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I want to show that if $X$ is a separable Banach space, then there exist a Banach space $Y$ and an injective compact operator from $X$ to $Y$. I know that it's true, since I have seen this fact multiple times (e.g. here), but I don't know how to prove it.

I tried to find single space $Y$ for every $X$ expliciliy, but haven't succeed. Also as a main example of $X$ I considered $C[0,1]$, and I wasn't able to find space $Y$ and compact injective map even for this particular case. I also tried to consider embedding of $X$ into it's topological dual space $X'$, but that also didn't work.

Mr Ynoplanetashka
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  • For $X=C[0,1]$, you can take the map that sends $u$ to the solution of the bvp $-v''=u$, $v(0)=v(1)=0$, and $Y=X$. – daw Dec 05 '24 at 14:16
  • And any separable $X$ embeds into $C(0,1)$ isometrically – Evangelopoulos Foivos Dec 05 '24 at 14:21
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    For future reference, the article Goldberg and A.H. Kruse, The Existence of Compact Linear Maps Between Banach Spaces. Proc. A.M.S. 13 (1962), 808-811 (mentioned in a MO post linked to the one you gave as an example) gives criteria for the existence of given classes of compact operators, including compact injective ones but also compact operators with dense range. – Bruno B Dec 05 '24 at 14:28

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Every separable Banach space embeds isometrically into $\ell^\infty$. The multiplication operator $T \colon \ell^\infty \to c_0$ given by $$(x_n)\mapsto \left (\frac{x_n}{n}\right)$$ is compact and injective.

Evangelopoulos Foivos
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