# Finding the spectrum of the composition of a projection with a multiplication operator

In reading a paper on numerical quadrature I've come across a result that is proved in a manner that is very clever:

Let $X \subset \mathbb{C}$ be a compact, convex set. If $U$ is a finite-dimensional subspace of $L^2(X)$, $P_U: L^2(X) \rightarrow U$ is the orthogonal projection onto $U$, and $T_U: U \rightarrow L^2(X)$ is the multiplication operator $\varphi(z) \mapsto z\varphi(z)$ restricted to $U$. Then the spectrum of $P_U \circ T_U$ is contained in $X$.

The proof is quite clever. Notice that $P_U \circ T_U$ is a map $U \rightarrow U$; it has only point spectrum. Suppose $P_UT_U\varphi = \lambda \varphi$ with $\lambda \notin X$. Then using the geometric assumptions on $X$, we can always find a $\mu \notin X$ such that $|\lambda - \mu| > |x - \mu|$ for all $x \in X$. But then $P_U T_U \varphi - \mu \varphi = (\lambda - \mu) \varphi$, from which we can deduce that there is an $x_0 \in X$ such that $|x_0 - \mu| > | \lambda - \mu|$, a contradiction.

This is all well and good, but I'm interested in knowing whether there is a more functional-analytic proof for this result, something that does not involve the trick with $\mu$. The primary motivation for this is to attempt to extend some of the results in the paper. One of the most basic results in the theory is that any multiplication operator $M_f: L^2(X) \rightarrow L^2(X)$ has spectrum equal to the essential range of $f$, if $f \in L^{\infty}(X)$. I have been trying to use this fact, since $T_U$ is just $M_z$ restricted to $U$, but I haven't been successful so far. Am I doomed to have to use this geometric trick, or is there another approach?

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This is very interesting. May I ask what is the paper you are reading? –  Markus Schweighofer Oct 24 '12 at 9:55
Spectra of Multiplication Operators as a Numerical Tool by Vioreanu and Rokhlin. –  Christopher A. Wong Oct 24 '12 at 18:14
You've got to use convexity somewhere, otherwise the claim is false. This means that some geometry is inevitable. –  fedja Oct 24 '12 at 18:50
@fedja, What is a counterexample? –  Christopher A. Wong Oct 25 '12 at 1:55
$X=\{-1,1\}$, $U$ is the space of constant functions. The operator in question becomes $0$ then. (I took the counting measure on $X$, of course) –  fedja Oct 25 '12 at 10:16

I would like to record an alternative method for proving the claim. First, there is a corollary of the Toeplitz-Hausdorff theorem:

Let $A$ be a bounded normal operator on a Hilbert space. Then the closure of its numerical range, $W(A)$, given by all values $\langle Ax, x \rangle$ with $\|x\|= 1$, is equal to the convex hull of $\sigma(A)$.

Now, using the definitions of $P_U$ and $T_U$ from my question, notice that if $\lambda \in P_U \circ T_U$, then for the eigenfunction $\varphi$, $P_U \varphi = \varphi$ since $T_U$ is the multiplication operator restricted to $U$, so $P_U T_U \varphi = \lambda \varphi$, from which we obtain $\langle T_U \varphi, \varphi \rangle = \lambda \|\varphi\|^2$, hence $\lambda \in W(M_z)$, where $M_z$ is the multiplication operator by $z$. $M_z$ is a normal operator, because $z$ and $\bar{z}$ commute; thus the corollary from above yields

$$\lambda \in \sigma(P_U \circ T_U) \Longrightarrow \lambda \in \mathrm{ch}(\sigma(M_z)) = \mathrm{ch}(X)$$

Thus, if $X$ is itself convex, this implies that $\lambda \in X$.

While also simple, this approach is of course significantly less elementary than the method of proof used in the paper, but nevertheless perhaps somebody else besides myself may find it of interest.

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