TITLE: No Nonzero multiplication operator is compact
QUESTION [8 upvotes]: 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.

REPLY [6 votes]: 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.