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.
 A: 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.
