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If $H$ and $K$ are Hilbert spaces,show that if $T:H\longrightarrow K$ is a compact operator and $\{e_{n}\}$ is any orthonormal sequence in $H$ then $\|Te_{n}\|\to0$.Is the converse true?


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I guess $H$ is a Hilbert space. What is $K$? Use the fact that the sequence $\{e_n\}$ is weakly convergent to $0$ and the fact that a compact operator transform a weakly convergent sequence in a convergent sequence (for the norm). –  Davide Giraudo Sep 21 '11 at 9:45
could you explain that why $\{e_{n}\}\to 0$ weakly? –  Vahid Sep 21 '11 at 9:50
This really reads like a homework question; so perhaps it would be good to consult the FAQ: meta.math.stackexchange.com/questions/1803/… –  Matthew Daws Sep 21 '11 at 9:52
If the Hilbert space $H$ is separable, we can show that $\lim_{n\to\infty}\langle e_n,v\rangle=0$ for all $v$ with $v=\sum_{k=1}^N\alpha_ke_k$. We can conclude that $\lim_{n\to\infty}\langle e_n,v\rangle=0$ for all $v$, since the vectors of the form $\sum_{k=1}^N\alpha_ke_k$, $N\in\mathbb N,\alpha_k\in\mathbb C$ is dense in $H$. –  Davide Giraudo Sep 21 '11 at 9:56
@DavideGiraudo: Another way to see this is that by Bessel's inequality, $\sum_{n=1}^\infty |\langle e_n, v \rangle|^2 \le ||v||^2 < \infty$. Since the series converges, its terms must go to 0. –  Nate Eldredge Sep 21 '11 at 12:44

3 Answers 3

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:

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

  2. 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.

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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}} $.

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As people have suggested, since $T$ is a compact operator, it is in particular a bounded linear operator. Since the orthonormal basis ${e_n}$ is weakly convergent (there is one proof here : http://en.wikipedia.org/wiki/Weak_convergence_(Hilbert_space)#Weak_convergence_of_orthonormal_sequences ), we get $\parallel T(e_n) \parallel \rightarrow 0.$ The converse would not be true since this applies to any bounded operator.

I would like to comment that a somewhat relevant result is the spectral theorem for compact symmetric operators: T is a compact symmetric operator on a separable Hilbert space $\mathscr{H}$ iff it has an orthornormal basis with eigenvalues $\lambda \rightarrow 0$ as $k \rightarrow \infty.$

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No, this cannot be right. If $H=K$ and $T$ is the identity operator then $||T e_n|| = 1$ for all $n$ and does not go to 0. You can't ignore the compactness assumption. –  Nate Eldredge Nov 17 '11 at 3:39

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