# Let’s Read: Sendov’s conjecture in high degree, part 4: details of case one

In this proof we (finally!) finish the proof of case one.

As usual, we throughout fix a nonstandard natural ${n}$ and a complex polynomial of degree ${n}$ whose zeroes are all in ${\overline{D(0, 1)}}$. We assume that ${a}$ is a zero of ${f}$ whose standard part is ${1}$, and assume that ${f}$ has no critical points in ${\overline{D(a, 1)}}$. Let ${\lambda}$ be a random zero of ${f}$ and ${\zeta}$ a random critical point. Under these circumstances, ${\lambda^{(\infty)}}$ is uniformly distributed on ${\partial D(0, 1)}$ and ${\zeta^{(\infty)}}$ is almost surely zero. In particular,

$\displaystyle \mathbf E \log\frac{1}{|\lambda|}, \mathbf E \log |\zeta - a| = O(n^{-1})$

and ${\zeta}$ is infinitesimal in probability, hence infinitesimal in distribution. Let ${\mu}$ be the expected value of ${\zeta}$ (thus also of ${\lambda}$) and ${\sigma^2}$ its variance. I think we won’t need the nonstandard-exponential bound ${\varepsilon_0^n}$ this time, as its purpose was fulfilled last time.

Last time we reduced the proof of case one to a sequence of lemmata. We now prove them.

1. Preliminary bounds

Lemma 1 Let ${K \subseteq \mathbf C}$ be a compact set. Then

$\displaystyle f(z) - f(0), ~f'(z) = O((|z| + o(1))^n)$

uniformly for ${z \in K}$.

Proof: It suffices to prove this for a compact exhaustion, and thus it suffices to assume

$\displaystyle K = \overline{D(0, R)}.$

By underspill, it suffices to show that for every standard ${\varepsilon > 0}$ we have

$\displaystyle |f(z) - f(0)|, ~|f'(z)| \leq C(|z| + \varepsilon)^n.$

We first give the proof for ${f'}$.

First suppose that ${\varepsilon < |z| \leq R}$. Since ${\zeta}$ is infinitesimal in distribution,

$\displaystyle \mathbf E \log |z - \zeta| \leq \mathbf E \log \max(|z - \zeta|, \varepsilon/2) \leq \log \max(|z|, \varepsilon/2) + o(1);$

here we need the ${\varepsilon/2}$ and the ${R}$ since ${\log |z - \zeta|}$ is not a bounded continuous function of ${\zeta}$. Since ${\varepsilon < |z|}$ we have

$\displaystyle \mathbf E \log |z - \zeta| \leq \log |z| + o(1)$

but we know that

$\displaystyle -\frac{\log n}{n - 1} - \frac{1}{n - 1} \log |f'(z)| = U_\zeta(z) = -\mathbf E \log |z - \zeta|$

so, solving for ${\log |f'(z)|}$, we get

$\displaystyle \log |f'(z)| \leq (n - 1) \log |z| + o(n);$

we absorbed a ${\log n}$ into the ${o(n)}$. That gives

$\displaystyle |f'(z)| \leq e^{o(n)} |z|^{n-1}.$

Since ${f'}$ is a polynomial of degee ${n - 1}$ and ${f}$ is monic (so the top coefficient of ${f'}$ is ${n}$) this gives a bound

$\displaystyle |f'(z)| \leq e^{o(n)} (|z| + \varepsilon)^{n - 1}$

even for ${|z| \leq \varepsilon}$.

Now for ${f}$, we use the bound

$\displaystyle |f(z) - f(0)| \leq \max_{|w| < |z|} |f'(w)|$

to transfer the above argument. $\Box$

2. Uniform convergence of ${\zeta}$

Lemma 2 There is a standard compact set ${S \subseteq \overline{D(0, 1)}}$ and a standard countable set ${T \subseteq \overline{D(0, 1)} \setminus \overline{D(1, 1)}}$ such that

$\displaystyle S = (\overline{D(0, 1)} \cap \partial D(1, 1)) \cup T,$

all elements of ${T}$ are isolated in ${S}$, and ${||\zeta - S||_{L^\infty}}$ is infinitesimal.

Tao claims

$\displaystyle \mathbf P(|\zeta - a| \geq \frac{1}{2m}) = O(n^{-1})$

where ${m}$ is a large standard natural, which makes no sense since the left-hand side should be large (and in particular, have positive standard part). I think this is just a typo though.

