Is this pre-Hilbert space complete?

Question

We are given a Hilbert space $$ B(\mathbb{C})= \{ f:\mathbb{C}\to \mathbb{C}\mid f\text{ is holomorphic and }\int_{\mathbb{C}}|f(z)|^2e^{-|z|^2}dz < \infty \}$$

The inner product is given by

$$\langle f\mid g\rangle=\frac{1}{\pi}\int_{\mathbb{C}}\overline{f(z)}g(z)e^{-|z|^2}dz$$

Now consider the space

$$ B(\mathbb{C}\setminus \mathbb{R})= \{ f:\mathbb{C}\setminus \mathbb{R}\to \mathbb{C}\mid f\text{ is holomorphic and }\int_{\mathbb{C}}|f(z)|e^{-|z|^2}dz < \infty\}$$

with the same inner product. By the identity theorem, the restriction map $\rho:B(\mathbb{C})\to B(\mathbb{C}\setminus \mathbb{R})$ in injective, and clearly preserves the inner product. Furthermore, the inclusion is proper since $\frac{1}{\sqrt{z}}=e^{-\log(z)/2}\in B(\mathbb{C}\setminus \mathbb{R})\setminus B(\mathbb{C})$, where we choose the principal branch of $\log$. Indeed, $z^n\in B(\mathbb{C}\setminus \mathbb{R})$ for all $n>-1$ and in this case

$$\langle z^n\mid z^n\rangle=\Gamma(n+1)$$

The question is: Is $B(\mathbb{C}\setminus \mathbb{R})$ complete? I suspect that it is not, but haven't been able to show it. I tried the sequence $f_n(z)=z^{-1+1/n}$ since the limit $z^{-1}$ is not in the space. But I'm having trouble showing that the sequence is Cauchy. If it does turn out to be complete, can we generalize this to $B(U)$ for $U\subseteq \mathbb{C}$ open?

Any help would be greatly appreciated!

Answer 1

We can show the following pretty ‘easily’:

Let $\Omega\subset\Bbb{C}^n$ be open, $dz$ be the $2n$-dimensional Lebesgue measure on $\Bbb{C}^n$, and $\rho:\Omega\to (0,\infty)$ a measurable function, and let define the positive measure $d\mu=\rho\,dz$. Suppose that for every compact set $K\subset\Omega$, we have $\operatorname{ess inf}\limits\limits_{z\in K}\rho(z)>0$ (i.e ignoring measure zero effects, the function has positive infimum… this is obviously the case if $\rho$ is continuous). Then, for any compact set $K\subset \Omega$, and any open set $U$ with compact closure and $K\subset U\subset\overline{U}\subset \Omega$, there is a constant $A=A_{K,U}>0$ such that for all holomorphic functions $f$ on $\Omega$, we have \begin{align} \sup_{z\in K}|f(z)|&\leq A_{K,U}\|f\|_{L^2(\Omega,d\mu)}. \end{align} In other words, the restriction map $f\mapsto f|_K$ is continuous from $L^2(\Omega,d\mu)\cap \text{holomorphic functions on $\Omega$}\to C(K)$ (continuous functions on $K$ with sup norm).

Your exponential weight clearly satisfies this assumption (since it is continuous).

As a simple consequence of this inequality, you can show that the space $\mathcal{B}(\Omega,d\mu)$ of holomorphic functions which lie in $L^2(\Omega,d\mu)$ is a closed subspace of $L^2(\Omega,d\mu)$, and hence is complete. Or, you could just directly show completeness of $\mathcal{B}(\Omega,d\mu)$ (start with a Cauchy-sequence, use completeness of $\Bbb{C}$ to define the candidate limit function, then using uniform convergence on compact subsets, show the limit is holomorphic, and then by Cauchyness and Fatou’s lemma, show this limit is in $L^2(d\mu)$).

The proof relies on Cauchy’s integral formula, Cauchy-Schwarz, and a standard ‘averaging trick’ in analysis to control a given quantity by an integral (you can view it as a form of applying the integral mean-value theorem, but there’s no need). Let me just provide the proof when $n=1$ so the notation is cleaner. I’ll use $d\zeta$ to mean contour integrals, and $dz$ for the planar Lebesgue measure.

So, fix $K,U$ as in the theorem, and let $\delta=\frac{1}{2}\text{dist}(K,U^c)$; this is positive since $K$ is compact and contained in an open set $U$. Fix $0<r_1<r_2<\delta$. Then, for any holomorphic function $f$ in $L^2(d\mu)$ (i.e $f\in B(\Omega,d\mu)$) and any $z_0\in K$ and any $r\in [r_1,r_2]$, we have that the closed ball $\overline{B}(z_0,r)$ lies inside $U\subset\Omega$, so by Cauchy’s theorem, \begin{align} f(z_0)&=\frac{1}{2\pi i}\int_{|\zeta-z_0|=r}\frac{f(\zeta)}{\zeta-z_0}\,d\zeta=\frac{1}{2\pi}\int_0^{2\pi}f(z_0+re^{i\theta})\,d\theta, \end{align} so by taking the absolute-value square and Cauchy-Schwarz with the function $f$ and the constant function $1$, we have \begin{align} |f(z_0)|^2&\leq \frac{1}{(2\pi)^2}\left(\int_0^{2\pi}1^2\,d\theta\right)\left(\int_0^{2\pi}|f(z_0+re^{i\theta})|^2\,d\theta\right)=\frac{1}{2\pi}\int_0^{2\pi}|f(z_0+re^{i\theta})|^2\,d\theta. \end{align} Now, integrate from $r_1$ to $r_2$ to get \begin{align} (r_2-r_1)|f(z_0)|^2&=\frac{1}{2\pi}\int_{r_1}^{r_2}\int_0^{2\pi}|f(z_0+re^{i\theta})|^2\,d\theta\,dr\\ &=\frac{1}{2\pi}\int_{r_1\leq |z-z_0|\leq r_2}\frac{|f(z)|^2}{|z-z_0|}\,dz\\ &= \frac{1}{2\pi}\int_{r_1\leq |z-z_0|\leq r_2}\frac{|f(z)|^2}{|z-z_0|\rho(z)}\,d\mu(z)\\ &\leq \frac{1}{2\pi}\cdot\frac{1}{r_1\cdot \left[\operatorname{ess inf}\limits\limits_{z\in \overline{U}}\rho(z)\right]}\int_{r_1\leq |z-z_0|\leq r_2}|f(z)|^2\,d\mu(z)\\ &\leq \frac{1}{2\pi}\cdot\frac{1}{r_1\cdot \left[\operatorname{ess inf}\limits\limits_{z\in \overline{U}}\rho(z)\right]}\|f\|_{L^2(\Omega,d\mu)}^2 \end{align} All these constants ($r_1,r_2$, the essential infimum) depend only on $K$ and $U$, but not on $f,z_0$. So, we can take square root, and supremum over $z_0\in K$ to deduce \begin{align} \sup_{z\in K}|f(z)|&\leq A_{K,U}\|f\|_{L^2(\Omega,d\mu)}. \end{align} The $n$-dimensional case is just more notation: rather than integrating over circles, you integrate over polydiscs, but the idea remains the same: Cauchy’s integral formula, Cauchy-Schwarz, a little averaging radially, and then use the assumption on $\rho$ to change the measure from Lebesgue to $d\mu$.

Also, a simple note: rather than Cauchy-Schwarz, we could have used Holder’s inequality in order to prove that for $p\neq 2$, these spaces of holomorphic functions give Banach spaces.