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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 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 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 at 8:24

1 Answer 1

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+100

There is an old paper by Ball where he discusses the abstract connections between weak solutions and semigroup orbits. He proves that the mild solution is essentially always the weak solution for strongly continuous semigroups.

Theorem 3.1.7 from the monograph by Courtain and Zwart is also proving this.

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MO won't let me release the bounty yet, but I will asap. –  Michael Tinker Jan 20 at 21:13

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