Let $(H_k)_{k=0}^\infty$ denote the normalized Hermite polynomials that fulfill $$ \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty H_k(x) H_j(x) e^{-\frac{x^2}{2}} = \delta_{k,j} . $$ Every function that is square integrable with respekt to the Gaussian measure has an expansion in terms of the Hermite polynomials, i.e. $$ f(x) = \sum_{k=0}^\infty \hat{f}_k H_k(x), $$ where $\hat{f}_k = \frac{1}{\sqrt{2 \pi}} \int_{\mathbb{R}} f(x) H_k(x) e^{-\frac{x^2}{2}}$ .

I am looking for a function for which the Hermite coefficients decay exponentially, i.e. there exists $\omega \in (1,\infty)$ such that $$ \sum_{k=0}^\infty \omega^k \hat{f}_k^2 < \infty. $$ but $$ \sum_{k=0}^\infty (\omega+\varepsilon)^k \hat{f}_k^2 = \infty \quad \text{ for all } \varepsilon > 0 . $$

Is this possible? Thanks very much!

  • 2
    $\begingroup$ Is is enough to take the function defined by the coefficients $\widehat{f}_k = \frac{1}{2^k(k^2+1)}$, for instance. $\endgroup$ – Jack D'Aurizio Jan 9 '17 at 18:10

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