# Compact operator

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?

thanks.

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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

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 $$\left\| \sum_{i=1}^{n} \lambda_{i} \mathbf{v}_{i} \right\|_{\mathcal{H}} < \epsilon_{n},$$ where $$\lambda_{i} = \left\{ \begin{array}{ll} \dfrac{\langle \mathbf{x}_{k},\mathbf{v}_{i} \rangle}{\| \mathbf{v}_{i} \|_{\mathcal{H}}^{2}} &\text{, if \| \mathbf{v}_{i} \|_{\mathcal{H}} > 0 }; \\ 0 &\text{, if \| \mathbf{v}_{i} \|_{\mathcal{H}} = 0 }. \end{array} \right.$$ Then set $$\alpha_{n+1} := k \quad \text{and} \quad \mathbf{v}_{n+1} := \mathbf{x}_{k} - \sum_{i=1}^{n} \lambda_{i} \mathbf{v}_{i}.$$

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 $$\forall l \in \mathbb{N}: \quad \| \mathbf{v}_{l} - (\mathbf{x}_{n_{k_{l}}} - \mathbf{x}) \|_{\mathcal{H}} < \frac{1}{l}.$$ 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, $$\lim_{l \rightarrow \infty} \langle T(e_{n_{k_{l}}}),y_{0} \rangle = \langle y_{0},y_{0} \rangle > 0.$$ This contradicts the fact that $(T(e_{n}))_{n \in \mathbb{N}}$ is weakly convergent to $0_{\mathcal{K}}$.

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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.

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