I'm trying to prove the following proposition taken from [Bercovici, Hari & Voiculescu, Dan] Free Convolution of Measures with Unbounded Support paper (Section 5, Prop. 5.1). Before enunciate the proposition let's define the $Upper\,cone$ with $\alpha>0$ width as $$ \Gamma_\alpha= \Big\{z\in\mathbb{C}\,:\,Im(z)>0\quad\text{&}\quad|Re(z)|<\alpha\,Im(z)\Big\}. $$ And recall the Cauchy transform given a probability measure $\mu$ on $\mathbb{R}$ being $$ G_\mu(z)=\int_\mathbb{R}\,\frac{1}{z-t}\,d\,\mu(t),\quad\forall z\notin\mathbb{R}. $$
Proposition 5.1: Let $G:\mathbb{C}^+\to\mathbb{C}^-$ be an analytic function. Then the following assertions are equivalent. $$ (i)\,\text{There exists a probability measure}\,\mu\text{ on }\mathbb{R}\text{ such that }G_\mu=G\,\text{ in }\,\mathbb{C}^+ $$ $$ (ii)\,\text{For each }\alpha>0 \text{ we have }\lim_{|z|\to\infty,\,\,z\in\Gamma_\alpha}zG(z)=1 $$ $$ (iii)\text{ We have } \lim_{y\to\infty}iyG(iy)=1 $$
I have already proved $(i)\Rightarrow\,(ii)\Rightarrow(iii)$. I got stucked trying $(iii)\Rightarrow(i)$.
First I define $\tilde{G}=-G$. So $\tilde{G}:\mathbb{C^+}\to\mathbb{C^+}$ and is still analytic. Next I'm using a representation for Nevanlinna functions that says that for every analytic function $f:\mathbb{C}^+\to\mathbb{C^+}$ we have the following representation $$ f(z)=az+b+\int_\mathbb{R}\,\frac{1+uz}{u-z}\,d\tau(u)\quad\forall z\in\mathbb{C^+}, $$ where $a>0,b\in\mathbb{R}$ are constants and $\tau(u)\in BV(\mathbb{R})$ is non-decreasing. So, since $\tilde{G}$ happens to be a Nevanlinna function we get that $$ G(z)=-az-b+\int_\mathbb{R}\,\frac{1+uz}{z-u}\,d\,\tau(u)\quad z\in\mathbb{C^+}. $$ Now I'm supposed to give a probability measure on $\mathbb{R}$, which I'm trying by denoting $d\mu(t)$ specifically, where a $d\tau(u)$ is involved. I have tried a series of variable changes trying $-az-b+\int_\mathbb{R}\,\frac{1+uz}{z-u}\,d\,\tau(u)=G_\mu(z)$ to hold, but haven't succeeded yet. Another thing I'm also missing is how $(iii)$ could help, since I've already examine $iyG(iy)$ and seems to doesn't give something clear about finding a probability measure. Any help would be appreciated.
UPDATE: Found that $$ \frac{1}{u-z}-\frac{u}{1+u^2}=\frac{1+uz}{(u-z)(1+u^2)} $$ so $$ \Big(\frac{1}{u-z}-\frac{u}{1+u^2}\Big)(1+u^2)=\frac{1+uz}{(u-z)} $$ so $$ \int_\mathbb{R}\frac{1+uz}{(u-z)}\,d\,\tau (u) = \int_\mathbb{R}\Big(\frac{1}{u-z}-\frac{u}{1+u^2}\Big)(1+u^2)\,d\,\tau (u) $$ So it seems that $d\mu(u)=(1+u^2)\,d\,\tau (u)$ should be the probability measure I'm looking for. In this way $$ G(z)=-az-b+G_\mu (z)-\int_\mathbb{R}\,u\,d\,\tau (u) $$ So now it suffices to show that $az+b+\int_\mathbb{R}\,u\,d\tau (u)=0$, which happens iff $a=0$ and $b=-\int_\mathbb{R}\,u\,d\tau (u)$. So now I think here is where the fact that $iyG(iy)\to 1$ helps. What do you think? Any help would be appreciated.
I got it. First $$ iyG(iy)=iy\Big(-iay-b+\int_\mathbb{R}\frac{1}{iy-u}d\mu(u)-\int_\mathbb{R}ud\tau(u)\Big) $$ $$ =iy\Big(-iay-b+\int_\mathbb{R}\frac{u}{y^2+u^2}d\mu(u)+i\int_\mathbb{R}\frac{y}{y^2+u^2}d\mu(u)-\int_\mathbb{R}ud\tau(u)\Big) $$ $$ =ay^2-iby+i\int_\mathbb{R}\frac{yu}{y^2+u^2}d\mu(u)-\int_\mathbb{R}\frac{y^2}{y^2+u^2}d\mu(u)-i\int_\mathbb{R}yu\,d\tau(u) $$ $$ =A+iB, $$ where $$ A_y=y^2\Big(a-\int_\mathbb{R}\frac{1}{y^2+u^2}\,d\tau(u)\Big) $$ $$ B_y=-y\Big(b+\int_\mathbb{R}u\,d\tau(u)-\int_\mathbb{R}\frac{u}{y^2+u^2}\,d\tau(u)\Big). $$ And since $|iyG(iy)|$ is bounded for all $y\ge 0$, so do are $|A_y|,|B_y|$, for all $y\ge 0$. Thus $$ \lim_{y\to\infty}\frac{1}{|y|^2}\,|A_y|=0 $$ $$ \lim_{y\to\infty}\frac{1}{|y|}\,|B_y|=0 $$ iff $$ a-\lim_{y\to\infty}\frac{1}{y^2}\int_\mathbb{R}\frac{1}{1+(u/y)^2}\,d\tau(u)=0 $$ $$ b+\int_\mathbb{R}\,u\,d\tau(u)-\lim_{y\to\infty}\frac{1}{y^2}\int_\mathbb{R}\frac{u}{1+(u/y)^2}\,d\tau(u)=0, $$ where $\frac{u}{1+(u/y)^2}$ and $\frac{1}{1+(u/y)^2}$ are bounded, thus dominated, for every $y\ge 0$, so $$ \lim_{y\to\infty}\frac{1}{y^2}\int_\mathbb{R}\frac{u}{1+(u/y)^2}\,d\tau(u)=0 $$ $$ \lim_{y\to\infty}\frac{1}{y^2}\int_\mathbb{R}\frac{1}{1+(u/y)^2}\,d\tau(u)=0. $$ Thereby $a=0$ and $b=-\int_\mathbb{R}\,u\,d\tau(u)$. So $G(z)=G_\mu(z)$, for all $z\in\mathbb{C}^+$.
That $\mu(u)$ is a probability measure follows from the fact that $\lim_{y\to\infty}|iyG(iy)|=1$, because $$ \lim_{y\to\infty}|iyG(iy)|=1 $$ iff $$ \lim_{y\to\infty}\Big|-\int_\mathbb{R}\frac{1}{1+(u/y)^2}\,d\mu(u)+i\frac{1}{y}\int_\mathbb{R}\frac{u}{1+(u/y)^2}\,d\mu(u)\Big|=1 $$ iff $$ \lim_{y\to\infty}\int_\mathbb{R}\frac{1}{1+(u/y)^2}\,d\mu(u) + \lim_{y\to\infty}\frac{1}{y}\int_\mathbb{R}\frac{u}{1+(u/y)^2}\,d\mu(u) =1 $$ iff $$ \int_\mathbb{R}\,d\mu(u)=1 $$