Let $L: L^2(\mathbb{R}) \supseteq Dom(L) \rightarrow L^2(\mathbb{R})$ be a densely defined closed operator. Assume that the resolvent admits an integral kernel (Greens function) $G$, i.e. for $z\in \mathbb{C}\setminus \sigma(L)$ and every $f\in L^2(\mathbb{R})$ the resolvent satisfies


$$ ((L-z)^{-1}f)(x) = \int\_{\mathbb{R}} G(z,x,y) f(y) dy. $$


We denote by $g$ the diagonal Greens function


$$ g(z,x) := G(z,x,x). $$



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If this is in general not true, how is it in the case of Sturm-Liouville operators


$$ L = -\Delta + V, \ Dom(L)= H^2 (\mathbb{R}),$$


where $V \in C^\infty (\mathbb{R}, \mathbb{R})\cap L^\infty (\mathbb{R}, \mathbb{R})$?



How to determine the spectrum from the diagonal Green's function
For the Sturm–Liouville operators with discrete spectrum, $G(z,x,y)$ is the Stieltjes transform of the discrete measure $$\mu\_{x,y}(ds) = \sum\_{n = 1}^\infty \varphi\_n(x) \overline{\varphi\_n(y)} \delta\_{-\lambda\_n}(ds) ,$$ that is, $$G(z,x,y) = \int\_\mathbf{R} \frac{1}{z - s} \, \mu\_{x,y}(ds) = \sum\_{n = 1}^\infty \frac{\varphi\_n(x) \overline{\varphi\_n(y)}}{z - \lambda\_n} \, .$$ So yes, $G(z,x,x)$ indeed determines the spectrum: it is a meromorphic function with a simple pole at $\lambda\_n$ with residue $|\varphi\_n(x)|^2$, unless $\varphi\_n(x) = 0$. This means that the spectrum is determined by the values of $G(z,x,x)$ for a single point $x$, provided that $\varphi\_n(x) \ne 0$ for all $n$.


In the continuous spectrum case (which you are interested in) things are slightly more complicated: $G(z,x,y)$ is again a Stieltjes transform of a measure $\mu\_{x,y}(ds)$. This time it is no longer a discrete measure, but it is supported in the spectrum of $L$. In principle, the set of discontinuities of $G(z,x,x)$ need not be closed in this case. However, the closure of this set is indeed the support of the Stieltjes measure $\mu\_{x,x}$. And the union of all supports of $\mu\_{x,x}$ will exhaust the spectrum of $L$.


I do not know good references for these claims. Properties of Stieltjes transforms are discussed in a lovely book *Bernstein functions* by Schilling, Song and Vondraček. The properties related to supports of $\mu\_{x,x}$ which I used above can be derived from generalised eigenfunction expansion of the heat kernel, but I am sure there is a simpler way to do this. Finally, the above approach seems to be pretty general, but my guess would be that some further restrictions on $L$ are necessary. At the very least, we need finiteness and continuity of $G(z,x,y)$ in $x$ and $y$ near the diagonal.


