[proofplan] The [heat equation](/page/Heat%20Equation) is verified by direct differentiation of the Gaussian density. The initial condition is proved by testing against a compactly supported smooth function and changing variables $y=x+\sqrt t\,z$. After this substitution the transition density becomes the standard normal density, and dominated convergence sends $\varphi(x+\sqrt t\,z)$ to $\varphi(x)$. [/proofplan]

[step:Differentiate the Gaussian density in the forward variables] Fix $x\in\mathbb R$. For $t>0$ and $y\in\mathbb R$, write \begin{align*} p(t,x,y)=\frac{1}{\sqrt{2\pi t}}\exp\left(-\frac{(y-x)^2}{2t}\right). \end{align*} This function is smooth in $(t,y)$ on $(0,\infty)\times\mathbb R$ because it is a product and composition of smooth functions with $t>0$. Differentiate with respect to $t$: \begin{align*} \partial_t p(t,x,y) &=p(t,x,y)\left(-\frac{1}{2t}+\frac{(y-x)^2}{2t^2}\right). \end{align*} Differentiate with respect to $y$: \begin{align*} \partial_y p(t,x,y) &=-\frac{y-x}{t}p(t,x,y). \end{align*} Differentiating once more in $y$ and using the product rule gives \begin{align*} \partial_{yy}p(t,x,y) &=-\frac{1}{t}p(t,x,y)-\frac{y-x}{t}\partial_y p(t,x,y)\\ &=-\frac{1}{t}p(t,x,y)+\frac{(y-x)^2}{t^2}p(t,x,y)\\ &=p(t,x,y)\left(-\frac{1}{t}+\frac{(y-x)^2}{t^2}\right). \end{align*} Therefore \begin{align*} \frac12\partial_{yy}p(t,x,y) =p(t,x,y)\left(-\frac{1}{2t}+\frac{(y-x)^2}{2t^2}\right) =\partial_t p(t,x,y). \end{align*} [/step]

[step:Test the density against a compactly supported smooth function] Let $\varphi\in C_c^\infty(\mathbb R)$ be a [test function](/page/Test%20Function). Define the standard normal density \begin{align*} \gamma:\mathbb R&\to(0,\infty)\\ z&\mapsto \frac{1}{\sqrt{2\pi}}\exp\left(-\frac{z^2}{2}\right). \end{align*} Using the substitution $y=x+\sqrt t\,z$, the one-dimensional Lebesgue measure transforms as $d\mathcal L^1(y)=\sqrt t\,d\mathcal L^1(z)$. Hence \begin{align*} \int_{\mathbb R}p(t,x,y)\varphi(y)\,d\mathcal L^1(y) &=\int_{\mathbb R}\frac{1}{\sqrt{2\pi t}}\exp\left(-\frac{(y-x)^2}{2t}\right)\varphi(y)\,d\mathcal L^1(y)\\ &=\int_{\mathbb R}\frac{1}{\sqrt{2\pi t}}\exp\left(-\frac{z^2}{2}\right)\varphi(x+\sqrt t\,z)\sqrt t\,d\mathcal L^1(z)\\ &=\int_{\mathbb R}\gamma(z)\varphi(x+\sqrt t\,z)\,d\mathcal L^1(z). \end{align*} For every fixed $z\in\mathbb R$, continuity of $\varphi$ gives \begin{align*} \varphi(x+\sqrt t\,z)\to\varphi(x)\qquad\text{as }t\to0^+. \end{align*} Also, \begin{align*} |\gamma(z)\varphi(x+\sqrt t\,z)|\le \|\varphi\|_\infty\gamma(z), \end{align*} and $\gamma\in L^1(\mathbb R,\mathcal B(\mathbb R),\mathcal L^1)$ with \begin{align*} \int_{\mathbb R}\gamma(z)\,d\mathcal L^1(z)=1. \end{align*} The [Dominated Convergence Theorem](/theorems/4) therefore yields \begin{align*} \lim_{t\to0^+}\int_{\mathbb R}p(t,x,y)\varphi(y)\,d\mathcal L^1(y) &=\int_{\mathbb R}\gamma(z)\varphi(x)\,d\mathcal L^1(z)\\ &=\varphi(x)\int_{\mathbb R}\gamma(z)\,d\mathcal L^1(z) =\varphi(x). \end{align*} This is precisely convergence to $\delta_x$ in the sense of distributions. [/step]