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This answer gives a proof of the fact that if $g\in L^\infty(0,1)$ and the multiplication operator $T_g:L^2(0,1)\rightarrow L^2(0,1)$ is compact, then $g=0$ almost everywhere:

We show that if $g$ is not the equivalence class of the null function, then $M_g$ is not compact. Let $c>0$ such that $\lambda(\{x,|g(x)|>c\})>0$ (such a $c$ exists by assumption). Let $S:=\{x,|g(x)|>c\}$, $H_1:=L^2[0,1]$, $H_2:=\{f\in H_1, f=f\chi_S\}$. Then $T\colon H_2\to H_2$ given by $T(f)=T_g(f)$ is onto. Indeed, if $h\in H_2$, then $T(h\cdot \chi_S \cdot g^{—1})=h\cdot\chi_S=h$.

As $H_2$ is a closed subspace of $H_1$, it's a Banach space. This gives, by the open mapping theorem that $T$ is open. It's also compact, so $T(B(0,1))$ is open and has compact closure. By Riesz theorem, $H_2$ is finite dimensional.

But for each $N$, we can find $N+1$ disjoint subsets of $S$ which have positive measure, and their characteristic functions will be linearly independent, which gives a contradiction.

I’m interested in the part in bold. My question is, why does the fact that $T(B_1)$ is open and its closure is compact imply that $H_2$ is finite-dimensional? The answer cites Riesz’s theorem, but that theorem just says that a Banach space whose closed unit ball is compact must finite dimensional. Why does the fact that the closure of the image of the open unit ball under $T$ is compact imply that the closure of the open unit ball is compact?

Or is there a mistake in this proof?

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$T(B(0,1))$ contains some open ball $B(0,r)$ around $0$ in $H_2$ (because it is open and contains $0$). So the closure of $B(0,r)$ is contained in the closure of $T(B(0,1))$ which is compact since $T$ is a compact operator. If a closed ball around $0$ in $H_2$ is compact then $H_2$ is finite dimensional.

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