A Question on Compact Operators 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.
 A: Let $ (e_{n})_{n \in \mathbb{N}} $ be an orthonormal sequence in $ \mathcal{H} $. As a consequence of Bessel's Inequality, $ (e_{n})_{n \in \mathbb{N}} $ is weakly convergent to $ 0_{\mathcal{H}} $. It follows that
\begin{align}
\forall y \in \mathcal{K}: \quad &\lim_{n \rightarrow \infty} \langle e_{n},{T^{*}}(y) \rangle = 0, \\
&\lim_{n \rightarrow \infty} \langle T(e_{n}),y \rangle = 0.
\end{align}
Therefore, $ (T(e_{n}))_{n \in \mathbb{N}} $ is weakly convergent to $ 0_{\mathcal{K}} $.
Now, assume for the sake of contradiction that $ (T(e_{n}))_{n \in \mathbb{N}} $ does not converge in norm to $ 0_{\mathcal{K}} $. Then there exists an $ \epsilon > 0 $ and a subsequence $ (e_{n_{k}})_{k \in \mathbb{N}} $ of $ (e_{n})_{n \in \mathbb{N}} $ such that $ \| T(e_{n_{k}}) \|_{\mathcal{K}} \geq \epsilon $ for all $ k \in \mathbb{N} $. As $ (e_{n_{k}})_{k \in \mathbb{N}} $ is bounded in norm, by the compactness of $ T $ as an operator, there exists a subsequence $ (e_{n_{k_{l}}})_{l \in \mathbb{N}} $ of $ (e_{n_{k}})_{k \in \mathbb{N}} $ such that $ (T(e_{n_{k_{l}}}))_{l \in \mathbb{N}} $ converges to some limit in $ \mathcal{K} $. Call this limit $ y_{0} $. Clearly, $ y_{0} \neq 0_{\mathcal{K}} $. Therefore,
\begin{equation}
\lim_{l \rightarrow \infty} \langle T(e_{n_{k_{l}}}),y_{0} \rangle = \langle y_{0},y_{0} \rangle > 0.
\end{equation}
This contradicts the fact that $ (T(e_{n}))_{n \in \mathbb{N}} $ is weakly convergent to $ 0_{\mathcal{K}} $.
A: 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.
A: Here is the second installment that answers the converse in the affirmative. The result is not easy to establish. One method of proof uses the spectral calculus for self-adjoint operators, but this is like cracking a nut with a sledgehammer. I provide a softer approach below, which exploits the geometric properties of Hilbert spaces.
Lemma 1
Every bounded sequence in a Hilbert space contains a weakly
convergent subsequence.
Proof
This follows from the reflexivity of Hilbert spaces. Q.E.D.
Lemma 2
Every weakly convergent sequence in a Hilbert space is bounded.
Proof
This follows from the Uniform Boundedness Principle. Q.E.D.
Definition 1
Let $ \mathcal{H} $ be a Hilbert space, and let $ C $ be a fixed
collection of sequences in $ \mathcal{H} $. Given a sequence $
(\mathbf{x}_{n})_{n \in \mathbb{N}} $ in $ \mathcal{H} $, we say
that $ (\mathbf{x}_{n})_{n \in \mathbb{N}} $ can be approximated by
$ C $ if for every sequence $ (\epsilon_{n})_{n \in \mathbb{N}} $ of
positive real numbers, there exists a $ (\mathbf{c}_{n})_{n \in
\mathbb{N}} \in C $ such that $ \| \mathbf{c}_{n} - \mathbf{x}_{n}
\|_{\mathcal{H}} < \epsilon_{n} $ for all $ n \in \mathbb{N} $.
Definition 2
Let $ \mathcal{H} $ be a Hilbert space. We denote by $
\mathbf{BOS}(\mathcal{H}) $ the set of all bounded orthogonal
sequences in $ \mathcal{H} $.
Lemma 3
Let $ \mathcal{H} $ be a Hilbert space, and let $
(\mathbf{x}_{n})_{n \in \mathbb{N}} $ be a weak null-sequence in $
\mathcal{H} $. Then $ (\mathbf{x}_{n})_{n \in \mathbb{N}} $ contains
a subsequence that can be approximated by $
\mathbf{BOS}(\mathcal{H}) $.
Proof
Let $ (\mathbf{x}_{n})_{n \in \mathbb{N}} $ be a weak null-sequence
in $ \mathcal{H} $. Fix a sequence $ (\epsilon_{n})_{n \in
\mathbb{N}} $ of positive real numbers. We inductively define a new
sequence $ (\mathbf{v}_{n})_{n \in \mathbb{N}} $ in $ \mathcal{H} $
and an increasing sequence $ (\alpha_{n})_{n \in \mathbb{N}} $ of
positive integers as follows:


*

*Set $ \alpha_{1} := 1 $ and $ \mathbf{v}_{1} := \mathbf{x}_{1} $.

