From the “This can’t be new” department

Here is a fundamentally obvious observation which must be well-known, but which seems to be overlooked in the three or four texts of functional analysis I’ve looked at (for instance, here). This came up during my Spectral Theory class). Consider a (bounded) multiplication operator M_g acting on

$H=L^2(X,\mu),$

where (X,μ) is a finite measure space, i.e., we have

$M_g(\varphi)=g\varphi\text{ for all } \varphi \in H,$

where the multiplier g is a measurable bounded function. A basic question is to understand the spectrum of M_g, which is the set

$\sigma(M_g)=\{\lambda\in\mathbf{C}\,\mid\, (M_g-\lambda)\text{ is invertible}\},$

where “invertible” is meant in the Banach algebra of bounded/continuous linear maps from H to itself (which means the same as “bijective” here because M_g-λ is continuous and H is a Hilbert space, so by the Closed Graph Theorem for instance, its set-theoretic inverse, if it exists, is continuous).

The intuitive answer is that the spectrum should be the image of g in C; however, this can’t be right in general, because the spectrum must be closed, and there is no reason that g(X) be closed. So the standard correct answer is that

$\sigma(M_g)=\mathrm{Essim}(g),$

the essential range (or image) of g, defined by

$\lambda\in \mathrm{Essim}(g)\text{ if and only if } \mu(\{x\,\mid\, |g(x)-\lambda|<\epsilon\})>0\text{ for all }\epsilon>0.$

This is a fine answer as it goes, but there is a feeling that one should minimize the number of definitions involving too many epsilons and quantifiers, if possible. And the observation is that this is perfectly possible here: this definition exactly means that

$\sigma(T)=\mathrm{Supp}\ g_*(\mu)$

is the support of the image measure; we recall that this measure is defined by

$g_*(\mu)(A)=\mu(\{x\in X\,\mid\, g(x)\in A\})$

for any measurable set A in C, and that a point λ belongs to its support if and only if any of its open neighborhoods, say the discs with radius ε>0 around it, have (strictly) positive measure, which exactly translates to

$(g_*\mu)(\{z\,\mid\, |z-\lambda|>\epsilon\})=\mu(\{x\,\mid\, |g(x)-\lambda|<\epsilon\})>0,\text{ for all } \epsilon>0,$

namely, this is equivalent to λ belonging to the spectrum of M_g, as claimed.

This coincidence means in particular that there is no need to try to remember how to prove that this essential range is closed in C: everyone knows that the support of something has to be closed…

Another point where this is useful is in the definition of the functional calculus for these multiplication operators: suppose

$f\,:\, \sigma(M_g)\rightarrow \mathbf{C}$

is given and is continuous. The (continuous) functional calculus defines a bounded operator f(M_g) in a natural way, which means that if f is a polynomial, this operator is the “obvious” one. Since one checks immediately by induction and linearity that

$f(M_g)=M_{f\circ g}$

if f is a polynomial, it is natural to expect the same formula to hold in general. The problem with this, is that it is not necessarily the case that

$g(X)\subset \sigma(M_g),$

since g is only measurable (and bounded), so the composition is not immediately well-defined. But the point is of course that

$\mu(\{x\,\mid\, g(x)\notin \mathrm{Supp}(g_*(\mu))\})=(g_*\mu)(\mathbf{C}-\mathrm{Supp}(g_*\mu))=0$

(using the characterization of the support as the complement of the largest open set with zero measure), so the composition makes sense “almost everywhere”.

In this interpretation, the spectral mapping theorem

$\sigma(f(M_g))=f(\sigma(M_g))$

becomes the fact that

$\mathrm{Supp}((f\circ g)_*(\mu))=f(\mathrm{Supp}\ g_*(\mu))$

(which makes sense because f is continuous and the support is compact here, so its image under f is closed, as it should be); this can be checked directly, of course.

Note: The reason why multiplication operators are important is, of course, that the spectral theorem shows that any normal operator on a Hilbert space is unitarily equivalent to an operator of this type.

From the “This can’t be new” department

Published by

Kowalski