Suppose one has a bounded operator $A:\ell_{2}\to\ell_{2}$, and suppose its matrix in the standard basis is given by $a_{ij}$. Let $k\in\mathbb{N}$ and suppose that:
- If $|i-j|>k$ then $a_{ij}=0$.
- For every $|t|\leq k$ one has $a_{i,i+t}\to 0$ as $i\to\infty$
Then $A$ is compact. Let me explain where I'm having trouble. Let $x^{n}\in\ell_{2}$ be a bounded sequence. Then $(Ax^{n})_i=\sum_{j=-k}^{k}a_{i,i+j}x^{n}_{i+j}$, a sum of a constant amount of elements, each of which goes to $0$. That's nice. Let's try to make it cauchy.
$||Ax^{n}-Ax^{m}||^{2}=\sum_{i=1}^{\infty}\left(\sum_{j=-k}^{k}a_{i,i+j}(x^{n}_{i+j}-x^{m}_{i+j})\right)^{2}$.
I'm not sure how to control this sum. For large $i$ the summands are small, (say $|a_{i,i+j}|<\epsilon$) but I can't really choose a subsequence of $x^{n}$ so that "something" happens for large $i$... For small $i$ I can choose a subsequence of $x^{n}$ so that $(x^{n}-x^{m})_{i}$ is controlled, but then, that would depend on $\epsilon$, and I need to choose "one" subsequence of $x^{n}$, I can't choose different subsequences for different $\epsilon$. Any hint\reference or comment on my approach would be greatly appreciated.