Consider $(t, x)\in [0,T]\times (\mathbb{R}^d,d\mu)$, where the measure $d\mu(x)=K^{-1}\exp(-U(x))dx$ is a reasonable Gibbs measure (it satisfies a Poincaré or log-Sobolev inequality. One can, for simplicity, start with the Gaussian $d\mu(x)=(2\pi)^{-\frac{d}{2}}\exp(-\dfrac{|x|^2}{2})dx$). We define Sobolev spaces $$L^2(I\times \mu)=\{f:\ \int_{\mathbb{R}^d}\int_0^T f(t,x)^2 dtd\mu(x)<\infty\}$$ and $$H^1(I\times\mu)= \{f:\ f,\partial_t f, \nabla_x f\in L^2(I\times \mu)\}$$ and $H^2(I\times\mu)$ in a similar way. Now, given any $f\in H^1(I\times\mu)$, can we find $u\in H^2(I\times \mu)$, such that $$u(t=0,\cdot)=u(t=T,\cdot)=0$$ and $$\partial_t u(t=0,\cdot)=f(0,\cdot), \ \partial_t u(t=T,\cdot)=f(t=T,\cdot),$$ such that $$\|u\|_{H^2(I\times\mu)} \le C\|f\|_{H^1(I\times\mu)}?$$
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$\begingroup$ One way to see it is by trace theorem, but I struggle to find a suitable definition for $H^\frac{1}{2}(\mu)$ (aside from being the trace space of $H^1(I\times\mu)$). Even if we do, a constructional proof seems nontrivial. The biggest obstacle is that the Radon-Nikodym derivative of measures is unbounded under translation, so a proof by convolution kernel seems undesirable. I have also thought about using Markov semigroup for Langevin (for example, $u(t, x)=\int_0^t P_{t-s}f(s, x)ds$, or $\partial_t u +\nabla_x^*\nabla_x u=f(t, x), u(t=0,x)=0$), but $\partial_{tt} u$ might lack regularity. $\endgroup$– b9c7d65gCommented Nov 14, 2019 at 15:44
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