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

I'm trying to check a bound on Stieltjes transform of a probability measure, that's given in equation (2.92) on P. 170 in Terence Tao's notes "Topics in Random Matrix Theory". Denote the Stieltjes transform of the probability measure $$\mu$$ by $$s_{\mu}(z)$$. Then the bound mentioned in his notes is:

$$zs_{\mu}(z)= 1 + o_{\mu}(z)$$ as $$z= x+iy \to \infty$$ so that $$|\frac{x}{y}|$$ is bounded. N.B. here "$$o_{\mu}(z)$$" is a notation used to denote $$o(z)$$ but with highlighting the fact that $$z=x+iy\to \infty$$ with $$|x/y|$$ bounded, and that the convergence rate depends on $$\mu$$.

But all I'm getting, at least under a special case, is: under the same condition of convergence, mentioned just now, $$zs_{\mu}(z)= -1 + o_{\mu}(z)$$, which I demonstrate below.

For the special case that I'll treat, just assume that: $$\frac{x}{y}=K$$. But if you follow my computation below, you'll see that the end result wouldn't change in the limit if you assume $$|\frac{x}{y}|\leq K$$.

$$zs_{\mu}(z) = \int_{\mathbb{R}} \frac{z}{t-z}d\mu(t)=\int_{\mathbb{R}} \frac{z(t-\bar{z})}{|t-z|^2}d\mu(t)= \int_{\mathbb{R}}\frac{(tx - x^2 - y^2)+i(ty)} {(t-x)^2 + y^2 }d\mu(t)= \int_{\mathbb{R}}\frac{(\frac{x}{y}.\frac{t}{y}- (\frac{x}{y})^2-1) + i(\frac{t}{y})}{(\frac{x}{y})^2+1}d\mu(t)= -1 + \int_{\mathbb{R}}\frac{K+1}{K^2 + 1}\frac{t}{y}d\mu(t)= -1 + \frac{K+1}{K^2 + 1}.\frac{\mathbb{E}[\mathbb{I}]}{y}$$, where $$\mathbb{E}[\mathbb{I}]$$ is really the expectation of the identity function $$\mathbb{I}(t):=t$$ w.r.t. $$\mu$$. Assume it exists for now!

Note that, above, since $$z=x+iy \to \infty$$ but $$|x/y|$$ is bounded (actually I assumed that $$|x/y|$$ is constant to make things bit easy), we must have $$y \to \infty$$, yielding:

$$zs_{\mu}(z)= -1 + o_{\mu}(z)$$, disproving $$zs_{\mu}(z)= 1 + o_{\mu}(z)$$. Did I do something wrong in my calculation?

Also note that: if you take: $$\mu$$ to be the Dirac measure at $$0$$, i.e. $$\mu = \delta_0$$, then $$s_{\mu}(z)= -1/z$$, which does satisfy: $$zs_{\mu}(z)= -1 + o_{\mu}(z)$$, but not $$zs_{\mu}(z)= 1 + o_{\mu}(z)$$.

Thanks for taking a look!

• Tao defines $X = o(Y)$ by the requirement $|X|≤c(n)Y$. Note the modulus on the left side. I think the equation $s_\mu = \frac{1 + o_\mu (z)}{z}$ on P. 170 should accordingly be interpreted as $\vert z s_\mu (z) - 1 \vert = o_\mu (z)$. – Bruno Krams Dec 5 '19 at 18:47
• @BrunoKrams Thanks, but Tao writes on P.170: "...where $o_{\mu}(z)$ is an expression that, for any fixed $\mu$, goes to zero as $z\to\infty$ non-tangentially in the sense that $|Re(z)/Im(z)|$ is kept bounded, where the rate of convergence is allowed to depend on $\mu$". Did I miss something? Where did you find that definition of "o" in Tao's book? Also, does your comment contradict $zs_{\mu}(z) + 1 =o_{\mu}(z) ?$ Following Tao's notation of $o_{\mu}(z)$, I did interpret it exactly as you wrote: $|zs_{\mu}(z) - 1| =o_{\mu}(z)$. Still, the counterexamples I gave above are valid. – Mathmath Dec 5 '19 at 18:54
• The definition can be found on page 6. If the o-Notation is interpreted as I wrote, than what you gave is not a counterexample but in perfect agreement with the equation (2.92) from Tao's book. Also your calculation is - with slight modifications - a proof of (2.92) :) – Bruno Krams Dec 5 '19 at 19:06
• By the way on page 169 Tao himself gives the series expansion $$s_n(z) = - \frac{1}{z} - \frac{1}{z^2} \frac{1}{n} \operatorname{tr} M_n - ...$$ so it would be a very surprising mistake if he missed a sign in (2.92) on the next page. – Bruno Krams Dec 5 '19 at 19:12
• Sorry, I just looked at Page 6. He defined there something entirely different: he's talking about the dimension $n$ of random matrices there, which, in my question itself, is not relevant at all. My question concerns not random matrix, so you can ignore that definition of $o$, and consider that of $o_{\mu}$ that he writes on P.170, specially related to my questions. Also, could you please be so kind to point out why my calculation and counterexamples aligns with $zs_{\mu}(z)= 1 + o_{\mu}(z)$? In fact, I totally agree with what he wrote on P.169, because it validates my calculation. – Mathmath Dec 5 '19 at 19:19

I believe there is a minor typo in the definition of the Stieltjes transform on page 169 of those notes, namely instead of $$s_\mu(z):=\int_{\mathbb R}\frac{1}{x-z}d\mu(x),\qquad \Im z>0$$ the intended definition is $$s_\mu(z):=\int_{\mathbb R}\frac{1}{z-x}d\mu(x),\qquad \Im z>0.$$ 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 $$z_n\to\infty$$ non-tangentially, and let $$f_n(x)=(z_n-x)^{-1}$$. Then for all $$x\in\mathbb R$$ $$\lim_{n\to\infty}z_nf_n(x)=1,$$ thus by the dominated convergence theorem it follows that $$\lim_{n\to\infty}z_ns_{\mu}(z_n)=\lim_{n\to\infty}\int_{\mathbb R}z_nf_n(x)d\mu(x)=\int_{\mathbb R}\lim_{n\to \infty}z_nf_n(x)d\mu(x)=\int_{\mathbb R}d\mu(x)=1,$$ 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.
• (contd.) $\int_{\mathbb{R}}\frac{d\mu(t)}{t-z}$, because look at the equation right after "Whereas the moment method started from the identity (2.83), the Stieltjes transform method proceeds from the identity" on P.169: it'd be $\frac{1}{n}tr((zI - \frac{M_n}{\sqrt(n)})^{-1})$ had he defined the transform same as wiki's defintion (the one you used). Also in the analytic expression, see that the leading term is $\frac{-1}{z}$ in Tao's notes, which'd be $\frac{1}{z}$ , had he defined using the one you mentioned. But I guess we both agree that at least either way, there's a typo. – Mathmath Dec 9 '19 at 9:36