# Understanding a proof that a compact multiplication operator is zero

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?

$$T(B(0,1))$$ 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.