We consider the heat kernel $$ g :\mathbb R_{>0} \times \mathbb R^d \to \mathbb R, (t, x) \mapsto \frac{1}{(4\pi t)^{d/2}} \exp \bigg ( - \frac{|x|^2}{4t} \bigg ). $$
I would like to verify that
Theorem If $f:\mathbb R^d \to \mathbb R$ is a continuous function with compact support, then $$ \Delta (g(t, \cdot) *f) = \Delta g(t, \cdot) *f. $$
Could you have a check on my attempt?
Does the theorem hold if $f$ is bounded continuous without compact support?
Thank you so much for your elaboration!
We have $*$ is continuous bilinear and $\Delta = \sum_{k=1}^d \frac{\partial^2}{\partial x^2_k}$. So it suffices to prove that $$ \frac{\partial^2}{\partial x^2_k} (g(t, \cdot) *f) = \frac{\partial^2}{\partial x^2_k} g(t, \cdot) *f. $$
First, we will prove $$ \frac{\partial}{\partial x_k} (g(t, \cdot) *f) = \frac{\partial}{\partial x_k} g(t, \cdot) *f. \tag{1} $$
Fix $x \in \mathbb R^d$. Let $K$ be a bounded open subset of $\mathbb R$ that contains the support of $f$. Then $$ \begin{align*} \frac{\partial}{\partial x_k} (g(t, \cdot) *f) (x) &= \frac{\partial}{\partial x_k} \int_{\mathbb R^d} g(t, x-y) f(y) \, \mathrm d y \\ &= \frac{\partial}{\partial x_k} \int_{K} g(t, x-y) f(y) \, \mathrm d y. \end{align*} $$
Fix $\varepsilon \in (0, 1)$. Let $B(x, \varepsilon)$ be the open ball centered at $x$ with radius $\varepsilon$. By Leibniz integral rule, it suffices to prove $$ \sup \{g(t, z-y) |f(y)| : y \in K, z \in B(x, \varepsilon)\} < \infty. \tag{2} $$
We have $g(t, \cdot)$ is smooth and $f$ is continuous, so $(y, z) \mapsto g(t, z-y) |f(y)|$ is continuous. On the other hand, $\overline{K \times B(x, \varepsilon)}$ is compact, so (2) is indeed true. It remains to prove $$ \frac{\partial}{\partial x_k} \left ( \frac{\partial}{\partial x_k} g(t, \cdot) *f \right ) = \frac{\partial^2}{\partial x^2_k} g(t, \cdot) *f. \tag{3} $$
Notice that $g(t, \cdot)$ is smooth, so $\frac{\partial^2}{\partial x^2_k} g(t, \cdot)$ is smooth. Then (3) follows from the same argument that we used to prove (1). This completes the proof.
The heat kernel is so nice that the theorem holds for bounded measurable $f$.
We have $*$ is continuous bilinear and $\Delta = \sum_{k=1}^d \frac{\partial^2}{\partial x^2_k}$, so it suffices to prove that $$ \frac{\partial^2}{\partial x^2_k} (g(t, \cdot) *f) = \frac{\partial^2}{\partial x^2_k} g(t, \cdot) *f. $$
Notice that $$ \begin{align*} \frac{\partial}{\partial x_k} g(t, x) &= \frac{-x_k}{2t} g(t, x), \\ \frac{\partial^2}{\partial x^2_k} g(t, x) &= \frac{x_k^2-2t}{4t^2} g(t, x). \end{align*} $$
First, we will prove $$ \frac{\partial}{\partial x_k} (g(t, \cdot) *f) = \frac{\partial}{\partial x_k} g(t, \cdot) *f, $$ or equivalently $$ \frac{\partial}{\partial x_k} \int_{\mathbb R^d} g(t, x-y) f(y) \, \mathrm d y = \int_{\mathbb R^d} \frac{\partial}{\partial x_k} g(t, x-y) f(y) \, \mathrm d y. \tag{1} $$
Fix $x \in \mathbb R^d$ and $\varepsilon \in (0, 1)$. Let $B(x, \varepsilon)$ be the open ball centered at $x$ with radius $\varepsilon$. By Leibniz integral rule, it suffices to prove that there is an integrable map $\theta:\mathbb R^d \to \mathbb R$ such that $$ \bigg| f(y)\frac{\partial}{\partial z_k} g(t, z-y) \bigg | \le \theta (y) \quad \forall z \in B(x, \varepsilon), y \in \mathbb R^d. \tag{2} $$
Indeed, $$ \begin{align*} \bigg| f(y)\frac{\partial}{\partial z_k} g(t, z-y) \bigg | &= \frac{|y_k-z_k| }{2t} g(t, y-z) |f(y)|\\ &\le \frac{(|z| + |y|) \|f\|_\infty}{2t} g(t, y-z) \\ &\overset{(\star)}{\le} 2^{d/2}e^{|z|^2/(4t)} \|f\|_\infty \frac{|z| + |y|}{2t} g(2t, y), \end{align*} $$ where $(\star)$ follows from inequality (2.3) of the paper Euler scheme for density dependent stochastic differential equations by Hao/Röckner/Zhang. Let $$ \alpha := 2^{d/2} \|f\|_\infty \sup_{z \in B(x, \varepsilon)} e^{|z|^2/(4t)} \frac{|z|+1}{2t} < \infty. $$
Let $\theta (y) := \alpha (1+|y|) g(2t, y)$ for $y \in \mathbb R^d$. Then $\|\theta\|_{L^1} < \infty$ and (2) is indeed true. It remains to prove $$ \frac{\partial}{\partial x_k} \left ( \frac{\partial}{\partial x_k} g(t, \cdot) *f \right ) = \frac{\partial^2}{\partial x^2_k} g(t, \cdot) *f, $$ or equivalently $$ \begin{align*} \frac{\partial}{\partial x_k} \int_{\mathbb R^d} \frac{\partial}{\partial x_k} g(t, x-y) f (y) \, \mathrm d y = \int_{\mathbb R^d} \frac{\partial^2}{\partial x^2_k} g(t, x-y) f (y) \, \mathrm d y. \end{align*} \tag{3} $$
Indeed, (3) follows from the same argument that we used to prove (1). This completes the proof.