Let $G(t, x) := \frac{1}{\sqrt{4 \pi t}} \exp\left( -\frac{x^2}{4 t }\right)$ for all $(t, x) \in (0, T) \times \mathbb{R}$ be the fundamental solution to the heat equation $\partial\_tu = \partial\_{xx}u$.


Evans PDE book (not verbatim but its clear from the context) states that
$$
\lim\_{\delta \downarrow 0} \int\_{\mathbb{R}} G(\delta,y) f(t-\delta, x-y)~\mathrm{d}y = f(x, y),
$$
if $f \in C\_c((0, T) \times \mathbb{R})$. I am wondering if this holds true a.e. if $f \in L^2((0, T); L^2(\mathbb{R})) \cap L^\infty((0, T), L^\infty(\mathbb{R}))$. My first impulse was to approximate $f$ by some $g\_\varepsilon \in C\_c^\infty((0, T) \times \mathbb{R})$ and to use triangle inequality a couple of times. Then, the only term that I am not sure about yet is
$$
\int\_{\mathbb{R}} G(\delta, y) \big(f(t-\delta, x-y) - g\_\varepsilon(t-\delta, x-y) \big)~\mathrm{d}y.
$$
So of course, for fixed $\delta$, this gets small as $\varepsilon \downarrow 0$. But does it get small uniformly in $\delta$? Remember, that afterwards I want to let $\delta \downarrow 0$?


I would be grateful for any input.



Does convolution with heat kernel converge to pointwise evaluation?
What you want cannot be true. Elements of $L^2((0,T); L^2(\mathbb{R}))\cap L^\infty((0,T),L^\infty(\mathbb{R}))$ can be discontinuous everywhere. Without doing any "time integration" you cannot get the "pointwise in time" limit.


For an example: let $k$ be a fixed, non-trivial, $C^\infty\_0(\mathbb{R})$ function. And define
$$ f(t,x) = \begin{cases}
 0 & t\not\in \mathbb{Q} \\
 k(x) & t \in \mathbb{Q}\end{cases}$$


This function is in the function space you specified, being a.e. equal to the 0 function. However, if $x$ is such that $k(x) \neq 0$, then for any $t$ the limit


$$ \lim\_{\delta \searrow 0} \int G(\delta,y) f(t-\delta,x-y) ~\mathrm{d}y $$


does not converge, as it has both $0$ and $k(x)$ as accumulation points.




---


Incidentally, this also shows that it is **not** true that if $g\_\epsilon$ is a smooth function approximating $f$ in the $L^2((0,T); L^2(\mathbb{R}))\cap L^\infty((0,T),L^\infty(\mathbb{R}))$ sense, then for fixed $\delta$ you can always make


$$ \int G(\delta,y) [ f(t-\delta, x-y) - g\_{\epsilon}(t-\delta,x-y) ] ~\mathrm{d}y $$


small just by taking $\epsilon$ sufficiently small. (Here you can take $g \equiv 0$.)


