Let the pdf of a multivariate normal distribution be
\begin{equation}
 p\_{Z}(\mathbf{z})=\frac{1}{\left(2\pi \sigma^2 \right)^{k/2}}\exp(-{\mathbf{z}}^{\text{T}}\mathbf{z}/2\sigma^2).
\end{equation}
Consider a continuous function $f(\mathbf{s})$ supported on a compact set $\mathcal{S}$. Then let
\begin{equation}
 I\_{\sigma}(\mathbf{s})=\int\_{\mathbb{R}^{k}} p\_{Z}(\mathbf{z}-\mathbf{s}) f(\mathbf{z}) d\Omega, \mathbf{s}\in \mathcal{S}.
\end{equation}
My question is whether $I\_{\sigma}(\mathbf{s})$ converges to $f(\mathbf{s})$ uniformly on $\mathcal{S}$ as $\sigma \to 0^{+}$. Can the condition of the compactness of $\mathcal{S}$ be relaxed? In addition, tight nonasymptotic lower and upper bounds on $I\_{\sigma}(\mathbf{s})$ are more welcome.



Asymptotic moment of a multivariate normal distribution
$\newcommand\si\sigma\newcommand\om\omega\newcommand\z{\mathbf z}\newcommand\R{\mathbb R}$Let us assume that your $d\Omega$ means $d\z$, so that
$$I\_\si(s):=\int\_{\R^k} p\_{Z}(z-s) f(z)\, dz,$$
and we shall assume that this definition holds for all $s\in\R^k$.


So,
$$I\_\si(s)=Ef(s+\si Z),$$
where $Z$ is a standard normal random vector in $\R^k$.


Let $\si\downarrow0$. Then for each $s\in\R^k$ we have $s+\si Z\to s$ in distribution, and [hence](https://en.wikipedia.org/wiki/Convergence_of_measures#Weak_convergence_of_measures) $I\_\si(s)=Ef(s+\si Z)\to f(s)$ for any bounded continuous function $f$.


Suppose now that for some bounded function $\om\colon[0,\infty)\to[0,\infty)$ we have $\om(0+)=0$ and $|f(x)-f(y)|\le\om(h)$ for any $x$ and $y$ in $\R^k$ such that $|x-y|\le h$; so, $\om$ is a modulus of continuity of $f$; here $|\cdot|$ denotes the Euclidean norm on $\R^k$ and on $\R$. (In particular, $f$ has such a modulus of continuity if $f$ is a continuous function with a compact support, or if $f$ is bounded and [Hölder continuous](https://en.wikipedia.org/wiki/H%C3%B6lder_condition).) Then for all $s\in\R^k$
$$|I\_\si(s)-f(s)|=|Ef(s+\si Z)-f(s)|\le E\om(\si Z)\to\om(0+)=0,$$
by dominated convergence, since $\si Z\to0$ in probability. It follows that, whenever a bounded modulus of continuity exists, $I\_\si(s)\to f(s)$ uniformly in all $s\in\R^k$.


Suppose now that $f$ is twice differentiable, with the second derivative bounded from above in the sense that for some real $b$ and all $s$ and $z$ in $\R^k$ we have $f''(s)(z,z)\le b|z|^2$. Then
$$f(s+\si Z)-f(s)\le\si f'(s)(Z)+ b\si^2|Z|^2/2.$$
Taking now the expectations, we get
$$I\_\si(s)-f(s)=Ef(s+\si Z)-f(s)\le bk\si^2/2.$$
Similarly, if $f$ is twice differentiable, with the second derivative bounded from below in the sense that for some real $a$ and all $s$ and $z$ in $\R^k$ we have $f''(s)(z,z)\ge a|z|^2$, then
$$I\_\si(s)-f(s)\ge ak\si^2/2.$$


