No Nonzero multiplication operator is compact Let $f,g \in L^2[0,1]$, multiplication operator $M_g:L^2[0,1] \rightarrow L^2[0,1]$ is defined by $M_g(f(x))=g(x)f(x)$. Would you help me to prove that no nonzero multiplication operator on $L^2[0,1]$ is compact. Thanks.
 A: Recall from the theory of Fourier series that the exponentials $h_n=e^{2\pi i n}$ form a basis for $L^2[0,1]$. This allows us to write $g(x)=\sum_1^\infty a_nh_n$, where the convergence is in the $L^2$ norm. Then $gh_k=\sum_1^\infty a_nh_{n+k}$, and in particular $\|gh_{k+j}-gh_k\|$ is at least $a_1$ (where we assume without loss of much generality that $1$ is the smallest $n$ such that $a_n$ is nonzero). This means no subsequence can be Cauchy, so the image of the unit ball cannot be pre compact. This means the operator cannot be compact. 
A: 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. 
