Is there a non-multiplication operator on $L^2([0,1])$? I just learned about multiplication operator on $L^2[0,1]$. These are defined as an operator of the form $M_f(h(t))=f(t)h(t)$ for all $h\in L^2[0,1]$ and for some $f\in L^{\infty}([0,1])$. Do these operators exhaust all operators on $L^2[0,1]$? That is, is there a non-multiplication operator on $L^2[0,1]$?
 A: 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 $S^*S=I$, $SS^*=I-E_{11}$.
A: If you want some example of non multiplicative operator, you have the differential operators, like
$$
(-\Delta)^{-s} f = \frac{C}{|x|^{1-s}} * f
$$
that are also continuous operators on $L^2([0,1])$ for any $s\in(0,1)$, and more generally, the operator of convolution by a nice function.These can be thought of as multiplication operator after taking the Fourier transform.
In general, one can write an operator as a pseudodifferential operator, so general operators can be thought of as operators of the form $f(x,\nabla)$.
