For any $f(x,t) : \mathbb{R}^2\to \mathbb{R}$, let $f_t := \frac{\partial f}{\partial t}$ and $f_{xx} := \frac{\partial^2 f}{\partial x^2}$ when those partial derivatives exist. Suppose $g$ is continuous on $\mathbb{R}$ and $g(x) = 0$ for any $x$ not in an interval $[a,b]$. Let $f(x,t) = \int_a^b \phi(x-y, t) g(y)dy$ for $t > 0$, where $\phi(x,t) = \frac{1}{\sqrt{4\pi t}}e^{-x^2/(4t)}$. Prove that $f_t (x,t) = f_{xx}(x,t)$ and that $f(x,0) = g(x)$ for all $x\in \mathbb{R}$, where $f(x,0) := \lim\limits_{t\to 0^+} \int_a^b \phi(x-y, t) g(y)dy$.

  For the initial proof, I think Leibniz's integration rule is useful. I know that for all $\epsilon > 0, \lim\limits_{t\to0}\int_{-\delta}^\delta \phi(x,t) dx = 1$, because the substitution $y = \frac{x}{\sqrt{2t}}$ shows that $\int_{-\delta}^{\delta}\phi(x,t) dx = \int_{-\delta/\sqrt{2t}}^{\delta/\sqrt{2t}} \phi(x,y)dy,$ which approaches $\int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}} e^{-x^2/2}dy = 1$ as $t\to 0$ (from above).  The Leibniz integral rule ensures that $f_t(x,t) = \int_a^b \phi_t(x-y, t) g(y)dy$ and using the fact that $\phi_t(x-y, t) = \phi_{xx}(x-y, t)$, one sees by applying the Leibniz integral rule twice that $f_{xx}(x,t) = f_t(x,t)$.  

But how can one show that $f(x,0) = g(x)$ for all $x\in \mathbb{R}$?

If one makes the nontrivial assumption that we can take a limit inside the integral, then I think the following might just work:

$\begin{align}f(x,0 ) &= \lim\limits_{t\to 0^+} \int_a^b \frac{1}{\sqrt{4\pi t}} e^{-(x-y)^2}{4t} g(y)dy\\ &= \lim\limits_{t\to 0^+} \int_{(a-x)/\sqrt{2t}}^{(b-x)/\sqrt{2t}} \frac{1}{\sqrt{2\pi}} e^{-z^2/2} f(\sqrt{2t} z + x) dz,\quad\text{ $z=(y-x)/\sqrt{2t})$}\\ &=\lim\limits_{t\to 0^+} \int_{(a-x)/\sqrt{2t}}^{(b-x)/\sqrt{2t}} \frac{1}{\sqrt{2\pi}} e^{-z^2/2} \lim\limits_{t\to 0^+} f(\sqrt{2t} z + x)dz\\ &= f(x) \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}}e^{-z^2}dz = f(x)\end{align}$

How can one justify the third last equality above?



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