Heat balls and their time slices

Assume Countable Choice. Let $n\ge 1$, and let $\mathrm{\Gamma }$ be the heat kernel of The heat kernel on ${\mathbb{R}}^n$ and its causal extension, with $\mathrm{\mid }x-y{\mathrm{\mid }}^2=⟨x-y,x-y⟩$ the Euclidean square of The Euclidean inner product $⟨x,y⟩={\sum }_{k<n}{x}_k{y}_k$ on ${\mathbb{R}}^n$. For a space-time point $P=(x,t)\in {\mathbb{R}}^n\times \mathbb{R}$ and $r>0$ the heat ball is $E_r(t,x)=E_r(P):=\{(y,s)\in {\mathbb{R}}^{n+1}:s<t,\mathrm{\Gamma }(x-y,t-s)\ge r^{-n}\}\cup \{(x,t)\}\mathrm{.}$ Write $\tau :=t-s$ for the time depth below the top point. Record the following facts.

(i) $E_r(t,x)$ is compact, and $(x,t)$ is its unique point of maximal time;

(ii) for $0<\tau <r^2\mathrm{/}(4\pi )$ the time slice is the closed ball $\{y:(y,t-\tau )\in E_r(t,x)\}=\bar{B}(x,{\rho }_n(\tau )),{\rho }_n(\tau ):=\sqrt{2n\tau \log \frac{r^2}{4\pi \tau }},$ the slice at $\tau =r^2\mathrm{/}(4\pi )$ is the single point $\{x\}$, and the slice is empty for $\tau >r^2\mathrm{/}(4\pi )$; the function ${\rho }_n$ is continuous on $(0,r^2\mathrm{/}4\pi )$ and ${\rho }_n(\tau )\to 0$ as $\tau \downarrow 0$;

(iii) ${\displaystyle \underset{0<\tau <r^2\mathrm{/}4\pi }{\sup }{\rho }_n(\tau )=\frac{r}{2}\sqrt{\frac{2n}{\pi e}}}$, attained at $\tau =r^2\mathrm{/}(4\pi e)$, so $E_r(t,x)\subseteq \bar{B}(x,\frac{r}{2}\sqrt{2n\mathrm{/}\pi e})\times [t-r^2\mathrm{/}4\pi ,t]$;

(iv) the boundary is $\mathrm{\partial }{E}_r(t,x)=\{(y,s):s<t,\mathrm{\Gamma }(x-y,t-s)=r^{-n}\}\cup \{(x,t)\}$, and on the level set $\mathrm{\Gamma }=r^{-n}$ the gradient of $\mathrm{\Gamma }$ is nowhere zero, so that level set is a $C^{\mathrm{\infty }}$ hypersurface.

• The slice formula is algebra. Taking $n$-th roots in $\mathrm{\Gamma }(x-y,\tau )=(4\pi \tau {)}^{-n\mathrm{/}2}{e}^{-\mathrm{\mid }x-y{\mathrm{\mid }}^2\mathrm{/}(4\tau )}\ge r^{-n}$ (The heat kernel on ${\mathbb{R}}^n$ and its causal extension) gives $\mathrm{\mid }x-y{\mathrm{\mid }}^2\le 2n\tau \log \frac{r^2}{4\pi \tau }$, whose right side is ${\rho }_n(\tau {)}^2$; the logarithmic factor is positive exactly for $0<\tau <r^2\mathrm{/}(4\pi )$, vanishes at $\tau =r^2\mathrm{/}(4\pi )$ (forcing $y=x$, so the slice there is the single point $x$), and is negative for $\tau >r^2\mathrm{/}(4\pi )$, where no point of the level set or of its superlevel set remains. This is the only reading of the inequality $\mathrm{\Gamma }\ge r^{-n}$ at the endpoint $\tau =r^2\mathrm{/}(4\pi )$; the enclosed region and its width are unaffected.

• Compactness (i) and the enclosure (iii). First, $E_r(t,x)$ is closed in ${\mathbb{R}}^{n+1}$: if points $(y_k,s_k)$ with $s_k<t$ and $\mathrm{\Gamma }(x-y_k,t-s_k)\ge r^{-n}$ converge to $(y_0,s_0)$ with $s_0<t$, then continuity of $\mathrm{\Gamma }$ gives $\mathrm{\Gamma }(x-y_0,t-s_0)\ge r^{-n}$; if $s_0=t$, then $t-s_k\to 0$ and the slice bound $\mathrm{\mid }x-y_k{\mathrm{\mid }}^2\le 2n(t-s_k)\log \frac{r^2}{4\pi (t-s_k)}\to 0$ forces $y_0=x$, so the limit is the top point (Interior, closure, boundary, limit point, isolated point and dense subset of a metric space). The enclosure in (iii) bounds $E_r(t,x)$; closed and bounded subsets of ${\mathbb{R}}^{n+1}$ are compact (Heine-Borel in ${\mathbb{R}}^n$: with the Euclidean metric a subset of ${\mathbb{R}}^n$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line, Open cover, subcover, compact metric space, and compact subset of a metric space). The maximisation of $\tau \log \frac{r^2}{4\pi \tau }$, whose substitution $u=4\pi \tau \mathrm{/}{r}^2$ reduces it to $\frac{r^2}{4\pi }u\log \frac{1}{u}$ with maximum $\frac{r^2}{4\pi e}$ at $u=1\mathrm{/}e$, follows by differentiating $u\log (1\mathrm{/}u)$: its derivative is $\log (1\mathrm{/}u)-1$, positive for $u<1\mathrm{/}e$ and negative for $u>1\mathrm{/}e$ (The natural logarithm has derivative 1/x and equals the integral from 1 to x of 1/t, Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm). Also $\tau \log (r^2\mathrm{/}(4\pi \tau ))\to 0$ as $\tau \downarrow 0$: writing $\tau =(r^2\mathrm{/}(4\pi ))e^{-v}$ reduces this to a constant times $v{e}^{-v}\to 0$ (The exponential dominates every fixed nonnegative integer power at $+\mathrm{\infty }$). The top point is the unique point of $E_r(t,x)$ with $s=t$, since the defining inequality requires $s<t$.

• The boundary (iv). On the level set $\mathrm{\Gamma }=r^{-n}$ one has $\mathrm{\Gamma }>0$. If the spatial gradient vanishes, ${\mathrm{\nabla }}_y\mathrm{\Gamma }(x-y,\tau )=0$, then $y=x$ by the explicit formula; at such a point the time derivative ${\mathrm{\partial }}_s\mathrm{\Gamma }(x-y,t-s)=-{\mathrm{\partial }}_{\tau }\mathrm{\Gamma }(0,\tau )$ is $-(-\frac{n}{2\tau })\mathrm{\Gamma }(0,\tau )\mathrm{\ne }0$ (Normalisation, parabolic scaling, heat equation and derivative bounds for the heat kernel). Hence the full gradient in $(s,y)$ never vanishes on the level set, which is therefore a $C^{\mathrm{\infty }}$ hypersurface of the open half-space $s<t$ by the implicit function theorem with higher regularity (The parametrized implicit function theorem with $C^k$ regularity), applied with a nonzero gradient component as the dependent coordinate. Nonzero gradient also gives points on both sides of each level point, so the displayed level is the boundary below time $t$. The adjoined top point is a boundary point, since $(x,t-\tau )\in E_r$ for small $\tau >0$ while no point at a time greater than $t$ belongs to $E_r$; it is not isolated.

• Choice. Countable Choice is the declared ambient hypothesis and is carried by the topology and calculus suppliers; the definition and the displayed computations select no sets.