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I notice a way of solving equations that goes roughly like this:

  1. Want to solve $Tu=v$, i.e. hoping to find "$T^{-1}(v)$".
  2. $T^{-1}$ might not exist "on the nose", so instead we consider $\int_0^\infty e^{-tT}(v) \,dt$, which is a heuristic/informal/non-rigorous version of $T^{-1}$.
  3. Understand $\int_0^\infty e^{-tT} \,dt$ as an operator, and say something about the equation.

A typical example is when $T$ is the Laplacian, its inverse is the Green function, and the exponentiated one is the heat kernel.

My questions are:

  1. In general, how to make this method rigorous?
  2. Are there examples other than heat kernel?

Thank you!

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    $\begingroup$ Now, concerning your question itself: what you are probably looking for is the theory of $C_0$-semigroups. If $-T$ generates a $C_0$-semigroup, it is not uncommon to denote this $C_0$-semigroup by $(e^{-tT})_{t \ge 0}$, and the integral $\int_0^s e^{tT}\,dt$ then exists (in the strong sense) for every $s \ge 0$. If the growth bound of the $C_0$-semigroup is strictly smaller than $0$, then $-T$ is bijective from its domain onto the entire Banach space, and $\int_0^s e^{-tT}\,dt$ does indeed converge (with respect to the operator norm) to $T^{-1}$ as $s \to \infty$. $\endgroup$
    – Jochen Glueck
    Commented Nov 9, 2019 at 9:34
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    $\begingroup$ There is such an approach and $T$ need not be linear. For $T$ linear look at Hille-Yosida theorem. This was generalized to $T$ nonlinear and maximal monotone. For example the theory works when $T$ is the (sub)gradient of a convex function. $\endgroup$
    – Liviu Nicolaescu
    Commented Nov 9, 2019 at 10:12
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    $\begingroup$ Ona compact manifold, the heat kernel determines the metric. $\endgroup$
    – Liviu Nicolaescu
    Commented Nov 10, 2019 at 9:27
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    $\begingroup$ @Student: Yes, I mean that the eigenvalues of $T$ should be located "on the right" rather than "on the left" if $-T$ is supposed to generate a $C_0$-semigroup. Details can be found in the Hille-Yosida theorem already mentioned by Liviu Nicolaescu. $\endgroup$
    – Jochen Glueck
    Commented Nov 11, 2019 at 7:06
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    $\begingroup$ On a compact Riemann manifold, the metric and its curvature are determined by the behavior as $t\searrow 0$ of the fourth order jets of the heat kernel along the diagonal. $\endgroup$
    – Liviu Nicolaescu
    Commented Nov 11, 2019 at 11:02

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