Proof: Since ${\zeta}$ was assumed far from ${a = 1 - o(1)}$ we have

$\displaystyle \zeta \in \overline{D(0, 1)} \setminus D(1, 1 - o(1)).$

We also have

$\displaystyle \mathbf E \log |\zeta - a| = O(n^{-1})$

so for every standard natural ${m}$ there is a standard natural ${k_m}$ such that

$\displaystyle \mathbf P(\log |\zeta - a| \geq \frac{1}{2m}) \leq \frac{k_m}{n}.$

Multiplying both sides by ${n}$ we see that

$\displaystyle \text{card } Z \cap K_m = \text{card } Z \cap \{\zeta_0 \in \overline{D(0, 1)}: \log |\zeta_0 - a| \geq \frac{1}{2m}\} \leq k_m$

where ${Z}$ is the variety of critical points ${f' = 0}$. Let ${T_m}$ be the set of standard parts of zeroes in ${K_m}$; then ${T_m}$ has cardinality ${\leq k_m}$ and so is finite. For every zero ${\zeta_0 \in Z}$, either

1. For every ${m}$,

$\displaystyle |\zeta_0 - a| < \exp\left(\frac{1}{2m}\right)$

so the standard part of ${|\zeta_0 - a|}$ is ${1}$, or

2. There is an ${m}$ such that ${d(\zeta_0, T_m)}$ is infinitesimal.

So we may set ${T = \bigcup_m T_m}$; then ${T}$ is standard and countable, and does not converge to a point in ${\partial D(1, 1)}$, so ${S}$ is standard and ${||\zeta - S||_{L^\infty}}$ is infinitesimal.

I was a little stumped on why ${S}$ is compact; Tao doesn’t prove this. It turns out it’s obvious, I was just too clueless to see it. The construction of ${T}$ forces that for any ${\varepsilon > 0}$, there are only finitely many ${z \in T}$ with ${|z - \partial D(1, 1)| \geq \varepsilon}$, so if ${T}$ clusters anywhere, then it can only cluster on ${\partial D(1, 1)}$. This gives the desired compactness. $\Box$

The above proof is basically just the proof of Ascoli’s compactness theorem adopted to this setting and rephrased to replace the diagonal argument (or 👏 KEEP 👏 PASSING 👏 TO 👏 SUBSEQUENCES 👏) with the choice of a nonstandard natural. I think the point is that, once we have chosen a nontrivial ultrafilter on ${\mathbf N}$, a nonstandard function is the same thing as sequence of functions, and the ultrafilter tells us which subsequences of reals to pass to.

3. Approximating ${f,f'}$ outside of ${S}$

We break up the approximation lemma into multiple parts. Let ${K}$ be a standard compact set which does not meet ${S}$. Given a curve ${\gamma}$ we denote its arc length by ${|\gamma|}$; we always assume that an arc length does exist.

A point which stumped me for a humiliatingly long time is the following:

Lemma 3 Let ${z, w \in K}$. Then there is a curve ${\gamma}$ from ${z}$ to ${w}$ which misses ${S}$ and satisfies the uniform estimate

$\displaystyle |z - w| \sim |\gamma|.$

Proof: We use the decomposition of ${S}$ into the arc

$\displaystyle S_0 = \partial D(1, 1) \cap \overline{D(0, 1)}$

and the discrete set ${T}$. We try to set ${\gamma}$ to be the line segment ${[z, w]}$ but there are two things that could go wrong. If ${[z, w]}$ hits a point of ${T}$ we can just perturb it slightly by an error which is negligible compared to ${[z, w]}$. Otherwise we might hit a point of ${S_0}$ in which case we need to go the long way around. However, ${S_0}$ and ${K}$ are compact, so we have a uniform bound

$\displaystyle \max(\frac{1}{|z - S_0|}, \frac{1}{|w - S_0|}) = O(1).$

Therefore we can instead consider a curve ${\gamma}$ which goes all the way around ${S_0}$, leaving ${D(0, 1)}$. This curve has length ${O(1)}$ for ${z, w}$ close to ${S_0}$ (and if ${z, w}$ are far from ${S_0}$ we can just perturb a line segment without generating too much error). Using our uniform max bound above we see that this choice of ${\gamma}$ is valid. $\Box$

Recall that the moments ${\mu,\sigma}$ of ${\zeta}$ are infinitesimal.

