TITLE: A Question on Compact Operators
QUESTION [12 upvotes]: Let $ \mathcal{H} $ and $ \mathcal{K} $ be Hilbert spaces, and let $ T: \mathcal{H} \to \mathcal{K} $ be a bounded linear operator. Show that if $ T $ is a compact operator, then
$$
\lim_{n \to \infty} \| T(e_{n}) \|_{\mathcal{K}} = 0
$$
for every orthonormal sequence $ (e_{n})_{n \in \mathbb{N}} $ in $ \mathcal{H} $. Is the converse of this statement true?
Thanks.

REPLY [4 votes]: Here is a topological proof of the converse that serves to complement Haskell Curry’s argument. It employs the fact that a totally bounded and closed subset of a complete metric space $ X $ is a compact subset of $ X $.
Let $ \mathcal{H} $ and $ \mathcal{K} $ be Hilbert spaces, and let $ T: \mathcal{H} \to \mathcal{K} $ be a bounded linear operator such that
$$
\lim_{n \to \infty} T(\mathbf{e}_{n}) = \mathbf{0}_{\mathcal{H}} \qquad (\spadesuit)
$$
for any orthonormal sequence $ (\mathbf{e}_{n})_{n \in \mathbb{N}} $ in $ \mathcal{H} $. We may assume, WLOG, that $ T \neq 0_{\mathscr{B}(\mathcal{H},\mathcal{K})} $. Fix $ \epsilon > 0 $, and choose $ S $ to be a maximal orthonormal (possibly empty) subset of
$$
\mathbb{S}(\mathcal{H}) \Big\backslash
T^{-1} \! \left[ \epsilon \cdot \overline{\mathbb{B}(\mathcal{K})} \right],
$$
where $ \mathbb{S}(\mathcal{H}) $ denotes the unit sphere in $ \mathcal{H} $ and $ \mathbb{B}(\mathcal{K}) $ the open unit ball in $ \mathcal{K} $.
Now, $ S $ is finite; if this were not the case, then $ S $ would contain an orthonormal sequence $ (\mathbf{e}_{n})_{n \in \mathbb{N}} $, thence
$$
\forall n \in \mathbb{N}: \quad
\| T(\mathbf{e}_{n}) \|_{\mathcal{K}} > \epsilon.
$$
A contradiction to $ (\spadesuit) $ would thus be obtained. Hence, $ \text{Span}(S) $ is a finite-dimensional subspace of $ \mathcal{H} $, which implies that $ \overline{\mathbb{B}(\text{Span}(S))} $ is a compact subset of $ \mathcal{H} $. Proceed to cover this compact subset of $ \mathcal{H} $ by finitely many open balls $ B_{1},\ldots,B_{N} $, each with radius $ \dfrac{\epsilon}{\| T \|} $. Then clearly
$$
\forall k \in \{ 1,\ldots,N \}: \quad
\text{Diam}(T[B_{k}]) \leq 2 \epsilon.
$$
Next, notice that by the maximality of $ S $, we have
$$
          \mathbb{S}(S^{\perp})
\subseteq T^{-1} \! \left[ \epsilon \cdot \overline{\mathbb{B}(\mathcal{K})} \right].
$$
Rewriting this as
$$
          T \! \left[ \mathbb{S}(S^{\perp}) \right]
\subseteq \epsilon \cdot \overline{\mathbb{B}(\mathcal{K})},
$$
we get
$$
          T \! \left[ \mathbb{B}(S^{\perp}) \right]
\subseteq \epsilon \cdot \overline{\mathbb{B}(\mathcal{K})},
$$
so
$$
\text{Diam} \! \left( T \! \left[ \mathbb{B}(S^{\perp}) \right] \right) \leq 2 \epsilon.
$$
Evidently, $ \left\{ \mathbb{B}(S^{\perp}) + B_{k} \right\}_{k = 1}^{N} $ covers $ \mathbb{B}(S^{\perp}) + \mathbb{B}(\text{Span}(S)) \supseteq \mathbb{B}(\mathcal{H}) $. Hence,
$$
          \overline{T[\mathbb{B}(\mathcal{H})]}
\subseteq \bigcup_{k = 1}^{N} \overline{T \! \left[ \mathbb{B}(S^{\perp}) + B_{k} \right]}
\subseteq \bigcup_{k = 1}^{N}
          \overline{T \! \left[ \mathbb{B}(S^{\perp}) \right] + T[B_{k}]}.
$$
As
$$
\forall k \in \{ 1,\ldots,N \}: \quad
     \text{Diam} \!
     \left( \overline{T \! \left[ \mathbb{B}(S^{\perp}) \right] + T[B_{k}]} \right)
\leq 4 \epsilon,
$$
we see that $ \overline{T[\mathbb{B}(\mathcal{H})]} $ is a totally bounded and closed subset of the complete metric space $ \mathcal{K} $, hence compact. Therefore, $ T $ is a compact operator.