In a paper dating back to 1990, Eijndhoven and Meyers [1] mention the following "elementary" upper bound for Hermite polynomials on the whole complex plane:
$$ \forall z \in \mathbb{C}, \forall n\in\mathbb{N},\;\; |H_n(z)| \leq 2^{\frac{n}{2}} \,\sqrt{n!}\; e^{\sqrt{2n}|z|}$$
They did not give the proof of this very useful inequality. They only gave a hint, which was to use the well-known, explicit polynomial sum — a nice piece of advice that left me stranded.
Strange as it may seem, I have not found it anywhere in reference books (Bateman, NIST, Gradshtein, Prudnikov and alii etc.).
The inequality plays an important part in the other proofs of the paper. Any idea of how to cleanly derive it?
I have a partial proof by induction that unfortunately only works in the complex plane outside of the disk of radius $\sqrt{\frac{n+1}{2}}$, what I am looking for is a proof valid in the whole complex plane. Also, there exist better bounds on the real and imaginary lines, so partial answers limited to these lines will not do.
(Historical edit)
For low to moderate values of $n$, this "elementary" upper bound is way better than it looks, at least for high real parts and low imaginary parts. The best uniform inequality in the complex plane I know of was given by P. Rusev in a paper of the Bulgarian academy of sciences [2]:
$$|H_n(z)| \leq (2e/\pi)^\frac{1}{4}(\Gamma(2n+1))^\frac{1}{2} (2n+1)^{-n/2-1/4} e^\frac{n}{2} e^{x^2}\cosh((2n+1)^\frac{1}{2}y), \text { where } z=x+iy$$
It is surprising to see that this "non elementary" estimate is in $e^{x^2}$ while the elementary one is only (asymptotically for $y$ fixed) in $e^{|x|}$. Rusev's bound is not much better for high imaginary parts and low real parts either. It may be better for higher values of $n$ and bounded $x$. Apparently the author did not know of Eijndhoven and Meyers' formula given 10 years earlier.
[1] Eijndhoven,S.J.L. and Meyers, J.L.H. "New Orthogonality Relations for the Hermite Polynomials and Related Hilbert Spaces", JOURNAL OF MATHEMATICAL
ANALYSIS AND APPLICATIONS 146, 89-98 (1990), eq. 1.2 p. 90
[2] P. Rusev "An Inequality for Hermite's Polynomials in the Complex Plane", Dokladi na b'lgarckata akademia na naukite/Comptes rendus de l'Académie bulgare des Sciences, 53,10 (2000).
Best Answer
EDIT: solution based on Conrad's hint (see above comment)
Starting from
$$H_{n}(z)=\sum _{m=0}^{\lfloor n/2\rfloor }(-1)^{m}{\dfrac {n!}{m!(n-2m)!}}(2z)^{n-2m}$$
then putting $k=n-2m$:
$$|H_{n}(z)| \leq \sum _{k\in I}{\dfrac {2^k\,n!}{ k!\left(\frac{n-k}{2}\right)!}}|z|^{k}$$ in which $I$ is the set of indices of same parity as $n$ and included in $\{0, ...,n\}$. Expanding now the RHS of the inequality to be proven, it suffices to show that, for all $k\in I$: $$\frac{\sqrt{n!}}{2^\frac{n}{2}} \leq \left(\frac{n-k}{2}\right)!\; \left(\frac{n}{2}\right)^\frac{k}{2} \;\;\;\;\;(1)$$ By Legendre's duplication formula: $$n! = \frac{n}{2}! \; \frac{2^n}{\sqrt{\pi}} \Gamma\left(\frac{n+1}{2}\right)$$ Also, in general for all $n$, $n\geq 2$, $$\left(\frac{n-1}{2}\right)! = \Gamma(\frac{n+1}{2}) < \Gamma(\frac{n}{2} + 1 ) = \left(\frac{n}{2}\right)!$$
as the Gamma function is increasing for corresponding argument values.
It follows that $$\frac{\sqrt{n!}}{2^\frac{n}{2}} \leq \left(\frac{n}{2}\right)!\; \pi^{-\frac{1}{4}} < \left(\frac{n}{2}\right)!$$ so that $(1)$ holds true for $k=0$ and $n\geq 2$ even. For $n=0$, $(1)$ is trivially true for $k=0$.