Since ${||\zeta - S||_{L^\infty}}$ is infinitesimal, and ${K}$ is a positive distance from any infinitesimals (since it is standard compact), we have

$\displaystyle |z - \zeta|, |z - \mu| \sim 1$

uniformly in ${z}$. Therefore ${f}$ has no critical points near ${K}$ and so ${f''/f'}$ is holomorphic on ${K}$.

We first need a version of the fundamental theorem.

Lemma 4 Let ${\gamma}$ be a contour in ${K}$ of length ${|\gamma|}$. Then

$\displaystyle f'(\gamma(1)) = f'(\gamma(0)) \left(\frac{\gamma(1) - \mu}{\gamma(0) - \mu}\right)^{n - 1} e^{O(n) |\gamma| \sigma^2}.$

Proof: Our bounds on ${|z - \zeta|}$ imply that we can take the Taylor expansion

$\displaystyle \frac{1}{z - \zeta} = \frac{1}{z - \mu} + \frac{\zeta - \mu}{(z - \mu)^2} + O(|\zeta - \mu|^2)$

of ${\zeta}$ in terms of ${\mu}$, which is uniform in ${\zeta}$. Taking expectations preserves the constant term (since it doesn’t depend on ${\zeta}$), kills the linear term, and replaces the quadratic term with a ${\sigma^2}$, thus

$\displaystyle s_\zeta(z) = \frac{1}{z - \mu} + O(\sigma^2).$

At the start of this series we showed

$\displaystyle f'(\gamma(1)) = f'(\gamma(0)) \exp\left((n-1)\int_\gamma s_\zeta(z) ~dz\right).$

Plugging in the Taylor expansion of ${s_\zeta}$ we get

$\displaystyle f'(\gamma(1)) = f'(\gamma(0)) \exp\left((n-1)\int_\gamma \frac{dz}{z - \zeta}\right) e^{O(n) |\gamma| \sigma^2}.$

Simplifying the integral we get

$\displaystyle \exp\left((n-1)\int_\gamma \frac{dz}{z - \zeta}\right) = \left(\frac{\gamma(1) - \mu}{\gamma(0) - \mu}\right)^{n - 1}$

whence the claim. $\Box$

Lemma 5 Uniformly for ${z,w \in K}$ one has

$\displaystyle f'(w) = (1 + O(n|z - w|\sigma^2 e^{o(n|z - w|)})) \frac{(w - \mu)^{n-1}}{(z - \mu)^{n - 1}}f'(z).$

Proof: Applying the previous two lemmata we get

$\displaystyle f'(w) = e^{O(n|z - w|\sigma^2)} \frac{(w - \mu)^{n-1}}{(z - \mu)^{n - 1}}f'(z).$

It remains to simplify

$\displaystyle e^{O(n|z - w|\sigma^2)} = 1 + O(n|z - w|\sigma^2 e^{o(n|z - w|)}).$

Taylor expanding ${\exp}$ and using the self-similarity of the Taylor expansion we get

$\displaystyle e^z = 1 + O(|z| e^{|z|})$

which gives that bound. $\Box$

Lemma 6 Let ${\varepsilon > 0}$. Then

$\displaystyle f(z) = f(0) + \frac{1 + O(\sigma^2)}{n} f'(z) (z - \mu) + O((\varepsilon + o(1))^n).$

uniformly in ${z \in K}$.

Proof: We may assume that ${\varepsilon}$ is small enough depending on ${K}$, since the constant in the big-${O}$ notation can depend on ${K}$ as well, and ${\varepsilon}$ only appears next to implied constants. Now given ${z}$ we can find ${\gamma}$ from ${z}$ to ${\partial B(0, \varepsilon)}$ which is always moving at a speed which is uniformly bounded from below and always moving in a direction towards the origin. Indeed, we can take ${\gamma}$ to be a line segment which has been perturbed to miss the discrete set ${T}$, and possibly arced to miss ${S_0}$ (say if ${z}$ is far from ${D(0, 1)}$). By compactness of ${K}$ we can choose the bounds on ${\gamma}$ to be not just uniform in time but also in space (i.e. in ${K}$), and besides that ${\gamma}$ is a curve through a compact set ${K'}$ which misses ${S}$. Indeed, one can take ${K'}$ to be a closed ball containing ${K}$, and then cut out small holes in ${K'}$ around ${T}$ and ${S_0}$, whose radii are bounded below since ${K}$ is compact. Since the moments of ${\zeta}$ are infinitesimal one has

