# prove differential properties of a function $f$

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?