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.
No Nonzero multiplication operator is compact
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functional-analysis
operator-theory
hilbert-spaces
compact-operators
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0Let $S = \{x:|g(x)| > 1/M\}$ where $M$ is sufficiently large so that $\mu(S) > 0$. Consider the set $X$ of functions supported on $S$ with $L_2$-norm at most $1$. Is $M_g(X)$ relatively compact? – 2012-11-13