$\displaystyle \int_\gamma (w - \mu)^{n-1} ~dw = \frac{(z - \mu)^n}{n} - \frac{\varepsilon^n e^{in\theta}}{n} = \frac{(z - \mu)^n}{n} - O((\varepsilon + o(1))^n).$

Here we used ${\varepsilon < 1}$ to enforce

$\displaystyle \varepsilon^n/n = O(\varepsilon^n).$

By the previous lemma,

$\displaystyle f'(w) = (1 + O(n|z - w|\sigma^2 e^{o(n|z - w|)})) \frac{(w - \mu)^{n-1}}{(z - \mu)^{n - 1}}f'(z).$

Integrating this result along ${\gamma}$ we get

$\displaystyle f(\gamma(0)) = f(\gamma(1)) - \frac{f'(\gamma(0))}{(\gamma(0) - \mu)^{n-1}} \left(\int_\gamma (w - \mu)^{n-1} ~dw + O\left(n\sigma^2 \int_\gamma|\gamma(0) - w| e^{o(n|\gamma(0) - w|)}|w - \mu|^{n-1}~dw \right) \right).$

Applying our preliminary bound, the previous paragraph, and the fact that ${|\gamma(1)| = \varepsilon}$, thus

$\displaystyle f(\gamma(1)) = f(0) + O((\varepsilon + o(1))^n),$

we get

$\displaystyle f(z) = f(0) + O((\varepsilon + o(1))^n) - \frac{f'(z)}{(z - \mu)^{n-1}} \left(\frac{(z - \mu)^n}{n} - O((\varepsilon + o(1))^n) + O\left(n\sigma^2 \int_\gamma|z - w| e^{o(n|z - w|)}|w - \mu|^{n-1}~dw \right)\right).$

We treat the first term first:

$\displaystyle \frac{f'(z)}{(z - \mu)^{n-1}} \frac{(z - \mu)^n}{n} = \frac{1}{n} f'(z) (z - \mu).$

For the second term, ${z \in K}$ while ${\mu^{(\infty)} \in K}$, so ${|z - \mu|}$ is bounded from below, whence

$\displaystyle \frac{f'(z)}{(z - \mu)^{n-1}} O((\varepsilon + o(1))^n) = O((\varepsilon + o(1))^n).$

Thus we simplify

$\displaystyle f(z) = f(0) + O((\varepsilon + o(1))^n) + \frac{1}{n} f'(z) (z - \mu) + \frac{f'(z)}{(z - \mu)^{n-1}} O\left(n\sigma^2 \int_\gamma|z - w| e^{o(n|z - w|)}|w - \mu|^{n-1}~dw \right).$

It will be convenient to instead write this as

$\displaystyle f(z) = f(0) + O((\varepsilon + o(1))^n) + \frac{1}{n} f'(z) (z - \mu) + O\left(n|f'(z)|\sigma^2 \int_\gamma|z - w| e^{o(n|z - w|)} \left|\frac{w - \mu}{z - \mu}\right|^{n-1}~dw \right).$

Now we deal with the pesky integral. Since ${\gamma}$ is moving towards ${\partial B(0, \varepsilon)}$ at a speed which is bounded from below uniformly in “spacetime” (that is, ${K \times [0, 1]}$), there is a standard ${c > 0}$ such that if ${w = \gamma(t)}$ then

$\displaystyle |w - \mu| \leq |z - \mu| - ct$

since ${\gamma}$ is going towards ${\mu}$. (Tao’s argument puzzles me a bit here because he claims that the real inner product ${\langle z - w, z\rangle}$ is uniformly bounded from below in spacetime, which seems impossible if ${w = z}$. I agree with its conclusion though.) Exponentiating both sides we get

$\displaystyle \left|\frac{w - \mu}{z - \mu}\right|^{n-1} = O(e^{-nct})$

which bounds

$\displaystyle f(z) = f(0) + O((\varepsilon + o(1))^n) + \frac{1}{n} f'(z) (z - \mu) + O\left(n|f'(z)|\sigma^2 \int_0^1 te^{-(c-o(1))nt} ~dt\right).$

