The distributional identity says $\sum_{k=0}^\infty \frac{1}{\sqrt{\pi} 2^k k!}e^{-x^2/2-y^2/2}H_k(x)H_k(y)=\delta(x-y)$.
For my understanding, that means for any nice function $g$, \begin{equation*} \lim_{N\to\infty}\Big[\int_{\mathbb{R}} g(y) \Big(\sum_{k=0}^N \frac{1}{\sqrt{\pi} 2^k k!}e^{-x^2/2-y^2/2}H_k(x)H_k(y)\Big) d y\Big]=g(x) \end{equation*}
Does it imply anything pointwise about the infinite sum, in particular, can we say that for any $x\neq y$, $\sum_{k=0}^\infty \frac{1}{\sqrt{\pi} 2^k k!}e^{-x^2/2-y^2/2}H_k(x)H_k(y)=0$? (Or, even if it does not imply this, do we know whether the infinite sum above converges to $0$ for $x\neq y$ or not?)