A mild heat solution need not be classical at the initial time

By [F1] the finite-interval indicator belongs to $L^p(\mathbb{R})$ for every $1\le p\le \mathrm{\infty }$. For each $1\le p<\mathrm{\infty }$, [F3] gives the mild curve $u\in C([0,T];L^p)$ with $u(0)=[f]$, and [F4] gives a smooth representative for every positive time.

By [F2] and Gaussian scaling, $u(0,t)={\int }_0^1\mathrm{\Gamma }(y,t)dy={\int }_0^{1\mathrm{/}\sqrt{t}}\mathrm{\Gamma }(z,1)dz$. As $t\downarrow 0$, dominated convergence (Dominated convergence) and evenness with unit mass [F5] give $u(0,t)\to 1\mathrm{/}2$, whereas the specified representative has $f(0)=1$. Thus the initial condition is not attained pointwise for that representative.

This failure is not removable by changing $f$ only on a null set. Any continuous representative of $[f]$ would be identically one on $(0,1)$ and zero on $(-\mathrm{\infty },0)$: otherwise continuity would give a nondegenerate interval of disagreement, whose measure is positive by the box measure in [F1]. The two one-sided limits at zero would then be one and zero, a contradiction. Hence the initial class has no continuous representative at all.

The mild curve from step 1.1 therefore cannot have a jointly continuous classical extension to time zero with its prescribed initial class. Positive-time smoothness and finite-$p$ norm convergence do not supply corner or initial-time continuity, and step 1.2 also gives the explicit failure of pointwise attainment for the chosen representative. No $L^{\mathrm{\infty }}$ norm convergence at zero is claimed.