Gaussian regularization preserves finite-action continuity equations

The Gaussian second moment is finite: in one dimension $x^2{e}^{-x^2\mathrm{/}(2\epsilon )}$ is bounded on $[-1,1]$ and, outside it, is bounded by a constant times $e^{-x^2\mathrm{/}(4\epsilon )}$, which is integrable; product integration gives $\int \mathrm{\mid }z{\mathrm{\mid }}^2{\rho }_{\epsilon }(z)dz=d\epsilon$. Hence ${\mu }_t\ast {\rho }_{\epsilon }$ has finite second moment by $\mathrm{\mid }y+z{\mathrm{\mid }}^2\le 2\mathrm{\mid }y{\mathrm{\mid }}^2+2\mathrm{\mid }z{\mathrm{\mid }}^2$. For bounded continuous f, $y\mapsto \int f(y+z){\rho }_{\epsilon }(z)dz$ is bounded and continuous by dominated convergence, so narrow continuity of ${\mu }_t$ gives narrow continuity of ${\mu }_t^{\epsilon }$. By [F1], $a\in L^1(0,T)$ and $N$ is a Borel null set. Use $\bar{v}$ from the statement; because $\bar{v}=v$ $M$-almost everywhere, the distributional equation and action are unchanged. Set $g(t)=\sqrt{a(t)}$ for $t\mathrm{\notin }N$ and $g(t)=0$ on $N$. Holder gives $\int \mathrm{\mid }{\bar{v}}_t\mathrm{\mid }d{\mu }_t\le g(t)$ for every t and $g\in L^1(0,T)$, so all integrals defining $j$ are finite for every t. By [F3], $p,j$ are Borel in $(t,x)$. Positivity of ${\rho }_{\epsilon }$ and total mass one give $p_t^{\epsilon }(x)>0$. For every component i and real r, $\{(t,x):v_{t,i}^{\epsilon }(x)>r\}=\{(t,x):j_{t,i}^{\epsilon }(x)-r{p}_t^{\epsilon }(x)>0\}$ is Borel by the sum and scalar-multiple closure in [F4]. Hence each component of $v^{\epsilon }$ is Borel. For every spatial multi-index, the Gaussian derivative is bounded by [F2], so differentiation under the integral is justified for $p$ and for $j$'s first derivatives by domination by a constant times $1+\mathrm{\mid }{\bar{v}}_t\mathrm{\mid }$. Thus $p_t^{\epsilon }$ is smooth in x and $j_t^{\epsilon }$ is continuously differentiable in x. Also $\int p_t^{\epsilon }dx=1$ by [F2] and Tonelli.

Fix a compact spatial ball K. Narrow continuity of ${\mu }_t$ and bounded continuity of each Gaussian derivative imply joint continuity of $p_t^{\epsilon }(x)$ and its spatial derivatives in $(t,x)$: for $t_n\to t$, $x_n\to x$, the spatial difference is uniformly bounded by a derivative bound times $\mathrm{\mid }{x}_n-x\mathrm{\mid }$, and the remaining integral converges by narrow continuity. On $[0,T]\times K$ the positive density therefore has a positive minimum. For every first spatial derivative, boundedness of the corresponding Gaussian derivative gives $\mathrm{\mid }{j}_t^{\epsilon }(x)\mathrm{\mid }+\mathrm{\mid }\mathrm{\nabla }{j}_t^{\epsilon }(x)\mathrm{\mid }\le C_{\epsilon }\int \mathrm{\mid }{\bar{v}}_t\mathrm{\mid }d{\mu }_t\le C_{\epsilon }g(t)$ and $\mathrm{\mid }\mathrm{\nabla }{p}_t^{\epsilon }(x)\mathrm{\mid }\le C_{\epsilon }$. The quotient rule and the positive lower bound for p show that $\mathrm{\mid }{v}_t^{\epsilon }(x)\mathrm{\mid }+\mathrm{\mid }\mathrm{\nabla }{v}_t^{\epsilon }(x)\mathrm{\mid }\le C_{K,\epsilon }g(t)$ on K. The right side is time-integrable, proving the local magnitude and Lipschitz bounds required for the Carathéodory flow.

For every t, the probability measure $p_t^{\epsilon }(x{)}^{-1}{\rho }_{\epsilon }(x-y)d{\mu }_t(y)$ averages ${\bar{v}}_t$ to $v_t^{\epsilon }(x)$. Cauchy--Schwarz for this probability gives $\frac{\mathrm{\mid }{j}_t^{\epsilon }(x){\mathrm{\mid }}^2}{p_t^{\epsilon }(x)}\le \int {\rho }_{\epsilon }(x-y)\mathrm{\mid }{\bar{v}}_t(y){\mathrm{\mid }}^{2}d{\mu }_t(y)\mathrm{.}$ Integrate in x and use Tonelli and $\int {\rho }_{\epsilon }(x-y)dx=1$ to obtain $\int \mathrm{\mid }{v}_t^{\epsilon }{\mathrm{\mid }}^2{p}_t^{\epsilon }dx\le \int \mathrm{\mid }{\bar{v}}_t{\mathrm{\mid }}^{2}d{\mu }_t$ for every t; since $\bar{v}=v$ $M$-almost everywhere, this is the asserted a.e. action inequality.

Since $\bar{v}=v$ $M$-almost everywhere, the distributional continuity-equation identity also holds with $\bar{v}$. Let $\phi \in C_c^{\mathrm{\infty }}((0,T)\times {\mathbb{R}}^d)$ and set $\psi (t,y)=\int \phi (t,x){\rho }_{\epsilon }(x-y)dx$. This function and its derivatives decay rapidly in y, but are not compactly supported. Choose smooth cutoffs ${\chi }_R$ equal to one on $B_R$, supported in $B_{2R}$, with $\mathrm{\mid }\mathrm{\nabla }{\chi }_R\mathrm{\mid }\le C\mathrm{/}R$, and apply the equation with $\bar{v}$ to ${\chi }_R(y)\psi (t,y)$. As $R\to \mathrm{\infty }$, dominated convergence [F5] sends the terms with ${\chi }_R$ to their uncut values: ${\mathrm{\partial }}_t\psi$ and ${\mathrm{\nabla }}_y\psi$ are bounded, while $\mathrm{\mid }v\mathrm{\mid }$ is integrable against M by [F1]. The extra term $\psi v\cdot \mathrm{\nabla }{\chi }_R$ tends to zero, bounded in absolute integral by $C\mathrm{\parallel }\psi {\mathrm{\parallel }}_{\mathrm{\infty }}{R}^{-1}\int \mathrm{\mid }v\mathrm{\mid }dM$. Thus $\int ({\mathrm{\partial }}_t\psi +{\mathrm{\nabla }}_y\psi \cdot v_t(y))d{\mu }_t(y)dt=0.$ Differentiating the convolution and integrating by parts in x gives ${\mathrm{\nabla }}_y\psi (t,y)=\int {\mathrm{\nabla }}_x\phi (t,x){\rho }_{\epsilon }(x-y)dx$. Fubini then rewrites the last identity as ${\int }_0^T\int ({\mathrm{\partial }}_t\phi (t,x)p_t^{\epsilon }(x)+{\mathrm{\nabla }}_x\phi (t,x)\cdot j_t^{\epsilon }(x))dxdt=0,$ which is exactly the continuity equation for density $p_t^{\epsilon }$ and velocity $v_t^{\epsilon }=j_t^{\epsilon }\mathrm{/}{p}_t^{\epsilon }$.