For $n=1$, one has to consider $k = 1$, and $(1)$ is also true (as an equality) as $\frac{1}{\sqrt{2}} = \left(\frac{1}{2}\right)^\frac{1}{2} $. For $n=3$ and $k=1$, inequality $(1)$ holds by the fact that $\frac{\sqrt{3}}{2 } < \sqrt{\frac{3}{2}}$.
By Legendre's duplication formula again, for all $n$ odd, $n \geq 3$:
$$\begin{align} \frac{\sqrt{n!}}{2^\frac{n}{2}} &< \Gamma\left(\frac{n}{2}+1\right) \pi^{-\frac{1}{4}}\\ &\leq \Gamma\left(\frac{n}{2}+\frac{1}{2}\right)\sqrt{\frac{1+\frac{n}{2}}{\sqrt{\pi}}}\quad \text{(by Gautschi's inequality)}\\ &< \Gamma\left(\frac{n}{2}+\frac{1}{2}\right)\sqrt{\frac{n}{2} } \end{align} $$
as $1 + x < \sqrt{\pi }\,x$ for $x \geq \frac{3}{2}$, which proves $(1)$ for all $n$ odd and $k=1$.
Fixing $n$ arbitrary and supposing that $(1)$ is true for $k\in I, k < n$, we now have for the next same-parity index:
$$\left(\frac{n-k}{2}-1\right)!\; \left(\frac{n}{2}\right)^{\frac{k}{2}+1 } = \frac{n}{n-k}\left(\frac{n-k}{2}\right)!\; \left(\frac{n}{2}\right)^{\frac{k}{2}} $$
which is higher than or equal to the RHS of $(1)$, thereby proving that $(1)$ holds true for all $k\in I$ by induction on $k$.
This finally proves Eijndhoven and Meyers' formula by Conrad's hint.
(edit 2)
A last note
What this proof shows is that Eijndhoven and Meyers' inequality can be refined as follows (see Approximations for the partial sums of exponential series (version: 2017-04-13)):
$$ |H_n(z)| \leq \sqrt{\frac{2^n}{n!}} \; \Gamma(n + 1, \sqrt{2n}|z|) \, e^{\sqrt{2n}|z|}$$
(previous try)
Ah, I may have one solution (but a bit convoluted, I am looking for a simpler one! It clearly is not the original Eijndhoven and Meyers solution).
Here it is, by induction. The inequality is trivial for $n=0$. Suppose it is true for $n$. Starting from:
$$ H_{n+1}'(z)=2(n+1) H_{n}(z)$$
and integrating along a radial path $\{\rho e^{i\theta_0}, \rho \in [0, \rho_0]\}$:
$$|H_{n+1}(z)|\leq \sqrt{2^{n+1}(n+1)!\frac{n+1}{n}}\left(e^{\sqrt{2n}|z|}-1\right)+ |H_{n+1}(0)|$$
For $n$ even, $H_{n+1}(0) = 0$ and it suffices to show that
$$\sqrt{\frac{n+1}{n}}\left( e^{\sqrt{2n}|z|} -1 \right) \leq e^{\sqrt{2(n+1)} |z|}$$
which straightforwardly results from the Maclaurin expansion of the exponential expression. For $n$ odd the convoluted part sets in. One defines for $x>0$:
$$f(x) = \frac{ \sqrt{2\Gamma(x+1)}}{\Gamma(\frac{x+1}{2})} - \sqrt{2^x (x+1)}$$
The variations of $f(x)$ are then analytically studied to show that $f(x) < 0$ for $x$ positive, whence
$$|H_{n+1}(0)| - \sqrt{2^{n+1}(n+1)!\frac{n+1}{n}} \leq \sqrt{2\,n!} f(n) < 0$$
which proves an inequality a little bit sharper than Eijndhoven and Meyers':
$$ \forall z \in \mathbb{C}, \forall n\in \mathbb{N^*},\;\; |H_{n+1}(z)| \leq 2^{\frac{n+1}{2}} \,\sqrt{(n+1)! \frac{n+1}{n}}\; e^{\sqrt{2n}|z|}$$
(edit) ... except for a $0$-centered disk of radius bounded by $\frac{1}{n\left(\sqrt{2(n+1)}-\sqrt{2n}\right)} ( < \frac{1}{2})$. This radius tends to $0$ for $n$ large, so that I only have a partial answer (though useful in its own right) using this method.