Logarithmic halfline heat mass loss at the pressure endpoint

The square of $w_0$ near zero is $y^{-1}(\log (1\mathrm{/}y){)}^{-2\alpha }$, integrable because $2\alpha >1$; its first power is also integrable. For $y,z>0$, $G_t(y-z)\ge G_t(y+z)$, proving positivity. Its integral in either positive variable is at most $1$. Thus the positive kernel is an $L^1$ contraction; Jensen's inequality in its subprobability measure, followed by integration in $y$, proves the $L^2$ contraction. Equivalently the latter follows by Young applied to the odd extension ${\tilde{w}}_0$ on the line, whose heat evolution remains odd; restricting the equal two halfline norms gives the same bound. The Gaussian derivative bounds justify differentiating the convolution for every $t>0$, and give $w_t=\nu w_{yy}$ and the zero boundary value. Approximate-identity convergence applied to ${\tilde{w}}_0$ gives strong $L^2$ initial convergence.

Fix a smooth $\beta :[0,\mathrm{\infty })\to [0,1]$ equal to $0$ on $[0,1]$ and $1$ on $[2,\mathrm{\infty })$. The explicit $w_{0,j}(y)=\beta (jy)w_0(y)$ are smooth and compactly supported away from zero and converge to $w_0$ in $L^2$ by dominated convergence. Evolve their odd extensions. Integration by parts on the line, with Gaussian tails, gives $\mathrm{\parallel }w(t){\mathrm{\parallel }}_2^2+2\nu {\int }_0^t\mathrm{\parallel }{w}_y{\mathrm{\parallel }}_2^2=\mathrm{\parallel }{w}_0{\mathrm{\parallel }}_2^2$ for each approximant. Applying this identity to differences makes the evolutions Cauchy in $L_t^{\mathrm{\infty }}{L}_y^2$ and their derivatives Cauchy in $L_{t,y}^2$. Their limit is the kernel evolution by its contraction, and its weak derivative is the derivative limit by integration against compactly supported tests. This gives the stated energy inequality. For a smooth halfline function vanishing at zero and decaying at infinity, the fundamental theorem gives $\mathrm{\parallel }w{\mathrm{\parallel }}_{\mathrm{\infty }}^2\le 2\mathrm{\parallel }w{\mathrm{\parallel }}_2\mathrm{\parallel }{w}_y{\mathrm{\parallel }}_2$, hence $\mathrm{\parallel }w{\mathrm{\parallel }}_4^4\le 2\mathrm{\parallel }w{\mathrm{\parallel }}_2^3\mathrm{\parallel }{w}_y{\mathrm{\parallel }}_2$. Time integration, the energy bound and Holder prove the $L^4$ assertion; approximation passes the inequality to this $w$.

Integrating the Gaussian kernel first in $y$ gives $M(t)={\int }_0^{\mathrm{\infty }}\mathrm{erf}(z\mathrm{/}(2\sqrt{\nu t}))w_0(z)dz$, where $\mathrm{erf}(r)=2{\pi }^{-1\mathrm{/}2}{\int }_0^r{e}^{-s^2}ds$. Dominated convergence gives $M(t)\to M_0$, and differentiation for positive time gives $m^{\mathrm{\prime }}(t)=-\frac{t^{-3\mathrm{/}2}}{2\sqrt{\pi \nu }}{\int }_0^{\mathrm{\infty }}z{e}^{-z^2\mathrm{/}(4\nu t)}{w}_0(z)dz\mathrm{.}$ After $z=\sqrt{t}r$, the prefactor is $t^{-3\mathrm{/}4}$. The integral on $1\le r\le 2$ is bounded below by $cL(t{)}^{-\alpha }$ for small $t$. For $r\le t^{-1\mathrm{/}4}$, $\log (1\mathrm{/}(\sqrt{t}r))\ge L(t)\mathrm{/}4$, so the integral is bounded above by $CL(t{)}^{-\alpha }{\int }_0^{\mathrm{\infty }}{r}^{1\mathrm{/}2}{e}^{-r^2\mathrm{/}(4\nu )}dr$. For $r>t^{-1\mathrm{/}4}$ the Gaussian tail is exponentially small; on the support of $\eta (\sqrt{t}r)$ the logarithmic factor is uniformly bounded above, which gives the same upper bound. This proves both derivative bounds. Integrating the upper bound from zero gives $\mathrm{\mid }m(t)\mathrm{\mid }\le C{t}^{1\mathrm{/}4}L(t{)}^{-\alpha }$.

The upper derivative bound is integrable. The lower bound gives ${\int }_0^{t_{\ast }}\mathrm{\mid }{m}^{\mathrm{\prime }}{\mathrm{\mid }}^{4\mathrm{/}3}\ge c{\int }_0^{t_{\ast }}{t}^{-1}L(t{)}^{-4\alpha \mathrm{/}3}dt=\mathrm{\infty }$, since $4\alpha \mathrm{/}3\le 1$. In contrast the squared critical-corrector expression is at most $C{t}^{-1}L(t{)}^{-2\alpha }$, which is integrable since $2\alpha >1$. Away from time zero every quantity is smooth, proving all stated time-integral conclusions.

Differentiating the tail integral gives $h_y=w$ and $h_t=\nu w_y=\nu h_{yy}$; the boundary term at infinity vanishes by Gaussian decay. Also $h(t,y)={\int }_0^{y}w(t,z)dz-M(t)$. Thus on $[0,B]$ its difference from $h_0$ is at most $\sqrt{B}\mathrm{\parallel }w(t)-w_0{\mathrm{\parallel }}_2+\mathrm{\mid }M(t)-M_0\mathrm{\mid }$, which tends to zero. Finally a positive separation between $y$ and $\mathrm{supp}{w}_0$ puts every differentiated Gaussian convolution under the bound $C_N{t}^{-N}{e}^{-c\mathrm{/}t}\mathrm{\parallel }{w}_0{\mathrm{\parallel }}_1$. These arguments prove the remaining assertions.