Problem: Let $l^{2}=\{(x_{1},x_{2},...)| x_{n} \in \mathbb{R}, \forall n,(x_{1}^{2}+x_{2}^{2}+...)<\infty\}$, $\|x\|=(x_{1}^{2}+x_{2}^{2}+...)^{\frac{1}{2}}$. Given $\lambda_{n}\in\mathbb{R}$($n=1,2,3...$), $\lim\limits_{n \rightarrow \infty} \lambda_{n}=0$.
Define $A:l^{2}\rightarrow l^{2}$ as follows: For $x=(x_{1},x_{2},...)\in l^{2}$, $Ax=(\lambda_{1}x_{1},\lambda_{2}x_{2},...)$.
Show that for a bounded sequence:$\{x^{(j)}\}_{j=1}^{\infty}$ in $l^{2}$, $\{Ax^{(j)}\}_{j=1}^{\infty}$ has a subsequence which converges in $l^{2}$.
I am new in fuctional analysis and I have no idea how to handle this problem. I am grateful to any help! Thanks a lot in advance!
There're several facts that you should use: 1) If the image of an operator is finite-dimensioned, then this operator is compact. 2) If a sequence of compact operators converges in operator norm, then the limit is also a compact operator.
Hence in your case you study a sequence of operators $A_k$ such that
$A_k(x) = (\lambda_1x_1,\dots,\lambda_kx_k,0,0,\dots)$. Apparently, $A_k$ is compact because the dimension of the image is at most $k$.
Then we need to prove that $\|A-A_k\|\to 0$. Take an arbitrary $x\in\ell_2$ with $\|x\|=1$. We write $\|(A-A_k)(x)\| \le \sup_{r>k} |\lambda_r|\cdot \|x\|_{\ell_2}$ (it's easy to see why it's true). This allows to say that $\| A-A_k \| \le \sup_{r>k} |\lambda_r|$. Then again, $\sup_{r>k} |\lambda_r|\to 0 $ as $k\to\infty$ because $\lambda_k\to 0$.
Thus, $A_k\to A$ in operator norm, and we can conclude that $A$ is compact.