Does convolution with heat kernel converge to pointwise evaluation?

Question

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.

Answer 1

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.

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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 $g \equiv 0$.)