The inhomogeneous heat Cauchy formula

In the mild setting $u=H_{\cdot }{u}_0+Df\in C([0,T];L^p({\mathbb{R}}^n))$ and $u(0)=u_0$: the curve $t\mapsto H_t{u}_0$ is continuous by the strong continuity and semigroup law of [F1], the curve $Df$ is continuous with $Df(0)=0$ by [F2], and $H_0{u}_0=u_0$; a sum of continuous curves is continuous.

In the bounded-continuous setting the same scalar formula is well defined, bounded and continuous with initial trace $u_0$. Boundedness: $\mathrm{\mid }u(t,x)\mathrm{\mid }\le \mathrm{\parallel }{u}_0{\mathrm{\parallel }}_{\mathrm{\infty }}+t\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$. Continuity and initial trace: the homogeneous term is continuous in $(t,x)$ for $t>0$ and converges to $u_0$ locally uniformly as $t\downarrow 0$ by [F3], while the potential can be written ${\int }_0^T{\mathbf{1}}_{\{\tau <t\}}({\mathrm{\Gamma }}_{\tau }\ast f(\cdot ,t-\tau ))(x)d\tau$. For $(x_j,t_j)\to (x,t)$, joint uniform continuity of $f$ gives convergence of the integrands at each fixed $\tau \mathrm{\ne }t$; they are bounded by $\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$, so Dominated convergence gives continuity. Its bound $t\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$ also gives uniform vanishing at zero; finally the scalar forced relation $u(t,x)=\int \mathrm{\Gamma }(x-y,t-s)u(s,y)dy+{\int }_s^t\int \mathrm{\Gamma }(x-y,t-\tau )f(y,\tau )dyd\tau$ for $0<s<t\le T$ follows from the semigroup law and Fubini-Tonelli applied to the real and imaginary positive and negative parts of the absolutely integrable iterated integrands (bounded by kernel masses times the data bounds), using Tonelli and Fubini for the completed product, with only almost-everywhere section measurability: the homogeneous term convolves to ${\mathrm{\Gamma }}_t\ast u_0={\mathrm{\Gamma }}_{t-s}\ast ({\mathrm{\Gamma }}_s\ast u_0)$ and the double integral splits at $s$. No differentiation of $f$ is used, so no $C^{1,2}$ claim is made here.

In the classical setting, $u$ is a classical solution with the stated data and is unique in the Gaussian growth class. Indeed $H_t{u}_0$ is, by [F3], $C^{\mathrm{\infty }}$ in positive time with ${\mathrm{\partial }}_t(H_t{u}_0)=\mathrm{\Delta }(H_t{u}_0)$ and initial data $u_0$, while the scalar potential $u^{\ast }$ of [F2] is $C^{1,2}$ with $u_t^{\ast }-\mathrm{\Delta }{u}^{\ast }=f$ and zero initial data; hence the scalar formula $u=H_{\cdot }{u}_0+u^{\ast }$ is $C^{1,2}$ on ${\mathbb{R}}^n\times (0,T]$ with $u_t-\mathrm{\Delta }u=f$ and $u(\cdot ,0)=u_0$. If $\tilde{u}$ is another classical solution with the same data and $\mathrm{\mid }\tilde{u}\mathrm{\mid }\le C{e}^{a\mathrm{\mid }x{\mathrm{\mid }}^2}$, then $w:=\tilde{u}-u$ is a $C^{1,2}$ solution of the homogeneous equation with zero initial data and Gaussian growth (the sum of the two growth bounds), so [F3] forces $w\equiv 0$ and $\tilde{u}=u$.

The mild curve of step 1.1 satisfies the forced relation: for $0<s<t\le T$, subtracting $H_{t-s}u(s)$ from $u(t)$ and using the semigroup law of [F1] together with the relation for $Df$ in [F2] and the additivity of the Bochner integral [F4] gives $u(t)-H_{t-s}u(s)=(H_t{u}_0-H_{t-s}{H}_s{u}_0)+(Df(t)-H_{t-s}Df(s))={\int }_s^t{H}_{t-\tau }f(\tau )d\tau$.

Mild uniqueness: if $w\in C([0,T];L^p({\mathbb{R}}^n))$ with $w(0)=u_0$ satisfies the relation, then $v:=w-u$ is continuous with $v(0)=0$ and satisfies $v(t)=H_{t-s}v(s)$ for all $0<s<t\le T$; the contraction bound of [F1] gives $\mathrm{\parallel }v(t){\mathrm{\parallel }}_p\le \mathrm{\parallel }v(s){\mathrm{\parallel }}_p$ for every $s\in (0,t)$, and letting $s\downarrow 0$ with continuity of $v$ at $0$ gives $\mathrm{\parallel }v(t){\mathrm{\parallel }}_p=0$. Hence $w=u$, and all clauses of the statement are proved; the only choices made are finitely many thresholds, and Countable Choice is inherited from the cited suppliers.