Gaussian data remain Gaussian under the heat flow

Comparing the two formulas, $f(x)=(2\pi {\sigma }^2{)}^{-n\mathrm{/}2}{e}^{-\mathrm{\mid }x{\mathrm{\mid }}^2\mathrm{/}(2{\sigma }^2)}=\mathrm{\Gamma }(x,{\sigma }^2\mathrm{/}2)$ for every $x$, because $(4\pi \cdot {\sigma }^2\mathrm{/}2{)}^{-n\mathrm{/}2}=(2\pi {\sigma }^2{)}^{-n\mathrm{/}2}$ and $4\cdot ({\sigma }^2\mathrm{/}2)=2{\sigma }^2$.

Hence $f\in L^1({\mathbb{R}}^n)$ with $\mathrm{\parallel }f{\mathrm{\parallel }}_1=1$ by unit mass in [F1], and $f\in L^{\mathrm{\infty }}({\mathbb{R}}^n)$ because $f$ is continuous with finite supremum $(2\pi {\sigma }^2{)}^{-n\mathrm{/}2}$ attained at $0$; so $f$ belongs to $L^1\cap L^{\mathrm{\infty }}$.

Since $f$ is bounded, [F2] gives $H_{t}f(x)=\int \mathrm{\Gamma }(x-y,t)f(y)dy$ for every $x$, and step 1.1 turns this into the convolution $({\mathrm{\Gamma }}_t\ast {\mathrm{\Gamma }}_{{\sigma }^2\mathrm{/}2})(x)={\mathrm{\Gamma }}_{t+{\sigma }^2\mathrm{/}2}(x)$ by the semigroup identity of [F1].

Substituting $s=t+{\sigma }^2\mathrm{/}2$ in the explicit formula of [F1] gives ${\mathrm{\Gamma }}_{t+{\sigma }^2\mathrm{/}2}(x)=(4\pi (t+{\sigma }^2\mathrm{/}2){)}^{-n\mathrm{/}2}{e}^{-\mathrm{\mid }x{\mathrm{\mid }}^2\mathrm{/}(4(t+{\sigma }^2\mathrm{/}2))}=(2\pi ({\sigma }^2+2t){)}^{-n\mathrm{/}2}{e}^{-\mathrm{\mid }x{\mathrm{\mid }}^2\mathrm{/}(2({\sigma }^2+2t))}$, which is the density of the centred Gaussian law with covariance $({\sigma }^2+2t)I_n$ by the first and second moments in [F3] with $s=t+{\sigma }^2\mathrm{/}2$.

Steps 1.1, 2.1, 2.2 and 3.1 show $f\in L^1\cap L^{\mathrm{\infty }}$ and identify the heat evolution pointwise with the centred Gaussian density of covariance $({\sigma }^2+2t)I_n$, which is the example.