$u(t,x)$ in $[0,\infty)\times\mathbb{R}$ such that
\begin{cases} \partial_tu-\partial_x^2u=0 \quad in~(x,t)\in(0,\infty)\times\mathbb{R} \\ u(0,x)=\phi(x) \quad for~x\in\mathbb{R} \end{cases} where $\phi:\mathbb{R}\to\mathbb{R}$ is bounded and piecewise continuous function
given this is a solution: $$u(t,x)=\int_{\mathbb{R}}{S(t,x-y)\phi(y)dy}$$ where $$S(t,x)=\frac{e^{-\frac{x^2}{4t}}}{\sqrt{4\pi t}}$$
which satisfy $$\int_{\mathbb{R}}{S(t,x)dx}=1$$
The question is:
prove: $$\lim_{\begin{smallmatrix} (t,x)\to(0,x_0) \\ t>0,~x\in\mathbb{R} \end{smallmatrix}}u(t,x)=\frac{1}{2}(\phi(x_{0+})+\phi(x_{0-}))~~~~~~~~\forall~x_0\in\mathbb{R}$$ where $\phi(x_{0+})=\lim_{x\to x_0^+}\phi(x)$, similarly for $\phi(x_{0-})$
The lecture note provided a similar example for $\phi(x)\in C(\mathbb{R})\cap L^{\infty}(\mathbb{R})$:
https://i.stack.imgur.com/P3FUW.png
https://i.stack.imgur.com/k5k31.png
my attemp:
To mimic the lecture note example, I try to prove
$$\forall \epsilon > 0,~\exists \delta >0:$$
$$x\in(x_0-\frac{\delta}{2},x_0+\frac{\delta}{2})~\implies~|u(t,x)-\frac{1}{2}\{\phi(x_{0+})+\phi(x_{0-})\}|~<~\epsilon$$
To start,
we see $S(t,x)$ is even with respect to x
$$i.e.~~~~S(t,x)=S(t,-x)$$
and $$S(t,x)>0$$
if fix $x$
$$1=\int_{\mathbb{R}}{S(t,\rho)d\rho}=\int_{\mathbb{R}}{S(t,y-x)d(y-x)}=\int_{\mathbb{R}}{S(t,x-y)dy}$$
then,
$$
|u(t,x)-\frac{1}{2}(\phi(x_{0+})+\phi(x_{0-})))| \\
=|\int_{\mathbb{R}}{S(t,x-y)\phi(y)dy}-\frac{1}{2}\{\phi(x_{0+})+\phi(x_{0-})\} \cdot \int_{\mathbb{R}}{S(t,\rho)d\rho}| \\
=|\int_{\mathbb{R}}{S(t,x-y)\cdot \frac{1}{2} \{\phi(y)+\phi(y)-\phi(x_{0+})-\phi(x_{0-})\}dy}| \\
\leq \frac{1}{2}\int_{\mathbb{R}}{|S(t,x-y)|\cdot|\{\phi(y)-\phi(x_{0+})+\phi(y)-\phi(x_{0-})\}|dy} \\
\leq \frac{1}{2} \int_{\mathbb{R}}{S(t,x-y)|\phi(y)-\phi(x_{0+}))|dy + \frac{1}{2} \int_{\mathbb{R}}{S(t,x-y)|\phi(y)-\phi(x_{0-})|dy}}
$$
Let me just consider the first intergal, since second part is just similar
To simplify, Define $g(t,x,y)=S(t,x-y)|\phi(y)-\phi(x_{0+})|$
$$\int_{\mathbb{R}}{g(t,x,y)dy} \\
=\int^{x_0-\delta}_{-\infty}{g(t,x,y)dy} +\int^{x_0}_{x_0-\delta}{g(t,x,y)dy} +\int_{x_0}^{x_0+\delta}{g(t,x,y)dy} +\int^{+\infty}_{x_0+\delta}{g(t,x,y)dy}
$$
As in lecture note, with piecewise continuity, we can prove
$$\int_{x_0}^{x_0+\delta}{g(t,x,y)dy} \leq \epsilon$$
and as $~~t \to 0^+$
$$
\int^{x_0-\delta}_{-\infty}{g(t,x,y)dy} \to 0 \\
\int^{+\infty}_{x_0+\delta}{g(t,x,y)dy} \to 0 \\
$$
However, $\int^{x_0}_{x_0-\delta}{g(t,x,y)dy}$ cannot be bounded by continuity. And I am unable to prove it $\to 0$, as range of y: $[x_0-\delta, x_0]$ overlap with range of x: $(x_0-\frac{\delta}{2},x_0+\frac{\delta}{2})$. If $x=y$, this intergal cannot be bounded, $$\lim_{t \to 0}{S(t,x-y)} \\
= \lim_{t \to 0}{S(t,0)} \\
= \lim_{t \to 0}{\frac{1}{\sqrt{4\pi t}}} \\
= \infty
$$