Checking a bound on the Stieltjes transform from Terence Tao's notes

This may be a little late but maybe this can help:

Though this is not part of the OP, it is mentioned in T. Tao's notes and I will show a short proof of some facts:

• Analyticity of the Stieltjes transform:

Let $mu$ be a complex measure on $(mathbb{R},mathscr{B}(mathbb{R})$, and suppose $operatorname{sup}(mu)=Omega$ so that $D=mathbb{C}setminusOmega$ is open.

Then, the map $f:Drightarrowmathbb{C}$ given by $$ begin{align} f(z)= int_Omegafrac{mu(domega)}{omega-z}tag{1}label{one} end{align} $$ is analytic. Moreover, if the closed ball $overline{B}(a;r)subset D$, then $$ begin{align} f(z)=sum^infty_{n=0}c_n(z-a)^n,qquad zin B(a;r)tag{2}label{two} end{align} $$ where $$ begin{align} c_n=int_Omegafrac{mu(domega)}{(omega-a)^{n+1}},qquad |c_n|leq frac{|mu|_{TV}}{r^{n+1}},qquad ninmathbb{Z}_+.tag{3}label{three} end{align} $$ If $R$ is the radius of convergence of $eqref{two}$, then $rleq R$. Here is a short proof:

If $overline{B}(a;r)subset D$, then $q:=inf_{omegainOmega}|omega-a|>r$, and so $$ begin{align} frac{|z-a|}{|omega-a|}leq frac{|z-a|}{q}leqfrac{r}{q}<1,qquad omegainOmega,quad zin B(a;r). end{align} $$ Hence, for any $zin B(a;r)$ fixed, the series $$ omegamapsto sum^infty_{n=0}frac{(z-a)^n}{(omega-a)^{n+1}} =frac{1}{omega-z} $$ converges absolutely and uniformly in $Omega$. By dominated convergence (to justify change of order of summation and integration) $$ f(z)=int_Omega frac{mu(domega)}{omega-z}= int_Omega sum^infty_{n=0}frac{(z-a)^n}{(omega-a)^{n+1}} ,mu(domega) = sum^infty_{n=0}c_n(z-a)^n, $$ where the $c_n$ satisfy $eqref{three}$. The last statement follows from the estimate $$begin{align} limsup_{nrightarrowinfty}sqrt[n]{|c_n|}leqlim_{nrightarrowinfty}frac{1}{r}sqrt[n]{frac{|mu|_{TV}}{r}}=frac{1}{r} end{align} $$

• Non tangential asymptotyics. Suppose $|z|rightarrowinfty$ over a region of the form $T_M={z=x+iyinmathbb{C}:|x|leq M|y|}setminusOmega$ for some fixed $M>0$. Then $$ Big|frac{z}{omega-z}Big|leqfrac{|z|}{||z|-|omega||}xrightarrow{|z|rightarrowinfty}1 $$ for each $omegainOmega$, and $$ Big|frac{z}{omega-z}Big|leqfrac{|x|+|y|}{|y|}leq M+1 $$ for all $omegainOmega$. Then, by dominated convergence, letting $zrightarrowinfty$ over the region $T_M$ $$ zf(z)=int_Omegafrac{z}{omega-z},mu(domega)xrightarrow{|z|rightarrowinfty}mu(Omega) $$

• If $mu$ has compact support, then the asymptotic may be improved considerably . Let $M=sup{|omega|:omegain Omega}$. Then $$ frac{1}{omega-z}=-frac{1}{z}frac{1}{1-tfrac{omega}{z}}=-frac{1}{z}sum^infty_{k=0}frac{omega^n}{z^n} $$ By dominated convergence

$$ f(z)=-frac{1}{z}sum^infty_{k=0}z^{-n}int_Omega omega^n,mu(domega) $$

Hence

$$zf(z)=mu(Omega) +z^{-1}int_Omega omega,mu(domega) +o_mu(z^{-1}) $$

I believe there is a minor typo in the definition of the Stieltjes transform on page 169 of those notes, namely instead of $$ s_mu(z):=int_{mathbb R}frac{1}{x-z}dmu(x),qquad Im z>0 $$ the intended definition is $$ s_mu(z):=int_{mathbb R}frac{1}{z-x}dmu(x),qquad Im z>0. $$ After fixing the sign mistake caused by swapping $x$ and $z$, this matches the usual definition and also allows a straightforward justification of $(2.92)$ on page 170 as follows. Let $z_n$ be any sequence of complex numbers with $Im(z_n)>0$ such that $z_ntoinfty$ non-tangentially, and let $f_n(x)=(z_n-x)^{-1}$. Then for all $xinmathbb R$ $$ lim_{ntoinfty}z_nf_n(x)=1, $$ thus by the dominated convergence theorem it follows that $$ lim_{ntoinfty}z_ns_{mu}(z_n)=lim_{ntoinfty}int_{mathbb R}z_nf_n(x)dmu(x)=int_{mathbb R}lim_{nto infty}z_nf_n(x)dmu(x)=int_{mathbb R}dmu(x)=1, $$ which in Tao's notation is equivalent to stating that $zs_{mu}(z)=1+o_{mu}(1)$, as claimed.

Note that the applicability of the dominated convergence theorem is justified in the exact same manner as in the observation $(2.91)$ on page 170, since the integrand satisfies $$ |z_nf_n(x)|leq frac{|z_n|}{|Im(z_n)|}leq 1. $$ In other words, we are actually using the special case of the dominated convergence theorem known as the bounded convergence theorem.