*For each $ n \in \mathbb{N} $, suppose that $ \alpha_{1},\ldots,\alpha_{n} $ and $ \mathbf{v}_{1},\ldots,\mathbf{v}_{n} $ have been defined. As $ (\mathbf{x}_{n})_{n \in \mathbb{N}} $ converges weakly to $ 0_{\mathcal{H}} $, we can choose a smallest positive integer $ k > \alpha_{n} $ such that
\begin{equation}
\left\| \sum_{i=1}^{n} \lambda_{i} \mathbf{v}_{i} \right\|_{\mathcal{H}} < \epsilon_{n},
\end{equation}
where
\begin{equation}
\lambda_{i} = \left\{
\begin{array}{ll}
\dfrac{\langle \mathbf{x}_{k},\mathbf{v}_{i} \rangle}{\| \mathbf{v}_{i} \|_{\mathcal{H}}^{2}} &\text{, if $ \| \mathbf{v}_{i} \|_{\mathcal{H}} > 0 $}; \\
0 &\text{, if $ \| \mathbf{v}_{i} \|_{\mathcal{H}} = 0 $}.
\end{array} \right.
\end{equation}
Then set
\begin{equation}
\alpha_{n+1} := k \quad \text{and} \quad \mathbf{v}_{n+1} :=
\mathbf{x}_{k} - \sum_{i=1}^{n} \lambda_{i} \mathbf{v}_{i}.
\end{equation}
Notice that $ (\mathbf{v}_{n})_{n \in \mathbb{N}} $ is the result of
applying the Gram-Schmidt orthogonalization procedure to $
(\mathbf{x}_{\alpha_{n}})_{n \in \mathbb{N}} $, which is a
subsequence of $ (\mathbf{x}_{n})_{n \in \mathbb{N}} $. Therefore, $
(\mathbf{v}_{n})_{n \in \mathbb{N}} $ is an orthogonal sequence. By
Lemma 2, $ (\mathbf{x}_{n})_{n \in \mathbb{N}} $ is bounded, so $
(\mathbf{v}_{n})_{n \in \mathbb{N}} \in \mathbf{BOS}(\mathcal{H}) $.
Finally, $ \| \mathbf{v}_{n} - \mathbf{x}_{\alpha_{n}}
\|_{\mathcal{H}} < \epsilon_{n} $ for all $ n \in \mathbb{N} $. Q.E.D.
Theorem
Let $ \mathcal{H} $ and $ \mathcal{K} $ be Hilbert spaces. Let $ T:
\mathcal{H} \rightarrow \mathcal{K} $ be a bounded linear operator
that maps every orthonormal sequence (hence every bounded orthogonal sequence) in $ \mathcal{H} $ to a strong
null-sequence in $ \mathcal{K} $. Then $ T $ is a compact operator.
Proof
Let $ (\mathbf{x}_{n})_{n \in \mathbb{N}} $ be a bounded sequence in
$ \mathcal{H} $. By Lemma 1, there exists a weakly convergent
subsequence $ (\mathbf{x}_{n_{k}})_{k \in \mathbb{N}} $ of $
(\mathbf{x}_{n})_{n \in \mathbb{N}} $. Let $ \mathbf{x} $ be the
weak limit of this subsequence. Clearly, $ (\mathbf{x}_{n_{k}} -
\mathbf{x})_{k \in \mathbb{N}} $ is then a weak null-sequence in $
\mathcal{H} $. By Lemma 3, there exists a subsequence $
(\mathbf{x}_{n_{k_{l}}} - \mathbf{x})_{l \in \mathbb{N}} $ of $
(\mathbf{x}_{n_{k}} - \mathbf{x})_{k \in \mathbb{N}} $ and a
sequence $ (\mathbf{v}_{l})_{l \in \mathbb{N}} \in
\mathbf{BOS}(\mathcal{H}) $ such that
\begin{equation}
\forall l \in \mathbb{N}: \quad \| \mathbf{v}_{l} -
(\mathbf{x}_{n_{k_{l}}} - \mathbf{x}) \|_{\mathcal{H}} <
\frac{1}{l}.
\end{equation}
Observe that $ T $ must map $ (\mathbf{v}_{l})_{l \in \mathbb{N}} $
to a strong null-sequence in $ \mathcal{K} $. Hence, by the
approximation property, we have $ \displaystyle \lim_{l \rightarrow
\infty} T(\mathbf{x}_{n_{k_{l}}} - \mathbf{x}) = 0_{\mathcal{K}} $.
In other words, $ (T(\mathbf{x}_{n}))_{n \in \mathbb{N}} $ contains
$ (T(\mathbf{x}_{n_{k_{l}}}))_{l \in \mathbb{N}} $ as a strongly
convergent subsequence. Therefore, $ T $ is a compact operator. Q.E.D.
