$Tx=\sum_{k=0}^{\infty}\lambda_k(x,e_k)e_k$ bounded and compact iff $\lambda_k\to 0$

Question

Let $H$ be a real Hilbert space and $e:\mathbb{N}\to H$ an orthonormal system. Let $\lambda\in l^{\infty}(\mathbb{R})$ be a bounded sequence and define $T:H \to H$ by $$Tx=\sum_{k=0}^{\infty}\lambda_k(x,e_k)e_k.$$

Show that

a) T is bounded with $||T||=||\lambda||_{l^{\infty}(\mathbb{R})}$ and

b) $T$ is compact iff $\lambda_k\to0$ as $k\to 0$.

I think I first need to show the operator is well-defined, i.e that the sum converges. I therefore look at the partial sums: |$\sum_{k=0}^{N}\lambda_k(x,e_k)e_k|\leq ||\lambda||_{l^{\infty}(\mathbb{R})} \sum_{k=0}^{N}||x||$ by Cauchy-Schwarz. But this is not good enough because I still have the sum I am not sure how to control it. The sum also reminded me of Bessel's inequality $$0\leq||x||^2-\sum_{k=0}^n |(x,e_k)|^2=||x-\sum_{k=0}^n (x,e_k)e_k||^2$$ since we have the bound $||\lambda||_{l^{\infty}} \sum_{k=0}^N (x,e_k)e_k$. If I have would have $\lambda \in l^2$ I could bound the partial sums by $$(\sum_{k=0}^n \lambda_k^2)^{1/2}\left(\sum_{k=0}^n|(x,e_k)|^2\right)^{1/2}\leq ||\lambda||_{l^2}||x||$$ but I only know $\lambda \in l^{\infty}$…

For part b) I know because $H$ is Hilbert T is compact iff $T$ is a limit of finite rank operators. So I define $T_N(x)=\sum_{k=0}^{N}\lambda_k(x,e_k)e_k$ and look at $||Tx-T_Nx||=||\sum_{k=N+1}^{\infty}\lambda_k(x,e_k)e_k||$. I was thinking this converges to $0$ because it is a tail sum of $Tx$ which converges. But this is obviously wrong as I didn't use $\lambda_k\to 0$.

Answer 1

Observe that $$\|Tx\|^2 = \sum_k (Tx,e_k)^2 = \sum_k \lambda_k^2 (x,e_k)^2 \le \|\lambda\|_\infty^2 \sum_k (x,e_k)^2 =\|\lambda\|_\infty^2 \|x\|^2.$$

As for compactness a similar calculation gives you that $$\|Tx - T_Nx\|^2 = \sum_{k \ge N+1} \lambda_k^2 (x,e_k)^2 \le \left( \sup_{k \ge N} \lambda_k^2 \right) \|x\|^2$$ so that $$\|T - T_N\| \le \sqrt{ \sup_{k \ge N} \lambda_k^2} \to 0$$ if and only if $\lambda_k \to 0$.

Answer 2

a)

We have:

$||T(x)||^2 = (T(x),T(x)) = (\sum_{k=0}^\infty \lambda_k (x,e_k)e_k, \sum_{n=0}^\infty \lambda_n(x,e_n)e_n) = \sum_{k=0}^\infty |\lambda_k|^2 |(x,e_k)|^2||e_k||^2 $

We can now bound every $|\lambda_k|^2 \le ||(\lambda_k)||^2$ and by Bessel inequality $\sum_{k=0}^\infty |(x,e_k)|^2 \le ||x||^2$, so that:

$||T(x)||^2 \le ||(\lambda_k)||^2 ||x||^2$

Which means $||T|| \le ||(\lambda_k)||$.

Moreover taking sequence $x_n = e_n$ we have:

$||T(x_n)||^2 = \sum_{k=0}^\infty |\lambda_k|^2 |(x,e_k)|^2 = |\lambda_n|^2$, so that $||T|| \ge |\lambda_n|$ for every $n\in \mathbb N$, which means $||T|| \ge ||(\lambda_k)||$

Conclusion: $||T|| = ||(\lambda_k)||$

b)

Take $ \lambda_k \to 0$ and let $T_N$ be such that $T_N(x) = \sum_{k=0}^N \lambda_k (x,e_k)e_k$

Then $T_N$ is compact (since finite dimensional operator), moreover $||(T-T_N)(x)||^2 = \sum_{k=N+1}^\infty |\lambda_k|^2 |(x,e_k)|^2 $.

Since $\lambda_k \to 0$, we can make that less than $\epsilon ||x||^2$ for any $\epsilon$ and that means $T_N \to T$ in operator norm. Limit of compact operators is compact.

In second direction: Assume contrary that $\lambda_k \not \to 0$, that there is some $\delta >0$ and some subsequence $|\lambda_{n_m}| > \delta$

Now take $x_m = e_m$ and note $||T(x_{s}) - T(x_{r})||^2 = || \lambda_s e_s - \lambda_r e_r ||^2 = ( \lambda_s e_s - \lambda_r e_r , \lambda_s e_s - \lambda_r e_r) = \lambda_s^2 + \lambda_r^2 \ge 2\delta^2 $ So that we cannot find convergent subsequence. That means for $T$ to be compact, we must have $\lambda_k \to 0$