Since ${c}$ is standard, it dominates the infinitesimal ${o(1)}$, so after shrinking ${c}$ a little we get a new bound

$\displaystyle f(z) = f(0) + O((\varepsilon + o(1))^n) + \frac{1}{n} f'(z) (z - \mu) + O\left(n|f'(z)|\sigma^2 \int_0^1 te^{-cnt} ~dt\right).$

Since ${n\int_0^1 te^{-cnt} ~dt}$ is exponentially small in ${n}$, in particular it is smaller than ${O(n^{-1})}$. Plugging in everything we get the claim. $\Box$

4. Control on zeroes away from ${S}$

After the gargantuan previous section, we can now show the “approximate level set” property that we discussed last time.

Lemma 7 Let ${K}$ be a standard compact set which misses ${S}$ and ${\varepsilon > 0}$ standard. Then for every zero ${\lambda_0 \in K}$ of ${f}$,

$\displaystyle U_\zeta(\lambda) = \frac{1}{n} \log \frac{1}{|f(0)|} + O(n^{-1}\sigma^2 + (\varepsilon + o(1))^n).$

Last time we showed that this implies

$\displaystyle U_\zeta(\lambda_0) = U_\zeta(a) + O(n^{-1}\sigma^2 + (\varepsilon + o(1))^n).$

Thus all the zeroes of ${f}$ either live in ${S}$ or a neighborhood of a level set of ${U_\zeta}$. Proof: Plugging in ${z = \lambda_0}$ in the approximation

$\displaystyle f(z) = f(0) + \frac{1 + O(\sigma^2)}{n} f'(z) (z - \mu) + O((\varepsilon + o(1))^n)$

we get

$\displaystyle f(0) + \frac{1 + O(\sigma^2)}{n} f'(\lambda_0) (\lambda_0 - \mu) = O((\varepsilon + o(1))^n).$

Several posts ago, we proved ${|f(0)| \sim 1}$ as a consequence of Grace’s theorem, so ${f(0)O((\varepsilon + o(1))^n) = O((\varepsilon + o(1))^n)}$. In particular, if we solve for ${f'(\lambda_0)}$ we get

$\displaystyle \frac{|f'(\lambda_0)}{n} |\lambda_0 - \mu| = |f(0)| (1 + O(\sigma^2 + (\varepsilon + o(1))^n).$

Using

$\displaystyle U_\zeta(z) = -\frac{\log n}{n - 1} - \frac{1}{n - 1} \log |f'(z)|,$

plugging in ${z = \lambda_0}$, and taking logarithms, we get

$\displaystyle -\frac{n - 1}{n} U_\zeta(\lambda_0) + \frac{1}{n} \log | \lambda_0 - \mu| = \frac{1}{n} \log |f(0)| + O(n^{-1}\sigma^2 + (\varepsilon + o(1))^n).$

Now ${\lambda_0 \in K}$ and ${K}$ misses the standard compact set ${S}$, so since ${0 \in S}$ we have

$\displaystyle |\lambda - \zeta|, |\lambda - \mu| \sim 1$

(since ${\zeta^{(\infty)} \in S}$ and ${\mu}$ is infinitesimal). So we can Taylor expand in ${\zeta}$ about ${\mu}$:

$\displaystyle \log |\lambda_0 - \zeta| = \log |\lambda_0 - \mu| - \text{Re }\frac{\zeta - \mu}{\lambda_0 - \mu} + O(\sigma^2).$

Taking expectations and using ${\mathbf E \zeta - \mu}$,

$\displaystyle -U_\zeta(\lambda_0) = \log |\lambda_0 - \mu| + O(\sigma^2).$

Plugging in ${\log |\lambda_0 - \mu|}$ we see the claim. $\Box$

I’m not sure who originally came up with the idea to reason like this; I think Tao credits M. J. Miller. Whoever it was had an interesting idea, I think: ${f = 0}$ is a level set of ${f}$, but one that a priori doesn’t tell us much about ${f'}$. We have just replaced it with a level set of ${U_\zeta}$, a function that is explicitly closely related to ${f'}$, but at the price of an error term.

