How to prove this limit Let $g:[0,\infty)\to\mathbb{R}$ with $g(0)=0$. For $x>0$ and $t>0$, define:
\begin{equation}
u(x,t) = \frac{x}{\sqrt{4\pi}} \int_0^t \frac{1}{(t-s)^{3/2}} e^{-\frac{x^2}{4(t-s)}} g(s) \,\mathrm{d}s.
\end{equation}
Show that $\lim\limits_{x\to0^+} u(x,t)=g(t)$.
I tried to split the integration from $0$ to $t-\delta$ and from $t-\delta$ to $t$ for some $\delta>0$. I think the limit of the integral from $0$ to $t-\delta$ vanishes, but what should I do for the second integral?
Thank you.
PS: This formula arises from the heat equation.

Let $f(t) = {1 \over \sqrt{\pi}}{ 1\over t \sqrt{t}} e^{-{1 \over t}} 1_{(0,\infty)}(t)$. Then $f(t) \ge 0$ for all $t$ and $f \in L^1(\mathbb{R})$. Using the substitution $t = {1 \over s^2}$ we can see that $\int f  = 1$.
We also see that $f$ is bounded.
Let $f_\epsilon(t) = {1 \over \epsilon} f({t \over \epsilon})$. Then $\{f_\epsilon\}_{\epsilon>0}$ form an approximation to the identity and we have
$\lim_{\epsilon \downarrow 0} (f_\epsilon * g) (t) = g(t)$ at every $t>0$ at
which $g$ is continuous.
Letting $\epsilon = {x^2 \over 4}$ and making the necessary substitutions yields
the desired result.
