The heat flow need not converge in supremum norm

The half-line datum $f$ is bounded and measurable, with $f(0)=1$, but ${\int }_{\mathbb{R}}\mathrm{\mid }f{\mathrm{\mid }}^p=\mathrm{\infty }$ for every finite $p$. Evenness and unit mass give $H_{t}f(0)={\int }_0^{\mathrm{\infty }}\mathrm{\Gamma }(y,t)dy=1\mathrm{/}2$, so its pointwise supremum distance is at least $1\mathrm{/}2$. This also gives an essential supremum bound: for any $0<\eta <1\mathrm{/}2$, continuity of $H_{t}f$ at $0$ supplies $\delta >0$ such that $\mathrm{\mid }{H}_{t}f(x)-1\mathrm{/}2\mathrm{\mid }<1\mathrm{/}2-\eta$ for $0<x<\delta$. On that interval $f(x)=1$, so $\mathrm{\mid }{H}_{t}f(x)-f(x)\mathrm{\mid }>\eta$. As the interval has positive measure and $\eta$ is arbitrary, $\mathrm{\parallel }{H}_{t}f-f{\mathrm{\parallel }}_{\mathrm{\infty }}\ge 1\mathrm{/}2$ for every $t>0$.

Uniform continuity is necessary for an actual supremum-norm convergence claim on bounded real data: for each fixed $t>0$, [F5] and the mean value theorem make $H_{t}g$ globally Lipschitz, hence uniformly continuous. If ${\sup }_x\mathrm{\mid }{H}_{t}g(x)-g(x)\mathrm{\mid }\to 0$, the sequence $H_{1\mathrm{/}(k+1)}g$ converges uniformly to $g$, which is uniformly continuous by [F5]. For convergence in the essential supremum norm the corresponding necessity concerns the class: the continuous differences $H_{1\mathrm{/}(k+1)}g-H_{1\mathrm{/}(\mathrm{\ell }+1)}g$ have equal supremum and essential supremum, so the sequence is uniformly Cauchy; real completeness gives a pointwise limit $v$. For any $\epsilon >0$, a uniform Cauchy bound $\mathrm{\mid }{H}_{1\mathrm{/}(k+1)}g(x)-H_{1\mathrm{/}(\mathrm{\ell }+1)}g(x)\mathrm{\mid }<\epsilon$ for all $x$ and sufficiently large $k,\mathrm{\ell }$, followed by $\mathrm{\ell }\to \mathrm{\infty }$, gives ${\sup }_x\mathrm{\mid }{H}_{1\mathrm{/}(k+1)}g(x)-v(x)\mathrm{\mid }\le \epsilon$. Thus the limit is uniform and $v$ is uniformly continuous by [F5]. Finally $\mathrm{\parallel }v-g{\mathrm{\parallel }}_{\mathrm{\infty }}\le {\sup }_x\mathrm{\mid }v-H_{1\mathrm{/}(k+1)}g\mathrm{\mid }+\mathrm{\parallel }{H}_{1\mathrm{/}(k+1)}g-g{\mathrm{\parallel }}_{\mathrm{\infty }}\to 0$. Thus that class has a uniformly continuous representative.

The interval datum $f_0$ has $\int \mathrm{\mid }{f}_0{\mathrm{\mid }}^p=1$ for every finite $p$ and $\mathrm{\parallel }{f}_0{\mathrm{\parallel }}_{\mathrm{\infty }}=1$. By positivity and step 1.1, $H_t{f}_0(0)={\int }_0^1\mathrm{\Gamma }(y,t)dy<1\mathrm{/}2$, since the omitted integral on $(1,\mathrm{\infty })$ is positive. Put $d=1-H_t{f}_0(0)>1\mathrm{/}2$ and $c=(d+1\mathrm{/}2)\mathrm{/}2$, so $1\mathrm{/}2<c<d$. Continuity at $0$ supplies $0<\delta <1$ with $\mathrm{\mid }{H}_t{f}_0(x)-H_t{f}_0(0)\mathrm{\mid }<d-c$ for $0<x<\delta$. There $f_0(x)=1$ and $\mathrm{\mid }{H}_t{f}_0(x)-1\mathrm{\mid }\ge d-\mathrm{\mid }{H}_t{f}_0(x)-H_t{f}_0(0)\mathrm{\mid }>c$. Hence $\mathrm{\parallel }{H}_t{f}_0-f_0{\mathrm{\parallel }}_{\mathrm{\infty }}\ge c>1\mathrm{/}2$ on a set of positive measure, and the pointwise supremum is also greater than $1\mathrm{/}2$.

For this same $f_0$, [F4] gives $\mathrm{\parallel }{H}_t{f}_0-f_0{\mathrm{\parallel }}_p\to 0$ for every $1\le p<\mathrm{\infty }$. Nevertheless each compact set containing $0$ has pointwise supremum error at least $\mathrm{\mid }{H}_t{f}_0(0)-f_0(0)\mathrm{\mid }>1\mathrm{/}2$, so locally uniform convergence fails there. The continuity hypothesis in the bounded-data theorem cannot be discarded.

Steps 1.1–3.1 preserve the exact half-line value $1\mathrm{/}2$ and supply compactly supported data in $L^{\mathrm{\infty }}\cap {\bigcap }_{1\le p<\mathrm{\infty }}{L}^p$ with essential and pointwise supremum errors bounded away from zero, while every finite-$p$ norm converges. Step 1.2 justifies the uniform-continuity qualification with the distinction between representatives and classes explicit. These witnesses refute the claimed endpoint extension and locally uniform convergence at the jump.