Let $b \in C[O,1]$ and define $(Bf)(x)=b(x)f(x)$ for $f \in L_2[0,1]$ show that $B$ is a compact operator in $L_2[0,1]$ iff $b \equiv 0$
The direction $\Leftarrow )$ is easy to prove because if $b \equiv 0$ then $(Bf)(x)=0$ i.e $Bf$ is the zero functional and that is compact. In the other way i not sure how use the continuity of b, i try supposing that $b(x)\neq 0$ but i dont see any contradiction. Any hint or suggestion i will very grateful.