# 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.

-
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 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.

-

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 $0_{\mathcal{K}}$.

-
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 $k \rightarrow \infty.$
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