Will this sequence of polynomials converge to a Hermite polynomial pointwise? While trying to solve this question my testing lead to an observation that I found interesting in its own right. Consider the linear transformation $L:P\to P$ from the space of polynomial functions $p\in\Bbb{R}[x]$ to itself defined by setting $L(p)=p+p'$, and its iterates
$$
L^n(p)=\sum_{i=0}^\infty\binom{n}{i}p^{(i)}.
$$
Write $G_{m,n}(x):=L^n(x^m)$. These polynomials are monic of degree $m$. If we then make the linear substitution
$$
H_{m,n}(x):=\frac1{n^{m/2}}G_{m,n}(\sqrt n x-n)
$$
we get another sequence of monic polynomials of degree $m$.

It seems to me that we have the limit
$$
\lim_{n\to\infty}H_{m,n}(x)=He_m(x),
$$
where the polynomial $He_m(x)$ is the so called probabilists' Hermite polynomial. Here the convergence can be thought of as either pointwise or in terms of the coefficients of the polynomials.


Can you prove this? Is it known?

The evidence that I have supports this very strongly, for we can calculate that
$$
\begin{aligned}
H_{1,n}(x)&=x,\\
H_{2,n}(x)&=x^2-1,\\
H_{3,n}(x)&=x^3-3x+\frac2{\sqrt{n}},\\
H_{4,n}(x)&=x^4-6x^2+\frac8{\sqrt{n}}x+3-\frac6n,\\
H_{5,n}(x)&=x^5-10x^3+\frac{20}{\sqrt{n}}x^2+\left(15-\frac{30}n\right)x+\frac{24}{n^{3/2}}-\frac{20}{\sqrt n}.
\end{aligned}
$$
Taking the limit as $n\to\infty$ is trivial here, and the results agree with $He_m(x)$.
Furthermore, the operator $L$ commutes with differentiation, so if we assume that the limit
$\tilde{H}_m(x)=\lim_{n\to\infty}H_{m,n}(x)$ exists as a polynomial for all $m$, then the chain rule gives us as a consequence of $Dx^m=mx^{m-1}$ that
$$
\tilde{H}_m'(x)=m\tilde{H}_{m-1}(x).
$$
This is one of properties listed of the probabilist's Hermite polynomials in that Wikipedia-article. If only we could determine the constant term, then this might lead to a proof by induction.

A follow-up question is related to the conjecture I made while trying to answer that other question. The answer by George Lowther shows that for any monic polynomial $p$ of degree $m$ the polynomials $L^n(p)$ have
$m$ distinct real roots for all large enough integers $n$.

If we number these zeros as
$z_{1,n}>z_{2,n}>\cdots>z_{m,n}$, then will the limits
$$
z_i=\lim_{n\to\infty}\frac{z_{i,n}+n}{\sqrt n}
$$
exist, and agree with the zeros of $He_m(x)$.

The evidence that I have for this is not as strong. It does look like the contribution of the leading term of $p$ in $L^n(p)$ will dominate the others for large $n$. However, I am very rusty at estimating the error terms, and don't have that result about the limits yet either :-(
 A: Yes, we do indeed have $H_{m,n}(x)\to He_m(x)$. More generally, for any monic degree $m$ polynomial $p$, setting $p_n=L^np$, then $n^{-m/2}p_n(\sqrt nx-n)\to He_m(x)$ as $n\to\infty$. As the probabilists' Hermite polynomial has distinct real roots, this implies that the zeros of $p_n(\sqrt nx-n)$ tend those of $He_m$.
The idea is to write
$$
p_{n+1}(x)=p_n(x)+p_n^\prime(x) = e^{-x}\frac{d}{dx}e^xp_n(x)
$$
which can then be inverted,
\begin{align}
p_n(x)&=\int_0^\infty e^{-t}p_{n+1}(x-t)dt\\
&=\mathbb{E}[p_{n+1}(x-X)]
\end{align}
where $X$ is a random variable with the exponential distribution (with rate parameter $1$), which has mean and variance equal to $1$. Then, we can iterate this procedure to get
$$
p(x)=\mathbb{E}[p_n(x-X_1-X_2-\cdots-X_n)]
$$
for independent exponentials $X_i$. By the central limit theorem, the random variables
$$
Z_n\equiv\frac{n-X_1-X_2-\cdots-X_n}{\sqrt n}
$$
converge in distribution to the standard normal distribution as $n\to\infty$. Then, setting $f_n(x)=n^{-m/2}p_n(\sqrt nx-n)$,
\begin{align}
\mathbb{E}[f_n(x+Z_n)]&=n^{-m/2}\mathbb{E}[p_n(\sqrt nx-X_1-X_2-\cdots-X_n)]\\
&=n^{-m/2}p(\sqrt nx)\\
&\to x^m
\end{align}
as $n\to\infty$. This suggests the result now. If $Z$ is a standard normal random variable, then the probabilists' Hermite polynomial satisfies the identity $\mathbb{E}[He_m(x+Z)]=x^m$. If we write $f_n(x)=He_m(x)+q_n(x)$ then, as $f_n$ and $He_m$ are both monic degree $m$ polynomials, $q_n$ is of degree less than $m$ and,
\begin{align}
\lim_{n\to\infty}\mathbb{E}[q_n(x+Z_n)]&=\lim_{n\to\infty}\mathbb{E}[f_n(x+Z_n)-He_m(x+Z_n)]\\
&=x^m-\mathbb{E}[He_m(x+Z)]\\
&=0.
\end{align}
Here, I applied the Central Limit Theorem1 and put in the limit $\mathbb{E}[He_m(x+Z_n)]\to\mathbb{E}[He_m(x+Z)]$. If $q_n(x)=\sum_{r=0}^{m-1}c^{(n)}_rx^r$, then the coefficient of $x^r$ in $\mathbb{E}[q_n(x+Z_n)]$ is
$$
c^{(n)}_r + \sum_{k=1}^{m-r-1}\binom{r+k}{k}c^{(n)}_{r+k}\mathbb{E}[Z_n^k],
$$
which has to tend to $0$ as $n\to\infty$. This immediately implies that $c^{(n)}_{m-1}\to0$. As the moment $\mathbb{E}[Z_n^k]$ converge to the moments of a standard normal, they are bounded. So, once we have shown that $\lim_{n\to\infty}c^{(n)}_{r_k}=0$ for each $k > 0$, it follows that $\lim_{n\to\infty}c^{(n)}_r=0$. So, by induction on $r$, $c^{(n)}_r\to0$ as $n\to\infty$, from which the limits $q_n\to0$ and $f_n\to He_m$ follow.

1 Above, I applied the Central Limit Theorem in the form $\mathbb{E}[g(Z_n)]\to\mathbb{E}[g(Z)]$ for polynomials $g$. This is not quite the form in which it is usually stated, where convergence in distribution is used (i.e., $g$ is a continuous bounded function). As the variables $X_n$ have finite moments, the moments of $Z_n$ do indeed converge to the moments of the normal $Z$ or, equivalently, $g$ can be taken to be any polynomial. I'll try to find a reference to the CLT in this form, although it is straightforward to prove (rather easier than the standard CLT).
