# (Ref req) Schrödinger heat kernel is a weak solution of parabolic Schrödinger equation

If we have nonnegative $V \in L^1_{\textrm{loc}}(\mathbb{R}^{n})$, then the operator $H = -\Delta + V$ can be defined on $L^{2}(\mathbb{R}^{n})$ via quadratic form methods. This is done by, for example, E.B. Davies in Heat Kernels and Spectral Theory. He also proves that $H$ is the infinitesimal generator of an ultracontractive symmetric Markov semigroup $e^{-Ht}$ which is given by integration against a kernel $p(x,y,t)$:

$$e^{-Ht}[f] = \int_{\mathbb{R}^{n}} p(x,y,t)f(y)\,dy \qquad f \in L^2(\mathbb{R}^{n})$$

Now I have seen several authors use, directly or indirectly, the fact that for $y$ fixed, $p(\cdot,y,t)$ is a weak solution of the equation $(\partial_{t} + H)u = 0$. But how exactly do we know this is true? I can't find anything about this in Davies.

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Why not apply the operator $L=\partial_t+H$ to the integral and obtain $\int Lp\cdot f =0$ for every test function $f$? – Piero D'Ancona Jan 18 '14 at 21:02
@PieroD'Ancona - thanks for the comment. I know that abstract Hilbert space theory gives us $Le^{-Ht}f = 0$ for any $f$; but how does this imply, for example, that $p(x,y,t)$ is strongly differentiable in $x$? – Michael Tinker Jan 18 '14 at 21:29
If $p$ solves a heat equation in weak sense, it gets some smoothness from the general theory of heat equation. But it should be easy to find references for this – Piero D'Ancona Jan 19 '14 at 8:24