# A specific projection and compactness on the Bargmann-Fock space

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).

• The limit is zero because the Taylor series of $e^{R^2}$ is smeared uniformly over about $R$ terms around the maximal one and your lacunary series picks up just one of those. Just note that the maximal term corresponds to the index $n=R^2$ (give or take one to get an integer) and that $R$ adjacent terms on both sides have comparable sizes after which you have Gaussian decay like $e^{-(m-n)^2/R}$. Basically, that's the Laplace asymptotic formula in the discrete setting. Jan 7 '12 at 22:09
• Sorry, it should be $e^{-c(m-n)^2/R^2}$. Jan 7 '12 at 22:10