The Hermite polynomials are given by:
$H_n(x)=(-1)^n e^{x^2} \dfrac{d^n}{dx^n}e^{-x^2}$
There is the proof that all the roots are real: https://math.stackexchange.com/a/104875/504137.
And I know the fact that they all are bounded i.e. all the roots lie in $(-\sqrt{4n+1}, \sqrt{4n+1})$.
But how to prove that?
Best Answer
A weaker bound allows a completely elementary proof.
By Rodrigues formula we know that all the zeroes of $H_n$ are simple and real: let us denote them as $\zeta_1,\ldots,\zeta_n$. By Vieta's formulas the quantity $\sum_{k=1}^{n}\zeta_k^2$ only depends on the coefficients of $H_n$, and by the recurrence relations for Hermite polynomials we get $$ \sum_{k=1}^{n}\zeta_k^2 =\frac{n(n-1)}{2},\qquad \sum_{k=1}^{n}\zeta_k^4 =\frac{n(n-1)(2n-3)}{4}, $$ $$ \sum_{k=1}^{n}\zeta_k^6 =\frac{n(n-1)(5n^2-17n+15)}{8},$$ $$\frac{7}{8}n(n-2)^4\leq\sum_{k=1}^{n}\zeta_k^8=\frac{n(n-1)(14n^3-79n^2+155n-105)}{16}\leq n\max_{k}|\zeta_k|^8 $$
$$\frac{7}{8}(n-1)^5\geq\sum_{k=1}^{n}\zeta_k^8=\frac{n(n-1)(14n^3-79n^2+155n-105)}{16}\geq 2\max_{k}|\zeta_k|^8 $$ so for any $n>2$ we have
$$\boxed{\frac{101}{112}(n-1)^{5/8}\geq \max_k |\zeta_k| \geq \frac{29}{30}\sqrt{n-2}.}$$
In order to prove the given upper bound one may use the integral representation for $H_n$ that comes from the residue theorem applied to the generating function: $$ H_n(x)=\frac{n!}{2\pi i}\oint_{\|z\|=\rho}\frac{e^{2xz-z^2}}{z^{n+1}}\,dz $$ Given some $x$ with a large absolute value ($\geq\sqrt{4n+1}$), it is enough to choose a suitable $\rho$ (close to $|x|$) and approximate the contour integral with decent accuracy, in order to show that its real part cannot be zero.
On the other hand, by quoting the answer of J.M. to the other question,
then applying the Gershgorin circle theorem, we immediately get $$ \max_k |\zeta_k| \leq \sqrt{2n-2} $$ for any $n>2$.