Is there a non-multiplication operator on $L^2([0,1])$?

The absolute majority of the bounded operators on $L^2[0,1]$ are not multiplication operators.

Let's see:

• $M_f$ is normal for all $f$. There are many many non-normal operators in $B(L^2[0,1])$.

• $M_f$ is never compact.

Those two give you scores of examples of operators that are not multiplication operators. Even more,

• the algebra $A$ of multiplication operators is not dense in $B(L^2[0,1])$ in any of the usual topologies (norm, sot, wot, etc.). This is easily seen from the fact that $A$ is maximal abelian.

Some concrete examples of non-normal operators: let $H$ be any Hilbert space. Fix an orthonormal basis $\{e_n\} _{n\in\mathbb N}$.

• Let $E_{12}$ be the linear operator induced by $E_{12}e_1=e_2$, $E_{12}e_n=0$ for $n\geq2$. Then $E_{12}^*=E_{21}$ and $E_{12}^*E_{12}=E_{22}$, $E_{12}E_{12}^*=E_{11}$.

• Similarly one can define $E_{kj} $ for $k\ne j$.

• The unilateral shift is the linear operator $S$ induced by $Se_n=e_{n+1}$. It satisfies $S^*S=I$, $SS^*=I-E_{11}$.