5. Fine control

We finish this series. If you want, you can let ${\varepsilon > 0}$ be a standard real. I think, however, that it will be easier to think of ${\varepsilon}$ as “infinitesimal, but not as infinitesimal as the term of the form o(1)”. In other words, ${1/n}$ is smaller than any positive element of the ordered field ${\mathbf R(\varepsilon)}$; briefly, ${1/n}$ is infinitesimal with respect to ${\mathbf R(\varepsilon)}$. We still reserve ${o(1)}$ to mean an infinitesimal with respect to ${\mathbf R(\varepsilon)}$. Now ${\varepsilon^n = o(1)}$ by underspill, since this is already true if ${\varepsilon}$ is standard and ${0 < \varepsilon < 1}$. Underspill can also be used to transfer facts at scale ${\varepsilon}$ to scale ${1/n}$. I think you can formalize this notion of “iterated infinitesimals” by taking an iterated ultrapower of ${\mathbf R}$ in the theory of ordered rings.

Let us first bound ${\log |a|}$. Recall that ${|a| \leq 1}$ so ${\log |a| \leq 0}$ but in fact we can get a sharper bound. Since ${T}$ is discrete we can get ${e^{-i\theta}}$ arbitrarily close to whatever we want, say ${-1}$ or ${i}$. This will give us bounds on ${1 - a}$ when we take the Taylor expansion

$\displaystyle \log|a| = -(1 - a)(1 + o(1)).$

Lemma 8 Let ${e^{i\theta} \in \partial D(0, 1) \setminus S}$ be standard. Then

$\displaystyle \log |a| \leq \text{Re } ((1 - e^{-i\theta} + o(1))\mu) - O(|\mu|^2 + \sigma^2 + (\varepsilon + o(1))^n).$

Proof: Let ${K}$ be a standard compact set which misses ${S}$ and ${\lambda_0 \in K}$ a zero of ${f}$. Since ${\zeta \notin K}$ (since ${S}$ is close to ${\zeta}$) and ${|a-\zeta|}$ has positive standard part (since ${d(a, S) = 1}$) we can take Taylor expansions

$\displaystyle -\log |\lambda_0 - \zeta| = -\log |\lambda_0| + \text{Re } \frac{\zeta}{\lambda_0} + O(|\zeta|^2)$

and

$\displaystyle -\log |a - \zeta| = -\log|a| + \text{Re } \frac{\zeta}{a} + O(|\zeta|^2)$

in ${\zeta}$ about ${0}$. Taking expectations we have

$\displaystyle U_\zeta(\lambda_0) = -\log |\lambda_0| + \text{Re } \frac{\mu}{\lambda_0} + O(\mathbf E |\zeta|^2)$

and similarly for ${a}$. Thus

$\displaystyle -\log |a| + \text{Re } \frac{\mu}{a} = -\log |\lambda_0| + \text{Re } \frac{\mu}{\lambda_0} + O(\mathbf E |\zeta|^2 + n^{-1}\sigma^2 + (\varepsilon + o(1))^n)$

since

$\displaystyle U_\zeta(\lambda_0) - U_\zeta(a) = O(n^{-1}\sigma^2 + (\varepsilon + o(1))^n).$

Since

$\displaystyle \mathbf E|\zeta|^2 = |\mu|^2 + \sigma^2$

we have

$\displaystyle -\log|\lambda_0| + \text{Re } \left(\frac{1}{\lambda_0} - \frac{1}{a}\right)\mu = -\log|a| + O(|\mu|^2 + \sigma^2 + (\varepsilon + o(1))^n).$

Now ${|\lambda_0| \leq 1}$ so ${-\log |\lambda_0| \geq 0}$, whence

$\displaystyle \text{Re } \left(\frac{1}{\lambda_0} - \frac{1}{a}\right)\mu \geq -\log|a| + O(|\mu|^2 + \sigma^2 + (\varepsilon + o(1))^n).$

Now recall that ${\lambda^{(\infty)}}$ is uniformly distributed on ${\partial D(0, 1)}$, so we can choose ${\lambda_0}$ so that

$\displaystyle |\lambda_0 - e^{i\theta}| = o(1).$

Thus

$\displaystyle \frac{1}{\lambda_0} - \frac{1}{a} = 1 - e^{-i\theta} + o(1)$

which we can plug in to get the claim. $\Box$

Now we prove the first part of the fine control lemma.

