Let $f : \mathbb{R} \to \mathbb{C}$ be a smooth and bounded function.

If we denote by $\{ H_n(x) \}$ the sequence of normalized Hermite polynomials, then the Hermite expansion of $f$ is defined as

\begin{equation} \sum_{n=0}^\infty d_n H_n(x) \end{equation} where \begin{equation} d_n=\frac{1}{\sqrt{2\pi}n!}\int_{\mathbb{R}}f(x)H_n(x)e^{-x^2/2}dx \end{equation}

According to Convergence of orthogonal polynomial expansions, the Hermite expansion converges uniformly to $f$ if $\lVert Hf \rVert_{L^2} < \infty$.

However, if I just assume that $f$ is smooth and bounded, can I still say something about convergence of its Hermite expansion?

I tried to look for references myself, but I could not find anything relevant.


2 Answers 2


According to the conclusion at the bottom of p. 603 of (say) Uspensky's paper, the Hermite expansion of $f$ converges pointwise to $f$ (under conditions much less restrictive that yours).

If $f$ is (say) bounded and locally Lipschitz, then the Hermite expansion of $f$ converges to $f$ uniformly on compact sets. This follows e.g. from Theorem 9.1.6 in the book *Orthogonal Polynomials" by Gábor Szegő (1939), by taking the integral in formula (9.1.17) there by parts, noting that the sine integral function is bounded, and then letting $\delta\downarrow0$ in (9.1.17).

Much more on this is said on pp. 250--251 of Szegő's book, starting from "From Theorem[...] 9.1.6 there follow the usual theorems on the convergence and the summability of [...] Hermite expansions" on p. 250, where, in particular, the mentioned result by Uspensky is briefly discussed.

  • $\begingroup$ So, in my case where $f$ is bounded and "smooth", the Hermite expansion of $f$ converges "uniformly" to $f$ on any compact interval. Is this right, I guess? $\endgroup$
    – Isaac
    Oct 23, 2023 at 14:17
  • 1
    $\begingroup$ @Isaac : Yes, this is correct. I have added a paragraph on this. $\endgroup$ Oct 23, 2023 at 14:57
  • $\begingroup$ And, if we have "global" Lipschitz continuity, then the convergence is uniform on while real line. Could you check my judgement once more? $\endgroup$
    – Isaac
    Oct 23, 2023 at 17:53
  • 2
    $\begingroup$ @Isaac : You cannot get the uniform convergence on the entire real line to a bounded nonzero function $f$ -- because the partial sums of the Hermite expansion of $f$ are nonzero polynomials and thus unbounded on the real line. If you have any additional questions, please ask them elsewhere. $\endgroup$ Oct 23, 2023 at 18:44
  • $\begingroup$ Thank you. I understand now. Could you perhaps answer the following question as well? mathoverflow.net/questions/457008/… $\endgroup$
    – Isaac
    Oct 23, 2023 at 23:13

Here's a simple and far-from-optimal condition guaranteeing uniform convergence.

Suppose for that there exists $C>0$ such that for any positive integer $n$ $\newcommand{\bR}{{\mathbb{R}}}$

$$ \int_\bR |f^{(n)}(x)| e^{-x^2/2} dx\leq C^n. $$

In this case the associated Hermite series is

$$ \sum_{n\geq 0} \frac{c_n}{n!} H_n(x), $$


$$ c_n=\frac{1}{\sqrt{2\pi}}\int_\bR f^{(n)}(x) e^{-x^2/2} dx. $$

and this converges to $f$ uniformly on compacts. This follows from known asymptotic estimates for Hermite polynomials.

For more precise results, you need to look at Gaussian-Sobolev spaces and the Ornstein-Uhlenbeck operator $H$.


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