I wanted to ask if my proof (sketch) of the following statement is correct. Namely, let $p>1$ and define $q= \frac{p}{p-1}$ we are given an operator $K : l^{p} \rightarrow l^{q}$ defined as $x \rightarrow \sum_{n = 1}^{\infty} a_{jk}x_k$ where $\sum_{j,k \in \mathbb{N}} a_{jk}^{q} < \infty$. In order to show that this map is compact, I wanted to proceed along the following steps
- Use reflexivity of $l^p$ to use $K$ compact iff $K$ maps weakly converging sequences to strongly converging ones.
- We find that the map $l_j (x) = \sum_{n = 1}^{\infty} a_{jk}x_k$ is in $(l_p)^*$ since it is bounded
- Then for any weakly converging sequence, $x^k \rightarrow x$, we now we use that $$\Vert K(x^m - x) \Vert = \left(\sum_{j\in \mathbb{N}} \left| l_j(x^m -x)\right|^{q}\right)^{1/q}$$ and can for every $\epsilon > 0$ find m (if necessary passing to a diagonal sequence) s.t for all $j \in \mathbb{N}$, $| l_j(x^m -x)| < \frac{\epsilon}{2^j}$ by weak convergence of $x^m$.
- Now reinserting this estimate into the norm and get that our sequence converges in the norm.
Is this proof correct? (maybe when producing the estimate I wonder whether $| l_j(x^m -x)| < \epsilon a_{jm}$ actually suffices which is definitely given?
Best Regards