct Sum

From ProofWiki

## Theorem

Let $\left({H_n}\right)_{n \in \N}$ be a sequence of Hilbert spaces.

Denote by $H = \displaystyle \bigoplus_{n=1}^\infty H_n$ their Hilbert space direct sum.

For each $n \in \N$, let $T_n \in B \left({H_n}\right)$ be a bounded linear operator.

Suppose that one has $\displaystyle \sup_{n \in \N} \left\Vert{T_n}\right\Vert < \infty$, where $\left\Vert{\cdot}\right\Vert$ signifies the operator norm.

Define $T \in B \left({H}\right)$ by:

- $\forall h = \left({h_n}\right)_{n \in \N}: Th = \left({T_n h_n}\right)_{n \in \N} \in H$

(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
- $\displaystyle \lim_{n \to \infty} \left\Vert{T_n}\right\Vert = 0$

## Proof

## Sources

- 1990: John B. Conway:
*A Course in Functional Analysis*... (previous) ... (next): $II.4$ Exercise $13$