Proof. We show that for , the polynomial has exactly d simple real roots — one near each root of . The proof proceeds in four steps.
Step 1 (Setup). Let
be the zeros of
, with minimum gap
(Lemma 2). From the Hermite expansion (
2):
since
. By the effective decay (Theorem 2),
where
. Write
with
.
To locate the zeros of
near
, apply the implicit function theorem: if the derivative
is bounded below and the residual (
3) is small relative to
, then
has exactly one simple zero within
of
.
Step 2 (Derivative bound). The derivative at
satisfies
Since
, the value
. The Plancherel–Rotach estimate [
8] gives
for bulk roots, so
for
small.
Step 3 (Bulk regime: ). In the bulk, the Plancherel–Rotach approximation [
8] gives for each
:
where
depends on
d and
k but is uniform over bulk roots. Summing the perturbation series (
3):
using
and the geometric decay of the sum in
. By the mean value theorem applied to
on
, the displacement
of the zero of
from
satisfies
This is less than whenever . Since , the condition is satisfied for . The exponent arises from the saddle-point relation , which is satisfied for with growing only as ; the binding constraint comes from Step 4.
Step 4 (Edge regime: ). Near the spectral edge, the Plancherel–Rotach approximation breaks down and Airy-function asymptotics apply [
8]. Write
with
close to 0 or
. The Airy estimate gives, for
:
where
is an absolute constant. The estimate (
5) no longer decays in
k as in the bulk, so all
terms in the sum contribute equally. The residual at
is bounded by:
Since
, the sum is exponentially large; however, the effective staircase coefficient bounds
(Theorem 2), so the sum converges geometrically and the exponential is replaced by a polynomial in
d:
for some absolute constant
. The key bound on
A is:
Lemma 4 (Staircase polynomial degree bound). uniformly in d and in .
Proof. Split the sum at . For : there are terms each with , but ; for the factor, so , giving total contribution . For : , which decays faster than any polynomial; the binomial coefficients are at most , but the exponential decay dominates. The leading contribution comes from k near , where for fixed , giving . The factor from the Airy correction at the edge contributes an extra , giving total exponent comfortably. Thus . □
The displacement bound becomes . This is less than whenever . Setting and solving: with . By Lemma 4, , so , and is independent of d; the threshold is genuinely .
Final step: stability of real zeros. We make the above approximation quantitative. Write
where
is the rescaled polynomial introduced above, and
is the error term.
Using the bulk and edge bounds we have the uniform estimate
for some absolute constant
.
On the other hand, it is classical that the Hermite polynomial
has
d simple real zeros
, all contained in
, and that there exists
such that
Combining (
6) and (
7), we see that if
for a sufficiently large constant
, then
It follows by a standard root stability argument (see, e.g., [
9, Ch. 6]) that each zero
of
perturbs to a unique real zero of
, and no nonreal zeros can appear. Therefore
, and hence
, has
d real zeros.
This proves hyperbolicity for all , completing the proof.
Step 5 (Conclusion). Taking , for every :
each root of has a corresponding simple root of within ;
these d roots are mutually separated (their neighbourhoods of radius are disjoint, since is the minimum root gap of );
has degree d, so these account for all its roots.
Hence — and therefore — is hyperbolic. □