# To which space does the derivative of a function in Fock space belong?

Let $$f : \mathbb C \to \mathbb C$$ be an entire function belonging to the Fock space $$F_\alpha^2$$, that is, $$\int_\mathbb{C} |f(z)|^2 e^{-\alpha|z|^2} \, dA(z)$$ with $$A$$ the Euclidean are measure. To which space does the complex derivative of $$f$$ belong, $$f' \in F_\beta^2$$ $$\beta > 0$$. Can we choose $$\beta = \alpha$$?

• I think the answer to your last question is no. If we could take $\beta=\alpha$ then differentiation would be a linear map $D$ from the Hilbert space $F^2_\alpha$ to itself. Then I think one can use the Closed Graph Theorem (for linear maps between Banach spaces) to show that $D$ is continuous with respect to the HSp norm, hence $D$ would have to be a bounded linear map wit hrespect to the HSp norm, but this is easily contradicted by considering $D(z^n)$ for larger and larger $n$. Mar 26 at 13:03
• Thank you! Could be choose an arbitrary $\beta > \alpha$ instead? Mar 26 at 13:13
• I think so, but I have not done the calculation. You can write down the $F^2_\alpha$ norm of $f$ in terms of its Taylor coefficients (just rewrite $dA(z)$ in polar co-ordinates and use known formulas for moments of the Gaussian) Mar 26 at 13:37
• There is a useful rule of thumb that differentiation "never" acts on an infinite dimensional Banach space (this can be formalised in various ways). One of many reasons why lcs's raise their ugly heads in functional analytic approaches to o.d.e.'s and p.d.e.'s. Mar 27 at 8:59

It is true for all $$\beta>\alpha$$. Use Cauchy estimate $$|f'(z)|\leq \frac{1}{2\pi}\int_{-\pi}^\pi |f(z+e^{it})|dt,$$ then Cauchy Schwarz, $$|f'(z)|^2\leq \frac{1}{2\pi}\int_{-\pi}^\pi|f(z+e^{it})|^2dt,$$ and substitute to your integral. You obtain $$\int_{\mathbf{C}}\int_{-\pi}^\pi e^{-\beta|z|^2}|f(z+e^{it})|^2dt\, dA_z.$$ Exchange the order of integration and make the change of the variable $$w=z+e^{it}$$ in the inner integral. Then use the fact that $$\exp(-\beta|w+e^{it}|^2)\leq \exp(-\alpha|w|^2)$$ when $$|w|$$ is sufficiently large.