Lemma 9 One has

$\displaystyle \mu, 1 - a = O(\sigma^2 + (\varepsilon + o(1))^n).$

Proof: Let ${\theta_+ \in [0.98\pi, 0.99\pi],\theta_- \in [1.01\pi, 1.02\pi]}$ be standard reals such that ${e^{i\theta_\pm} \notin S}$. I don’t think the constants here actually matter; we just need ${0 < 0.01 < 0.02 < \pi/8}$ or something. Anyways, summing up two copies of the inequality from the previous lemma with ${\theta = \theta_\pm}$ we have

$\displaystyle 1.9 \text{Re } \mu \geq \text{Re } ((1 + e^{-i\theta_+} + 1 + e^{-i\theta_-} + o(1))\mu) \geq \log |a| + O(|\mu|^2 + \sigma^2 + (\varepsilon + o(1))^n)$

since

$\displaystyle 2 + e^{-i\theta_+} + e^{-i\theta_-} + o(1) \leq 1.9.$

That is,

$\displaystyle \text{Re } \mu \geq \frac{\log|a|}{1.9} + O(|\mu|^2 + \sigma^2 + (\varepsilon + o(1))^n).$

Indeed,

$\displaystyle -\log |a| = (1 - a)(1 + o(1)),$

so

$\displaystyle \text{Re }\mu \geq -\frac{1 - a}{1.9 + o(1)} + O(|\mu|^2 + \sigma^2 + (\varepsilon + o(1))^n).$

If we square the tautology ${|\zeta - a| \geq 1}$ then we get

$\displaystyle |\zeta|^2 - 2a \text{Re }\zeta + a^2 \geq 1.$

Taking expected values we get

$\displaystyle |\mu|^2 + \sigma^2 - 2a \text{Re }\mu + a^2 \geq 1$

or in other words

$\displaystyle \text{Re }\mu \leq -\frac{1 - a^2}{2a} + O(|\mu|^2 + \sigma^2) = -(1 - a)(1 + o(1)) + O(|\mu|^2 + \sigma^2)$

where we used the Taylor expansion

$\displaystyle \frac{1 - a^2}{2a} = (1 - a)(1 + o(1))$

obtained by Taylor expanding ${1/a}$ about ${1}$ and applying ${1 - a = o(1)}$. Using

$\displaystyle \text{Re }\mu \geq -\frac{1 - a}{1.9 + o(1)} + O(|\mu|^2 + \sigma^2 + (\varepsilon + o(1))^n)$

we get

$\displaystyle -\frac{1 - a}{1.9 + o(1)} + O(|\mu|^2 + \sigma^2 + (\varepsilon + o(1))^n) \leq \text{Re }\mu \leq -(1 - a)(1 + o(1)) + O(|\mu|^2 + \sigma^2)$

Thus

$\displaystyle (1 - a)\left(1 + \frac{1}{1.9 + o(1)} + o(1)\right) = O(|\mu|^2 + \sigma^2 + (\varepsilon + o(1))^n).$

Dividing both sides by ${1 + \frac{1}{1.9 + o(1)} + o(1) \in [1, 2]}$ we have

$\displaystyle 1 - a = O(|\mu|^2 + \sigma^2 + (\varepsilon + o(1))^n).$

In particular

$\displaystyle \text{Re }\mu = O(|\mu|^2 + \sigma^2 + (\varepsilon + o(1))^n)(1 + o(1)) + O(|\mu|^2 + \sigma^2) = O(|\mu|^2 + \sigma^2 + (\varepsilon + o(1))^n).$

Now we treat the imaginary part of ${\text{Im } \mu}$. The previous lemma gave

$\displaystyle \text{Re } ((1 - e^{-i\theta} + o(1))\mu) - \log |a| = O(|\mu|^2 + \sigma^2 + (\varepsilon + o(1))^n).$

Writing everything in terms of real and imaginary parts we can expand out

$\displaystyle \text{Re } ((1 - e^{-i\theta} + o(1))\mu) = (\sin \theta + o(1))\text{Re } \mu + (1 - \cos \theta + o(1))\text{Re }\mu.$

Using the bounds

$\displaystyle (1 - \cos \theta + o(1))\text{Re }\mu, ~\log |a| = O(|\mu|^2 + \sigma^2 + (\varepsilon + o(1))^n)$

(Which follow from the previous paragraph and the bound ${\log |a| = O(1 - a)}$), we have

