5
$\begingroup$

Let $\mu$ be standard Gaussian measure on $\mathbb{C}^n$, i.e. $d\mu = \frac{1}{(2 \pi)^n} e^{-|z|^2/2}\,dz$, and fix $0 < p < 1$ (note carefully).

Suppose $g$ is holomorphic on $\mathbb{C}^n$ and satisfies $\int |g|^p\,d\mu < \infty$. Do there exist holomorphic polynomials $g_n$ such that $\int |g-g_n|^p\,d\mu \to 0$?

Stated another way, let $\mathcal{H}$ be the space of holomorphic functions on $\mathbb{C}^n$ and $\mathcal{P}$ be the holomorphic polynomials. Let $\mathcal{H}L^p(\mu) = \{ g \in \mathcal{H} : \int |g|^p\,d\mu < \infty\}$, equipped with the usual metric $d(f,g) = \int |f-g|^p\,d\mu$ (which of course is not a norm). Clearly $\mathcal{P} \subset \mathcal{H} L^p(\mu)$. Is $\mathcal{P}$ dense in $\mathcal{H} L^p(\mu)$?

Even a proof or counterexample for $n=1$ would shed some light.

This is true for $1 \le p < \infty$ and seems to be well known, but the proof I've seen does not go through for $p < 1$, since it needs $t^p$ to be a convex function. It goes like this. First you show that as $\theta \to 0$ we have $g(e^{i\theta} \cdot) \to g$ in $L^p(\mu)$; this part still works for $p < 1$. Now let $F_n$ be the Fejér kernel and set $g_n(z) = \int_{-\pi}^\pi F_n(\theta) g(e^{i \theta} z)\,d\theta$, which can be shown to be a polynomial of degree at most $n-1$. If $p \ge 1$ we can use Jensen's inequality on the probability measure $F(\theta) \,d\theta$ to show $$\int_{\mathbb{C}^n} |g(z)-g_n(z)|^p\,\mu(dz) \le \int_{\mathbb{C}^n} \int_{-\pi}^\pi F_n(\theta) \left|g(z)-g(e^{i\theta} z)\right|^p\,d\theta \,\mu(dz).$$ After interchanging the integrals, properties of the Fejér kernel imply this converges to $\lim_{\theta \to 0} \int_{\mathbb{C}^n} \left|g(z)-g(e^{i\theta} z)\right|^p\,d\mu = 0$ as previously argued. But for $p < 1$, Jensen's inequality goes the wrong way.

One idea is to try to find a dominating function for $|g-g_n|$ and use dominated convergence. For instance, $g_n$ is controlled by $\sup_{\theta} |g(e^{i \theta} z)|$, and maybe we could bound that in terms of $g$ or some related $L^p$ function. But I don't see how to do that.

$\endgroup$
2
$\begingroup$

Yes, they are dense.

The key result comes from a paper of R. Wallstén [1], in which Theorem 3.1 implies that the set $\mathcal{E}$ of functions of the form $$f(z) = \sum_{j=1}^m a_j e^{\langle z, w_j \rangle}, \qquad a_j \in \mathbb{C}, \, w_j \in \mathbb{C}^n$$ is dense in $\mathcal{H} L^p(\mu)$ for any $0 < p < 1$. Clearly $\mathcal{E} \subset \mathcal{H} L^1(\mu)$, and we already know that $\mathcal{P}$ is dense in $\mathcal{H} L^1(\mu)$. So given $g \in \mathcal{H} L^p(\mu)$ and $\epsilon > 0$, we may choose $f \in \mathcal{E}$ and $h \in \mathcal{P}$ with $$\int |g-f|^p\,d\mu < \epsilon, \qquad \int |f-h|\,d\mu < \epsilon.$$ But by Jensen's inequality, $\int |f-h|^p \,d\mu \le \left(\int |f-h|\,d\mu\right)^p$, so by the triangle inequality in $L^p$ we have $$\int |g-h|^p\,d\mu \le \int |g-f|^p\,d\mu + \int |f-h|^p \,d\mu \le \epsilon + \epsilon^p.$$

[1] Wallstén, Robert. The $S^p$-criterion for Hankel forms on the Fock space, $0<p<1$. Math. Scand. 64 (1989), no. 1, 123–132. MR 1036432 Open access via DigiZeitschriften (PDF)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.