## Tuesday, December 22, 2015

### Examples of the Stieltjes transformation

I give the definition and two examples of the Stieltjes transformation.
If $f$ is a function with support $I \subseteq \mathbb{R}$, then for $z \in\mathbb{C} - I$, the Stieltjes transformation $S(z)$ is
\label{eq:20151219a}
S(z) = \int_{-\infty}^{+\infty} \frac{f(t)}{t-z} dt
The inverse of the Stieltjes transformation is $$\label{eq:20151219b} f(z) = \lim_{ \epsilon \overset{ > }{\to} 0} \frac{ S(z+ i \epsilon) - S(z- i \epsilon)}{2 \pi i}$$ I deduce \eqref{eq:20151219b} with a sloppy calculation: \begin{equation*}S(z+ i \epsilon) - S(z- i \epsilon) = 2 i \epsilon \int_{-\infty}^{+\infty} f(t) dt \frac{1}{(t-z)^2 + \epsilon^2} \end{equation*} Because \begin{equation*} \lim_{ \epsilon \overset{ > }{\to} 0} \frac{\epsilon}{x^2 + \epsilon^2} = \pi \delta(x) \end{equation*} I get \begin{equation*} \lim_{ \epsilon \stackrel{>}{\to} 0} \left( S(z+ i \epsilon) - S(z- i \epsilon)\right) = 2 i \int_{-\infty}^{+\infty} f(t) dt\ \pi \delta(t - z) = 2 i \pi f(z) \end{equation*}

It seems that the Stieltjes transformation is often used in probability theory. I therefore take the examples from probability theory. Example 1
The simplest probability distribution is the uniform distribution. Thus I take $f(t) = 1$ if $t \in [ 0,a ]$ and $f(t) = 0$ otherwise. Then one can calculate that \begin{equation*} S(z) = \log \left( \dfrac{z-a}{z} \right) \stackrel{\mathrm{def}}{=} \mathrm{Log}(z-a) - \mathrm{Log} z \end{equation*} with $\mathrm{Log}$ the logarithm with branch cut on $\left] - \infty, 0 \right]$ and $\mathrm{Log} 1 = 0$. With this definition, $\log \left( \dfrac{z-a}{z} \right)$ has a branch cut on the interval $[0,a ]$. If one moves from the lower half plane over the branch cut to the upper half plane, the value of $\log \left( \dfrac{z-a}{z} \right)$ jumps with the amount $2 \pi i$.
Here is a graph of the function $\log \left( \dfrac{z-a}{z} \right)$
 $\mathrm{Re} \left( \log \left( \dfrac{z-a}{z} \right) \right)$ for $a = 1$
 $\mathrm{Im} \left( \log \left( \dfrac{z-a}{z} \right) \right)$ for $a = 1$
Hence,

• If $z \not\in [ 0, a]$, then $S(z)$ is continuous, thus $\displaystyle\lim_{ \epsilon \stackrel{>}{\to} 0} \left( S(z+ i \epsilon) - S(z- i \epsilon)\right) = 0$
• If $z \in [ 0, a]$, then $\displaystyle\lim_{ \epsilon \stackrel{>}{\to} 0} \left( S(z+ i \epsilon) - S(z- i \epsilon)\right) = 2 \pi i$
which is indeed formula \eqref{eq:20151219b}

Example 2
The second example is motivated by Wigner's semicircle distribution from random matrix theory. Take  $f(t) = \sqrt{ 1 - t^2}$ if $t \in [ -1 , 1 ]$ and $f(t) = 0$ otherwise. By integrating the contour integral around the branch cut, one can calculate that \begin{equation*} S(z) = \int_{-1}^1 \frac{\sqrt{1 - t^2}}{t -z} dt = \pi \left( \sqrt{z - 1}\sqrt{z + 1} - z \right) \end{equation*} Here, in the right hand side, the complex square root $\sqrt{z}$ is defined with branch cut on the negative real axis. It then follows that $\sqrt{z - 1}\sqrt{z + 1}$ has a branch cut on $[-1 , 1]$. Also, one can see that if $x \in [-1,1]$ and $z = x + i \epsilon$ with $\epsilon > 0$, then $\sqrt{z^2 - 1} = i \sqrt{1 - x^2 }$.
Here is a graph of the function $\sqrt{z - 1}\sqrt{z + 1} - z$
 $\mathrm{Re} \left( \sqrt{z - 1}\sqrt{z + 1} - z \right)$
 $\mathrm{Im} \left( \sqrt{z - 1}\sqrt{z + 1} - z \right)$
Hence,

• If $z \not\in [ -1, 1]$, then $S(z)$ is continuous, thus $\displaystyle\lim_{ \epsilon \stackrel{>}{\to} 0} \left( S(z+ i \epsilon) - S(z- i \epsilon)\right) = 0$
• If $z \in [-1, 1]$, then \begin{equation*} \lim_{ \epsilon \stackrel{>}{\to} 0} \left( S(z+ i \epsilon) - S(z- i \epsilon)\right) = \pi \left( i \sqrt{1 - x^2 } - (- i) \sqrt{1 - x^2 }\right) = 2 \pi i \sqrt{1 - x^2 } \end{equation*}
which is again formula \eqref{eq:20151219b}

Remark: I decided to write this post because the Stieltjes transformation is used in random matrix theory. I wanted therefore to understand the Stieltjes transformation on some easy examples.