User joshua isralowitz - MathOverflow

A specific projection and compactness on the Bargmann-Fock space

2012-01-07T20:34:32Z

Let $F_2$ be the Bargmann Fock space defined as the space of entire functions $f$ on $\mathbb{C}$ such that \begin{align*} \int_{\mathbb{C}} |f(z)|^2 e^{- |z|^2} dA(z) \end{align*} ($dA$ is just ordinary area measure, and I'm restricting to the one dimensional case for simplicity.)

The normalized reproducing kernel $k_z$ is the function $k_z(w) = e^{w \overline{z} - \frac{|z|^2}{2}}$.

I'm interested in an example of a bounded operator $S$ on $F_2$ where \begin{align*} \underset{|z| \rightarrow \infty}{\lim} \|S k_z\|_{F_2} = 0 \end{align*} and $S$ is non-compact on $F_2$.

The obvious try is the well known example that works for the Bergman space (see Axler/Zheng "Compact Operators via the Berezin Transform"):

\begin{align*} S (\sum_{k = 0} ^\infty a_k z^k) = \sum_{k = 0} ^\infty a_{2^k} z^{2^k}. \end{align*}

Trivially $S$ is non-compact as a self adjoint projection with infinite dimensional range. Computing the above limit for this $S$ gives the limit \begin{align*} \underset{|z| \rightarrow \infty}{\lim} e^{-|z|^2} \sum_{k = 0}^\infty \frac{|z|^{2^{k + 1}}}{(2^k)!} \end{align*}

I've tried a few things but I can't seem to show this limit is $0$ (and admittedly I'd like to use this example in a talk of mine very soon). Does anyone know if this is true, or if this is in the literature anywhere?

Are there other examples out there ($S$ needs not necessarily be self adjoint or a projection, $S$ just needs to satisfy the norm limit above being $0$ and $S$ is non-compact).

breaking frames when writing long proofs in beamer

2011-10-02T19:55:12Z

So I'm using beamer and I have a long proof that won't fit on the page. It seems allowframebreaks with \break or \framebreak doesn't work. Anyone have an idea how to break a frame with proof environment or theorem environment (the latter for curiosity)?

Invertibility of frame/sampling operator on Bargmann-Fock spaces

2011-05-21T20:22:01Z

Let $F_\alpha ^p (\mathbb{C}^n)$ for $1 < p < \infty$ and $\alpha > 0$ be the Bargmann-Fock space defined as the Banach space of entire functions $f$ such that $f(\cdot) e^{- \frac{\alpha}{2} |\cdot|^2} \in L^p(\mathbb{C}^n, dv) $ where $dv$ is ordinary Lebesgue volume measure and where $F_\alpha ^p (\mathbb{C}^n)$ is given it's natural Banach space norm.

For $\epsilon > 0$ small, treat $\epsilon \mathbb{Z}^{2n}$ canonically as a lattice in $\mathbb{C}^n$. It is well known that $\epsilon \mathbb{Z}^{2n}$ is a "sampling set" for $F_\alpha ^p (\mathbb{C}^n)$, so that there exists constants $C_1, C_2 > 0$ independent of $f \in F_\alpha ^p (\mathbb{C}^n)$ where

\begin{align} C_1 \sum_{\sigma \in \epsilon \mathbb{Z}^{2n} } |f(\sigma)|^p e^{- \frac{\alpha p |\sigma|^2}{2} } \leq \|f\|_ {F_\alpha ^p (\mathbb{C}^n) } ^p \leq C_2 \sum_{\sigma \in \epsilon \mathbb{Z}^{2n} } |f(\sigma)|^p e^{- \frac{\alpha p |\sigma|^2}{2} }. \end{align}

The question is then whether or not the "sampling operator" mapping $F_\alpha ^p (\mathbb{C}^n)$ to $F_\alpha ^p (\mathbb{C}^n)$ given by \begin{align} f \mapsto \sum_{ \sigma \in \epsilon \mathbb{Z}^{2n} } f(\sigma) e^{ \alpha (z \cdot \overline{\sigma} ) - \alpha |\sigma|^2 } \nonumber \end{align} is invertible for small enough $\epsilon$.

Elementary Hilbert space theory says that for $p = 2$ it is, but I don't find where in the literature it is shown for $p \neq 2$, or shown not to be invertible. It's easy to show that this operator is injective and has dense range, but proving surjectivity seems to no easy task.

trace norm inequality for positive matrices

2011-07-26T21:56:22Z

If $A, B$ are positive $n \times n$ complex matrices, $n$ some integer, then obviously \begin{equation*} \|ABA\|_\text{tr} = tr(ABA) = tr(A^2 B). \end{equation*}

But can we say there is a constant $C_n > 0$ depending only on $n$ where $\|ABA\|_\text{tr} \geq C_n \| A^2 B\|_\text{tr}$?

Note that it's easy to get the reverse:

$\|ABA\|_\text{tr} \leq \|A^2B\| _\text{tr} $

so it's the above inequality I really need.

More generally though, I'm guessing $\|AB\|_\text{tr}$

and $\|BA\|_\text{tr}$ are not equivalent (modulo a constant depending only on $n$)

Comment by Joshua Isralowitz

2011-08-18T11:38:57Z

Hey Nelson thanks. Actually I carefully checked Grochenig's "Describing functions: atomic decomposition vs. frames" and the result can be proven pretty easily by using his theory of coorbit spaces (the proof is more or less entirely in that paper, though when specializing to the Fock space, one gets a much shorter proof.) Interestingly, it appeared that the above question answered in the affirmative was a well known "folklore" theorem. I have no idea if something stronger (i.e. precisely for what \epsilon it's true, or if it's true for other lattices) is also part of the folklore though.

Comment by Joshua Isralowitz

2011-08-03T19:57:39Z

Thanks for answering my other question and for giving me that quick easy proof.

Comment by Joshua Isralowitz

2011-08-03T19:57:15Z

Thanks, yea that's a nice example.

Comment by Joshua Isralowitz

2011-07-26T23:35:45Z

Can you roughly describe what kind of matrices you tested out?

Comment by Joshua Isralowitz

2011-07-26T22:49:48Z

Sorry I should say that for that trace inequality I mean $A$ is any square matrix.

Comment by Joshua Isralowitz

2011-07-26T22:47:38Z

Right, but in my case you can also use the well known fact that $|\text{tr} A | \leq \text{tr} |A|$ to get that. It's the reverse of this (modulo a universal constant) that I really need.

Comment by Joshua Isralowitz

2011-07-26T22:10:00Z

Yep, sorry didn't catch that last post in time

Comment by Joshua Isralowitz

2011-07-26T22:09:40Z

For a positive matrix $A$, the singular values ARE the eigenvalues, so isn't $\text{tr} A = \|A\|_{\text{tr}}?$

Comment by Joshua Isralowitz

2011-07-26T22:06:39Z

Why is $\text{tr} (A^2 B)$ necessarily $\|A^2 B\|_{\text{tr}}$ if $A^2 B$ isn't even self adjoint?