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.