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 Mg 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 Mg, 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 Mg 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 Mg, 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(Mg) 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.

Published by

Kowalski

I am a professor of mathematics at ETH Zürich since 2008.

6 thoughts on “From the “This can’t be new” department”

  1. Dear Emmanuel,

    Thanks for putting this very nice observation in writing. My (somewhat vague) impression is that pushforward measures are a little neglected, and so it doesn’t surprise me that this is not a standard remark.

    Regards,

    Matthew

  2. It’s true that image measures are not used often in analysis textbooks at least. Of course they occur everywhere in probability as probability laws of random variables, but since the name is different, students may not make the connection.

  3. You are right, of course. On the other hand, I remember
    when I first realized that probability laws *were* pushforward measures. I had two reactions: the first was that this was a very clean way to think about what was going on; the second was surprise that they never seemed to be described this way.

  4. I’ll have a look in various books to see when pushforward of measures are introduced there, and if the link with probability is highlighted.

    (I vaguely remember that my integration teacher — who was a topologist — defined them quite early on, though he didn’t talk of probability; if my memory is correct, it was at that time that he said something like “certain mathematical objects are meant to be pushed forward, and other to be pulled back”, which was a first introduction to functoriality…)

  5. In your definition of “essential range”, I believe the first inequality sign is pointing in the wrong direction.

Leave a Reply to Kowalski Cancel reply

Your email address will not be published. Required fields are marked *