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.
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.
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}} $.
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.
