User joshua isralowitz - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T13:30:39Z http://mathoverflow.net/feeds/user/15280 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/85150/a-specific-projection-and-compactness-on-the-bargmann-fock-space A specific projection and compactness on the Bargmann-Fock space Joshua Isralowitz 2012-01-07T20:34:32Z 2012-01-07T20:34:32Z <p>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.)</p> <p>The normalized reproducing kernel $k_z$ is the function $k_z(w) = e^{w \overline{z} - \frac{|z|^2}{2}}$. </p> <p>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 $F_2$. </p> <p>The obvious try is the well known example that works for the Bergman space (see Axler/Zheng "Compact Operators via the Berezin Transform"):</p> <p>\begin{align*} S (\sum_{k = 0} ^\infty a_k z^k) = \sum_{k = 0} ^\infty a_{2^k} z^{2^k}. \end{align*} </p> <p>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*} </p> <p>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?</p> <p>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). </p> http://mathoverflow.net/questions/76998/breaking-frames-when-writing-long-proofs-in-beamer breaking frames when writing long proofs in beamer Joshua Isralowitz 2011-10-02T19:55:12Z 2011-10-02T19:55:12Z <p>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)?</p> http://mathoverflow.net/questions/65674/invertibility-of-frame-sampling-operator-on-bargmann-fock-spaces Invertibility of frame/sampling operator on Bargmann-Fock spaces Joshua Isralowitz 2011-05-21T20:22:01Z 2011-08-18T11:35:18Z <p>Let $F_\alpha ^p (\mathbb{C}^n)$ for $1 &lt; p &lt; \infty$ 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. </p> <p>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 </p> <p>\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} </p> <p>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 $\epsilon$. </p> <p>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. </p> http://mathoverflow.net/questions/71362/trace-norm-inequality-for-positive-matrices trace norm inequality for positive matrices Joshua Isralowitz 2011-07-26T21:56:22Z 2011-07-27T08:54:45Z <p>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*} </p> <p>But can we say there is a constant $C_n > 0$ depending only on $n$ where $\|ABA\|_\text{tr} \geq C_n \| A^2 B\|_\text{tr}$? </p> <p>Note that it's easy to get the reverse: </p> <p>$\|ABA\|_\text{tr} \leq \|A^2B\| _\text{tr}$ </p> <p>so it's the above inequality I really need. </p> <p>More generally though, I'm guessing $\|AB\|_\text{tr}$ </p> <p>and $\|BA\|_\text{tr}$ are not equivalent (modulo a constant depending only on $n$) </p> http://mathoverflow.net/questions/65674/invertibility-of-frame-sampling-operator-on-bargmann-fock-spaces/73117#73117 Comment by Joshua Isralowitz Joshua Isralowitz 2011-08-18T11:38:57Z 2011-08-18T11:38:57Z Hey Nelson thanks. Actually I carefully checked Grochenig's &quot;Describing functions: atomic decomposition vs. frames&quot; 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 &quot;folklore&quot; 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. http://mathoverflow.net/questions/71362/trace-norm-inequality-for-positive-matrices/71388#71388 Comment by Joshua Isralowitz Joshua Isralowitz 2011-08-03T19:57:39Z 2011-08-03T19:57:39Z Thanks for answering my other question and for giving me that quick easy proof. http://mathoverflow.net/questions/71362/trace-norm-inequality-for-positive-matrices/71370#71370 Comment by Joshua Isralowitz Joshua Isralowitz 2011-08-03T19:57:15Z 2011-08-03T19:57:15Z Thanks, yea that's a nice example. http://mathoverflow.net/questions/71362/trace-norm-inequality-for-positive-matrices Comment by Joshua Isralowitz Joshua Isralowitz 2011-07-26T23:35:45Z 2011-07-26T23:35:45Z Can you roughly describe what kind of matrices you tested out? http://mathoverflow.net/questions/71362/trace-norm-inequality-for-positive-matrices/71363#71363 Comment by Joshua Isralowitz Joshua Isralowitz 2011-07-26T22:49:48Z 2011-07-26T22:49:48Z Sorry I should say that for that trace inequality I mean $A$ is any square matrix. http://mathoverflow.net/questions/71362/trace-norm-inequality-for-positive-matrices/71363#71363 Comment by Joshua Isralowitz Joshua Isralowitz 2011-07-26T22:47:38Z 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. http://mathoverflow.net/questions/71362/trace-norm-inequality-for-positive-matrices Comment by Joshua Isralowitz Joshua Isralowitz 2011-07-26T22:10:00Z 2011-07-26T22:10:00Z Yep, sorry didn't catch that last post in time http://mathoverflow.net/questions/71362/trace-norm-inequality-for-positive-matrices Comment by Joshua Isralowitz Joshua Isralowitz 2011-07-26T22:09:40Z 2011-07-26T22:09:40Z For a positive matrix $A$, the singular values ARE the eigenvalues, so isn't $\text{tr} A = \|A\|_{\text{tr}}?$ http://mathoverflow.net/questions/71362/trace-norm-inequality-for-positive-matrices Comment by Joshua Isralowitz Joshua Isralowitz 2011-07-26T22:06:39Z 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?