I have posted ths on the Quant Finance page as it is part of a QF problem but realised I may get a swifter response here! Iam working on a problem where I have successfully reduced a version of Black Scholes to the Heat Equation and then shown the solution to be:
$$u(x,t)=\frac{1}{2\sqrt{t\pi}}\int_{-\infty}^\infty{f(\xi)e^{-\frac{(x-\xi)^2}{4t}}}d\xi$$
the integral is from -infinity to infinity
I now need to show that if $f(x)$ is continuous then $$\lim_{t\rightarrow 0+}u(x,t)=f(x)$$
I am not looking for the solution but guidance in where to start as I need to be able to complete this off my own back.
This is not a full answer, but just some thoughts that can be useful and are too much for a comment.
The LHS can be recognized as $\mathbb{E}f\left(X\right)$ where $X\sim Norm\left(x,t\sqrt{2}\right)$.
If $t\rightarrow0+$ then it tends to $X\sim Norm\left(x,0\right)$ wich means that $X$ tends to constant $x$ and $\mathbb{E}f\left(X\right)$ tends to $f\left(x\right)$ if $f$ is continuous.
I think you are dealing with weak convergence of distributions.
Maybe it is enough allready if $f$ is continuous at $0$.