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Let $(M,g)$ be a noncompact Riemannian manifold whose isometry group acts transitively on $M$, i.e. a (not necessarily normal) homogeneous space. Let $e_{\lambda}(x,y)$ be the integral kernel of $f \mapsto \int_{0}^{\lambda} dE_{\nu}(f)$ where $dE_{\nu}$ is the spectral measure of the (non-negative) Laplacian associated to $(M,g)$. Is there a relatively simple proof that

$$ \int_M |\nabla_x e_{\lambda}(x,y)|^2_g\, dy = \int_M e_{\lambda}(x,y) \cdot \Delta_x e_{\lambda}(x,y)\, dy \ ?$$

Note that the differentiation is in $x$ and the integration is in $y$. In particular, it's surely false for most non-homogeneous manifolds.

I ask for `relatively simple' because I know of a proof using the expression of the Laplacian as the generator of the heat semi-group. That proof seems to be overkill.

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