Let $(\lambda_{n})_{n \in \mathbb{N^{*}}}$ be a sequence of real numbers converging to $0$ and let $(u_{n})_{n \in \mathbb{N^{*}}}$ be an orthonormal family in a Hilbert space $H$. Define $T:H \rightarrow H$ by $$T(u) = \sum_{k=1}^{\infty} \lambda_{k}\langle u,u_{k}\rangle u_{k}$$
Prove that $T$ is compact. could anyone help me please in doing so?
Thanks!
First, I ask you to prove that $T$ is compact if and only if it satisfies the following: if $\{v\}, (v_{n})_{n \in \mathbb{N}} \subseteq H$ and $v_{n} \rightharpoonup v$, then $T(v_{n}) \to T(v)$. Here and henceforth the notation "$x_{n} \rightharpoonup x$" reads "$(x_{n})_{n \in \mathbb{N}}$ converges weakly to $x$."
Now suppose $\{v\}, (v_{n})_{n \in \mathbb{N}} \subseteq H$ and $v_{n} \rightharpoonup v$. Recall that $v_{n} \rightharpoonup v$ implies the sequence $(\|v_{n}\|)_{n \in \mathbb{N}}$ is bounded. Hence the sequence $(\|v_{n} - v\|)_{n \in \mathbb{N}}$ is bounded, by the triangle inequality. It remains to prove $0 = \lim_{n \to \infty} \|T(v_{n} - v)\|$. To do this, fix $N \in \mathbb{N}$ and observe \begin{align*} \|T(v_{n} - v)\|^{2} &= \sum_{j = 1}^{\infty} \lambda_{j}^{2} |\langle v_{n} - v, u_{j} \rangle|^{2} \\ &\leq \sum_{j = 1}^{N} \lambda_{j}^{2} |\langle v_{n} - v, u_{j} \rangle|^{2} + \sup\{\lambda_{j}^{2} \, \mid \, j \geq N + 1\} \|v_{n} - v\|^{2}. \end{align*} Since $v_{n} \rightharpoonup v$, we obtain $$\limsup_{n \to \infty} \|T(v_{n} - v)\|^{2} \leq \sup\{\lambda_{j}^{2} \, \mid \, j \geq N + 1\} \cdot \sup \{\|v_{n} - v\|^{2} \, \mid \, n \in \mathbb{N}\}.$$ From this, conclude that $\limsup_{n \to \infty} \|T(v_{n} - v)\|^{2} = 0$.
Edit: I'll do this without weak convergence. To show that $T$ is a compact operator, it suffices to show that if $(v_{n})_{n \in \mathbb{N}}$ is a bounded sequence, then $(T(v_{n})_{n \in \mathbb{N}}$ has a subsequence that converges in norm. Thus, assume $(v_{n})_{n \in \mathbb{N}}$ is a bounded sequence.
Step 1: Observe that, for each $j \in \mathbb{N}$, the sequence $(\langle v_{n},u_{j} \rangle)_{n \in \mathbb{N}}$ is bounded. Thus, there is a subsequence $(n^{(j)}_{k})_{k \in \mathbb{N}}$ such that $(\langle v_{n^{(j)}_{k}}, u_{j} \rangle)_{k \in \mathbb{N}}$ is a Cauchy sequence. (This follows from the Bolzano-Weierstrass Theorem.)
Step 2: By Step 1, we can apply a diagonalization argument to obtain a subsequence $(n^{*}_{k})_{k \in \mathbb{N}}$ such that $$\forall j \in \mathbb{N} \quad (\langle v_{n^{*}_{k}}, u_{j} \rangle)_{k \in \mathbb{N}} \, \, \text{is a Cauchy sequence.}$$
Step 3: For now, $N$ denotes a positive integer to be determined. If $k, \ell \in \mathbb{N}$, then \begin{align*} \|T(v_{n^{*}_{k}} - v_{n^{*}_{\ell}})\|^{2} \leq \sum_{j = 1}^{N} \lambda_{j}^{2} |\langle v_{n^{*}_{k}} - v_{n^{*}_{\ell}}, u_{j} \rangle|^{2} + \sup\{\lambda_{j}^{2} \, \mid \, j \geq N + 1\} \cdot \|v_{n^{*}_{k}} - v_{n^{*}_{\ell}}\|^{2} \end{align*} Now pick $\epsilon > 0$. First, choose $N \in \mathbb{N}$ such that $$\sup\{\lambda_{j}^{2} \, \mid \, j \geq N +1\} < \frac{\epsilon}{2} \cdot \left(\sup\{\|v_{n} - v_{m}\|^{2} \, \mid \, n, m \in \mathbb{N}\}\right)^{-1}.$$ This is possible since $\lim_{j \to \infty} \lambda_{j}^{2} = 0$.
For each $j \in \{1,2,\dots,N\}$, there is a $M_{j} \in \mathbb{N}$ such that $|\langle v_{n^{*}_{k}} - v^{n^{*}_{\ell}}, u_{j} \rangle|^{2} < \frac{\epsilon}{2N}$ if $k, \ell \geq M_{j}$. Therefore, our previous work and choice of the integers $N,M_{1},\dots,M_{n}$ shows that if $k, \ell \geq \max\{M_{1},\dots,M_{N}\}$, then $$\|T(v_{n^{*}_{k}} - v_{n^{*}_{\ell}})\|^{2} < \epsilon.$$ This proves $(T(v_{n^{*}_{k}}))_{k \in \mathbb{N}}$ is a Cauchy sequence. $H$ is complete so it must converge. Therefore, $(T(v_{n^{*}_{k}}))_{k \in \mathbb{N}}$ is a convergent subsequence of $(T(v_{n}))_{n \in \mathbb{N}}$, which is what we sought to find.