The heat evolution $H_t$ of initial data

Assume Countable Choice, let $n\ge 1$ and $1\le p\le \mathrm{\infty }$, and let $\mathrm{\Gamma }$ be the heat kernel of The heat kernel on ${\mathbb{R}}^n$ and its causal extension, with ${\mathrm{\Gamma }}_t:=\mathrm{\Gamma }(\cdot ,t)$ the $L^1$ function of unit norm supplied by Normalisation, parabolic scaling, heat equation and derivative bounds for the heat kernel. Convolution is that of Convolution of two functions on ${\mathbb{R}}^n$, and $L^p$ means the class space of The space $L^p(\mu )$ as the quotient by null functions.

For $t>0$ and $f\in L^p({\mathbb{R}}^n)$ define $H_{t}f$ to be the $L^p$ class of the function

$x⟼{\int }_{{\mathbb{R}}^n}\mathrm{\Gamma }(x-y,t)f(y)dy\mathrm{.}$

By Young's convolution inequality Young's convolution inequality under Countable Choice applied with the exponent triple $(p,1,p)$, which satisfies $1\mathrm{/}p=1\mathrm{/}p+1\mathrm{/}1-1$, this convolution is defined for almost every $x$ and belongs to $L^p$ with $\mathrm{\parallel }{H}_{t}f{\mathrm{\parallel }}_p\le \mathrm{\parallel }{\mathrm{\Gamma }}_t{\mathrm{\parallel }}_1\mathrm{\parallel }f{\mathrm{\parallel }}_p=\mathrm{\parallel }f{\mathrm{\parallel }}_p$; hence $H_{t}f$ is a well-defined element of $L^p$ satisfying the contraction bound. For $p=\mathrm{\infty }$ the integral converges absolutely for every $x$ because $\mathrm{\mid }\mathrm{\Gamma }(x-y,t)f(y)\mathrm{\mid }\le \mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\Gamma }(x-y,t)$ almost everywhere in $y$, with $\int \mathrm{\Gamma }(x-y,t)dy=1$, so it defines a bounded representative with $\mathrm{\parallel }{H}_{t}f{\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$.

The value depends only on the class of $f$: if $f=f^{\mathrm{\prime }}$ almost everywhere then the null set where they differ is carried by translation to a null set (Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation), so at every $x$ the two $y$-integrands agree almost everywhere. Their absolute convergence holds at the same points, and wherever they converge the integrals agree by Two integrable functions are equal almost everywhere exactly when all of their indefinite integrals agree. Each $H_t$ is complex-linear on $L^p$, by linearity of the integral of each representative.

Set $H_{0}f:=f$, the identity operator on $L^p$; the singular kernel formula is never evaluated at $t=0$.