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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.

Marcin Wnuk
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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 $1\le q<p<\infty$ is compact. Hence the answer to the question is NO for $X=l^p$ ($p<2$) or $Y=l^q$ ($2<q<\infty$).

daw
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  • Thanks a lot! I did not know Pitt's theorem and was all the time on the wrong track. I was also wondering - rather out of sheer curiosity - is there some nice characterization of Banach spaces for which such "double sided extension" is actually always possible? – Marcin Wnuk Jan 26 '22 at 09:29