I've seen a general proof of this property for continuous functions, but it can be generalized to functions with a jump discontinuity by averaging the jump. To that end, I'm attempting it for a particular case.
Want to show $\lim\limits_{t \to 0} u(x,t)=\tilde\phi(x) \ \forall x \in \mathbb{R}$ where:
- $u(x,t)=\int_{-\infty}^{\infty}\Phi(x-y,t)\phi(y) \ dy$
- $\Phi(x,t)=\frac{1}{\sqrt{4\pi \alpha t}}e^{\frac{-x^2}{4\alpha t}}$ the fundamental solution
- $\phi(x) = \begin{cases} 1, & \text{if $x\ge0$} \\ 0, & \text{if $x<0$} \end{cases}$
- $\tilde\phi(x) = \begin{cases} 1, & \text{if $x>0$} \\ 1/2, & \text{if $x=0$} \\ 0, & \text{if $x<0$} \end{cases}$
fix $x\in\mathbb{R}$
Want to show $\forall\epsilon>0 \ \exists\delta>0$ such that $t<\delta \implies |\int_{-\infty}^{\infty}\Phi(x-y,t)\phi(y) \ dy - \tilde\phi(x)|<\epsilon$
because $\int_{-\infty}^{\infty}\Phi(x-y,t) \ dy = 1$ and $\Phi(x-y,t)>0$ we can write
\begin{align} &= \left| \int_{-\infty}^{\infty}\Phi(x-y,t)\phi(y) \ dy - \tilde\phi(x)\right| \\ &= \left| \int_{-\infty}^{\infty}\Phi(x-y,t)\phi(y) \ dy - \tilde\phi(x)\int_{-\infty}^{\infty}\Phi(x-y,t) \ dy \ \right| \\ &= \left| \int_{-\infty}^{\infty}\Phi(x-y,t)\phi(y) \ dy - \int_{-\infty}^{\infty}\Phi(x-y,t)\tilde\phi(x) \ dy \ \right| \\ & \le \int_{-\infty}^{\infty}\Phi(x-y,t) \left| \phi(y) -\tilde\phi(x) \right| \ dy \end{align}
I'm not sure how to introduce $\delta$ or $\epsilon$.
In the continuous case, the integral was split into 2 cases:
$x$ and $y$ are close: continuity of $\phi$ was used to bound $\left| \phi(y) -\phi(x) \right|$ arbitrarily close to 0.
$x$ and $y$ are far: bound on $\phi$ was used to bound $\left| \phi(y) -\phi(x) \right|$. Then the integral of the fundamental solution, with the $x$-close-to-$y$-piece of the domain missing, went to $0$ as $t\to 0$.
I've noticed for this particular case $\left| \phi(y) -\tilde\phi(x) \right|\le 1$, so I could still make use of 2, but I don't know if 1 can be adapted.
I have a feeling this isn't what you're looking for because it doesn't seem to generalise, but the proof for $C^0$ functions works at all the continuity points, and at $x=0$ you can directly compute-
$$ u(0,t) = \int_{-\infty}^\infty \Phi(0-y)\phi(y)dy = \int_{0}^\infty\frac1{\sqrt{4\pi \alpha t}}e^{-y^2/4\alpha t} dy = \frac1{\sqrt{\pi}} \int_{0}^\infty e^{-z^2}dz =\frac12.$$ In particular, $u(0,t) \to \tilde\phi(0)$. A similar calculation leads to an explicit expression for any $x$ in terms of the error function: $ u(x,t)=\frac{1}{2}\left( 1+\mathrm{erf} \left( \frac{x}{\sqrt{4t}} \right) \right) $.