# Extending a non-compact operator to a non-compact operator between Hilbert spaces

functional-analysis

Let $$X,Y$$ be separable Banach spaces, and $$S: X \rightarrow Y$$ be a non-compact (continuous) operator. Can I always find two separable Hilbert spaces $$H_1, H_2$$ and continuous operators $$T: H_1 \rightarrow X, T_2: Y \rightarrow H_2$$ such that
$$T_2 S T_1: H_1 \rightarrow H_2$$
is non-compact?

I believe the answer is positive, but I cannot come up with a proof.

The most promising attempt relied on the fact that for each separable Banach space $$B$$ we can find Hilbert spaces $$H$$ and $$G$$ s.t. $$H$$ is densely embedded in $$B$$ and $$B$$ is densely embedded in $$G$$, see Lemma 4.48 in
https://arxiv.org/pdf/1607.03591.pdf and the answer of shalop in the following post Is every Banach space densely embedded in a Hilbert space?

Any help would be greatly appreciated.

Since separable Hilbert spaces are isomorph to $$l^2$$, we can reduce the question to $$H_1=H_2=l^2$$. According to Pitt's theorem, every linear and continuous operator from $$l^p$$ to $$l^q$$ with $$1\le q is compact. Hence the answer to the question is NO for $$X=l^p$$ ($$p<2$$) or $$Y=l^q$$ ($$2).