Let $u_0 \in L^2$ and $f \in L^2(0,T;H^{-1})$ and consider the solution $u \in L^2(0,T;H^1)\cap H^1(0,T;H^{-1})$ of $$u_t - \Delta u = f$$ $$u(0) = u_0$$ with some BC (eg. zero Dirichlet).
I am asking about parabolic smoothing, that the solution is in fact smooth. How does this work? Does it work for any data?
For example, if I test with $tu_t$ then I can obtain that $u \in C^0([\delta,T];H^1)$ for $\delta > 0$.
If the domain of $A=-\Delta$ is chosen so that it is positive and selfadjoint, then the spectral theorem for $A$ is $$ Au = \int_{0}^{\infty} \lambda dE(\lambda)u $$ with $u \in \mathcal{D}(A)$ iff $\int_{0}^{\infty}\lambda^{2}d\|E(\lambda)u\|^{2} < \infty$. More generally, $u \in \mathcal{D}(A^{s})$ for some $0 < s < \infty$ iff $\int_{0}^{\infty}\lambda^{2s}d\|E(\lambda)u\|^{2} < \infty$. The semigroup solution of the homogeneous equation ($u_{t}=-\Delta u$, $u(0)=u_{0}$) is the classical separation of variables type of solution $$ u(t) = \int_{0}^{\infty}e^{-t\lambda}dE(\lambda)u_{0}. $$ So you can see right away that $u(t) \in \mathcal{D}(A^{s})$ for all $s > 0$, for any $t > 0$. The continuity properties of $u$ at $t=0$ depend on the spatial smoothness of $u_{0}$. For example, if $u_{0}\in\mathcal{D}(A^{s/2})=H^{s}$ for some $s > 0$, then $u \in C(0,T; H^{s})$ by the Lebesgue dominated convergence theorem, because $$ \|A^{s/2}u_{0}\|^{2}=\int_{0}^{\infty}\lambda^{s}d\|E(t)u_{0}\|^{2} < \infty, $$ and because $$ \begin{align} \|u(t)-u_{0}\|_{H^{s}}^{2} & =\|A^{s/2}(u(t)-u_{0})\|^{2}+\|u(t)-u_{0}\|^{2} \\ & =\int_{0}^{\infty}|e^{-t\lambda}-1|\lambda^{s}d\|E(\lambda)u_{0}\|^{2} +\|u(t)-u_{0}\|^{2}. \end{align} $$ The inhomogeneous solution is written in terms of the semigroup $S(t)=\int_{0}^{\infty}e^{-t\lambda}dE(t)$, which acts as an integrating factor for the inhomogenous equation: $$ u(t) = S(t)u_{0} + \int_{0}^{t}S(t-t')f(t')\,dt',\;\; t \ge 0. $$ This is a little trickier to analyze, but the basic ideas are the same. The first term on the right is the homogenous solution mentioned previously. The second term is quite smooth in space when the integral is restricted to $[0,t-\epsilon]$ for some $0 < \epsilon < t$ because of the smoothing effects of $S(s)$ for $s > 0$. The remaining piece of the integral over $[t-\epsilon,t]$ influences the spatial smoothness of $u(t)$ at $t$, and that piece is, in turn, influenced by the behavior of $f(t')$ on $[t-\epsilon,t]$.