Mathematics Asked on December 1, 2021
Suppose that $mu$ is a Radon measure on $mathbb{R}$, and let $I(mu)$ be the Stieltjes transform of $mu$, i.e. $I(mu)(lambda)=int_{mathbb{R}}frac{1}{x-lambda},mathrm{d}mu(x)$ for $lambda$ non-real. If $I(mu)(lambda)$ happens to be equal to zero, would it imply that $mu=0$? If yes, why? I can’t find any theorem stating about it, so I ask for your help. Maybe it has something with injectivity to do with?
As I mentioned in an earlier comment, The imaginary part of the Stieltjes transform $I_mu$ (in your notation) is $pi P*mu$, where $P$ is the Poisson kernel.
A short but concise presentation if the Stieltjes integral in Barry Simon's Comprehensive Course in Analysis, Volume 3 (Harmonic Analysis) section 2.5. Amongst all the different cool things one can find there is the following Theorem:
Theorem: Let $mu$ be a finite measure on $mathbb{R}$ and $F_mu(z)=intfrac{mu(dx)}{x-z}$ be its Stieltjes transform. Suppose $$ mu(dx) = fcdotlambda(dx) + mu_s(dx)$$ be its Lebesgue decomposition (absolute and singular decomposition with respect the LEbeshue measure $lambda$. Then
a. $frac{1}{pi}int operatorname{Im},F_mu(x+ ivarepsilon)g(x),dx xrightarrow{varepsilonrightarrow0+}int g(x)mu(dx)$ for any $gin mathcal{C}_{00}(mathbb{R})$ (functions of compact support).
b. For $lambda$-a.a. $xin mathbb{R}$, $$frac{1}{pi}lim_{varepsilonrightarrow0+}operatorname{Im} F_mu(x+ivarepsilon)= f(x)$$
c. There exists a measurable set $Esubsetmathbb{R}$ such that $lambda(E)=0=mu_s(mathbb{R}setminus E)$ such that $$ lim_{varepsilonrightarrow0+}operatorname{Im} F_mu(x+ivarepsilon)= infty,qquad xin E $$
d. For all $x_0inmathbb{R}$,
$$ lim_{varepsilonrightarrow0+}varepsilonoperatorname{Im} F_mu(x_0+ivarepsilon)= mu({x_0}) $$
I hope this helps answer your question, or at least provides you with good reference that discusses interesting results and historical notes about the Stieltjes transform.
Answered by Oliver Diaz on December 1, 2021
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