Compactly supported nonzero terminal profiles are outside the heat range

Let $e_1$ be the first standard basis vector and put $g(s):=u(x+s{e}_1)$ for $s\in \mathbb{R}$. At a centre $c\in \mathbb{R}$, the expansion of [F1] about $x+c{e}_1$ converges absolutely in every direction, so substituting the displacement $\sigma e_1$ turns it into a one-variable power series ${\sum }_{k\ge 0}{a}_k{\sigma }^k$ that converges absolutely for every real $\sigma$ and sums to $g(c+\sigma )$; hence $g$ is real analytic on $\mathbb{R}$. For complex-valued $u$ the same argument is applied to the real and imaginary parts of $g$, which are real analytic by [F1].

Assume now that $u$ has compact support. Then $\{u\mathrm{\ne }0\}$ is bounded, so there is $R>0$ with $x+s{e}_1\mathrm{\notin }\{u\mathrm{\ne }0\}$ whenever $\mathrm{\mid }s\mathrm{\mid }>R+\mathrm{\mid }x\mathrm{\mid }$, and for those $s$ the definition of $g$ gives $g(s)=0$.

Suppose first that $u$ is real-valued and let $R$ be as in step 2.1. The zero set of $g$ contains the open interval $(R+\mathrm{\mid }x\mathrm{\mid },\mathrm{\infty })$, so the point $c:=R+\mathrm{\mid }x\mathrm{\mid }+1$ is an accumulation point, lying in the interval $I:=\mathbb{R}$, of the agreement set of $g$ and the zero function; both are real analytic on $I$ by step 1.1, so [F2] gives $g\equiv 0$ on $\mathbb{R}$, and evaluating at $s=0$ gives $u(x)=0$.

Suppose instead that $u$ is complex-valued with compact support. Then $\mathrm{Re}u$ and $\mathrm{Im}u$ are real analytic by [F1] and vanish outside the same bounded set, so step 3.1 applied to each of them gives $\mathrm{Re}u(x)=0$ and $\mathrm{Im}u(x)=0$; hence $u(x)=0$.

Since $x\in {\mathbb{R}}^n$ was arbitrary, steps 3.1 and 4.1 show that a compactly supported representative vanishes identically, so the class $H_{t}f$ is the zero class of $L^p({\mathbb{R}}^n)$; consequently no nonzero compactly supported element of $L^p({\mathbb{R}}^n)$ equals $H_{t}f$ for data and time as in the statement.

Finally assume only that $u$ vanishes almost everywhere outside a compact set $K$. If $x\mathrm{\notin }K$, choose $\rho >0$ with the ball $B(x,\rho )$ disjoint from $K$; then $u=0$ almost everywhere on the nonempty open set $B(x,\rho )$. If $u(x)\mathrm{\ne }0$, continuity of $u$ from [F1] would give $\mathrm{\mid }u\mathrm{\mid }>\mathrm{\mid }u(x)\mathrm{\mid }\mathrm{/}2>0$ on a smaller ball, so that this ball contains no point where $u$ vanishes, contradicting [F3]. Hence $u=0$ on ${\mathbb{R}}^n\setminus K$, so $u$ has compact support and step 5.1 applies.