Significance of multiplication operators in operator theory I just read the three versions of the spectral theorem, one of which is the unitary equivalence to a multiplication operator.
Now I asked myself two things


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*How significant are the multiplication operators? Some examples of questions we cannot answer via them for instance would be nice.

*Are there analogues of the Jordan form? Looking at the construction of spectral measures it looks like we could use this machinery on an arbitrary operator, not just as successfully.

You can think of this version of the spectral theorem as the best direct generalization of the spectral theorem for self-adjoint compact operators (or symmetric matrices, to make it even simpler). Recall the statement of this theorem: if $A$ is a self-adjoint compact operator on a Hilbert space $H$ then there exists an orthonormal basis $V$ of $H$ and a set of real numbers $\{\lambda_v\}_{v\in V}$ such that $Av=\lambda_vv$ for every $v\in V$. The spectrum of $A$ then is equal to the closure of $\{\lambda_v\}_{v\in V}$, with $0$ the only possible accumulation point.
Now let $\mu$ be the counting measure on $V$ (i.e. $\mu(B)$ is the number of elements in $B$ if $B$ is finite, and is equal to $+\infty$ otherwise) and consider the measure space $X=L^2(V,2^V,\mu)$. Introduce the multiplication operator $M$ on $X$ by $Mf(x)=\lambda_xf(x)$. This map is unitarily equivalent to $A$ - indeed, the map $U:X\to H:f\mapsto\sum_{v\in V}f(v)v$ is unitary and $U^*AU=M$. Note that in the finite dimensional case, a multiplication operator corresponds to a diagonal matrix.
It is not hard to see that my statement of the spectral theorem above is equivalent to the statement that $A$ is unitarily equivalent to some real multiplication operator on $L^2(V,2^V,\mu)$ for some counting measure $\mu$ on a space $V$. This makes the more general spectral theorem a direct generalization: the result is still true for some measure space, it just may not be a counting measure. This of course means that the spectrum may not consist only of eigenvalues, but multiplication operators are still in a sense simpler to understand.
