[proofplan] We show that any solution of the heat equation is $C^\infty$ by localising via a smooth cutoff, representing the localised function through [Duhamel's Principle](/theorems/55) as a convolution with the heat kernel against a smooth compactly supported source, and observing that the heat kernel is $C^\infty$ away from its singularity -- which is separated from the evaluation region by the support gap of the source term. [/proofplan] [step:Localise via a smooth space-time cutoff to reduce to a whole-space problem] Fix $(x_0, t_0) \in \Omega_T$ and choose a parabolic cylinder $Q := B(x_0, 2\rho) \times (t_0 - 4\rho^2, t_0] \subseteq \Omega_T$. Let $\zeta \in C^\infty(\mathbb{R}^n \times \mathbb{R})$ be a cutoff function satisfying $\zeta \equiv 1$ on $B(x_0, \rho) \times [t_0 - \rho^2, t_0]$, $\operatorname{supp}(\zeta) \subseteq Q$, and $0 \leq \zeta \leq 1$. Define $w := \zeta u$, extended by zero outside $Q$. Computing the heat operator applied to $w$ using the product rule: \begin{align*} \partial_t w - \Delta w &= \zeta(\partial_t u - \Delta u) + u\,\partial_t\zeta - 2\nabla\zeta \cdot \nabla u - u\,\Delta\zeta \\ &= u\,\partial_t\zeta - 2\nabla\zeta \cdot \nabla u - u\,\Delta\zeta =: F, \end{align*} where we used $\partial_t u - \Delta u = 0$. Since $u \in C^{2,1}(\Omega_T)$ and $\zeta \in C^\infty$, the source $F$ is smooth on $Q$. The support of $F$ is contained in $\operatorname{supp}(\nabla\zeta) \cup \operatorname{supp}(\partial_t\zeta)$, which lies in the annular region $Q \setminus (B(x_0, \rho) \times [t_0 - \rho^2, t_0])$. [/step] [step:Represent $w$ via Duhamel's principle as a space-time convolution with the heat kernel] The function $w$ solves $\partial_t w - \Delta w = F$ on $\mathbb{R}^n \times (0, T]$ with $w(\cdot, 0) = 0$ (since $\operatorname{supp}(\zeta) \subseteq (t_0 - 4\rho^2, t_0]$ and $w$ is extended by zero). By [Duhamel's Principle](/theorems/55): \begin{align*} w(x, t) = \int_0^t \int_{\mathbb{R}^n} \Phi(x - y, t - s)\,F(y, s) \, d\mathcal{L}^n(y) \, d\mathcal{L}^1(s). \end{align*} [/step] [step:Conclude $u \in C^\infty$ on the inner cylinder by differentiating the Duhamel representation] On the inner cylinder $B(x_0, \rho) \times [t_0 - \rho^2, t_0]$, $w = u$ since $\zeta = 1$ there. For $(x, t)$ in the inner cylinder and $(y, s) \in \operatorname{supp}(F)$, the pair $(x - y, t - s)$ is bounded away from the singularity $(0, 0)$ of $\Phi$: the support of $F$ lies in the annular region, which is disjoint from the inner cylinder. When $t - s > 0$, $\Phi(x - y, t - s)$ is $C^\infty$ in all variables. When $t - s = 0$, the integrand vanishes because $F$ is supported away from the temporal level $s = t$ in the inner cylinder. Since $F$ is bounded and compactly supported, and the integrand $(x, t) \mapsto \Phi(x - y, t - s)\,F(y, s)$ is $C^\infty$ in $(x, t)$ for $(x, t)$ in the inner cylinder, differentiation under the integral sign is justified by the dominated convergence theorem. Therefore $u = w$ is $C^\infty$ on $B(x_0, \rho) \times (t_0 - \rho^2, t_0]$. Since $(x_0, t_0) \in \Omega_T$ was arbitrary, every point has a neighbourhood on which $u$ is $C^\infty$, hence $u \in C^\infty(\Omega_T)$. [/step]