$\displaystyle (\sin \theta + o(1))\text{Im } \mu = O(|\mu|^2 + \sigma^2 + (\varepsilon + o(1))^n).$

Since ${T}$ is discrete we can find ${\theta}$ arbitrarily close to ${\pm \pi/2}$ which meets the hypotheses of the above equation. Therefore

$\displaystyle \text{Im } \mu = O(|\mu|^2 + \sigma^2 + (\varepsilon + o(1))^n).$

Pkugging everything in, we get

$\displaystyle 1 - a \sim \mu = O(|\mu|^2 + \sigma^2 + (\varepsilon + o(1))^n).$

Now ${|\mu|^2 = o(|\mu|)}$ since ${\mu}$ is infinitesimal; therefore we can discard that term. $\Box$

Now we are ready to prove the second part. The point is that we are ready to dispose of the semi-infinitesimal ${\varepsilon}$. Doing so puts a lower bound on ${U_\zeta(a)}$.

Lemma 10 Let ${I \subseteq \partial D(0, 1) \setminus S}$ be a standard compact set. Then for every ${e^{i\theta} \in I}$,

$\displaystyle U_\zeta(a) - U_\zeta(e^{i\theta}) \geq -o(\sigma^2) - o(1)^n.$

Proof: Since ${\lambda^{(\infty)}}$ is uniformly distributed on ${\partial D(0, 1)}$, there is a zero ${\lambda_0}$ of ${f}$ with ${|\lambda_0 - e^{i\theta}| = o(1)}$. Since ${|\lambda_0| \leq 1}$, we can find an infinitesimal ${\eta}$ such that

$\displaystyle \lambda_0 = e^{i\theta}(1 - \eta)$

and ${|1 - \eta| \leq 1}$. In the previous section we proved

$\displaystyle U_\zeta(a) - U_\zeta(\lambda_0) = O(n^{-1}\sigma^2) + (\varepsilon + o(1))^n).$

Using ${n^{-1} = o(1)}$ and plugging in ${\lambda_0}$ we have

$\displaystyle U_\zeta(a) - U_\zeta(e^{i\theta}(1 - \eta)) = o(\sigma^2) + O((\varepsilon + o(1))^n).$

Now

$\displaystyle \text{Re } \eta \int_0^1 \frac{dt}{1 - t\eta + e^{-i\theta}\zeta} = \log |1 - e^{-i\theta}\zeta| - \log|1 - \eta - e^{-i\theta}\zeta| = \log|e^{i\theta} - \zeta| - \log|e^{i\theta} - e^{i\theta}\eta - \zeta|.$

Taking expectations,

$\displaystyle \text{Re }\eta \mathbf E\int_0^1 \frac{dt}{1 - t\eta + e^{-i\theta}\zeta} = U_\zeta(e^{i\theta}(1 - \eta)) - U_\zeta(e^{i\theta}).$

Taking a Taylor expansion,

$\displaystyle \frac{1}{1 - t\eta - e^{-i\theta}\zeta} = \frac{1}{1 - t\eta} + \frac{e^{-i\theta}\zeta}{(1 - t\eta)^2} + O(|\zeta|^2)$

so by Fubini’s theorem

$\displaystyle \mathbf E\int_0^1 \frac{dt}{1 - t\eta + e^{-i\theta}\zeta} = \int_0^1 \left(\frac{1}{1 - t\eta} + \frac{e^{-i\theta}}{(1 - t\eta)^2}\mu + O(|\mu|^2 + \sigma^2)\right)~dt;$

using the previous lemma and ${\eta = o(1)}$ we get

$\displaystyle U_\zeta(e^{i\theta}(1 - \eta)) - U_\zeta(e^{i\theta}) = \text{Re }\eta \int_0^1 \frac{dt}{1 - t\eta} + o(\sigma^2) + O((\varepsilon + o(1))^n).$

We also have

$\displaystyle \text{Re } \eta \int_0^1 \frac{dt}{1 - t\eta} = -\log \frac{1}{e^{i\theta} - e^{i\theta}\eta} = U_0(1 - \eta)$

since ${0}$ is deterministic (and ${U_0(e^{i\theta} z) = U_0(z)}$, and ${U_0(1) = 0}$; very easy to check!) I think Tao makes a typo here, referring to ${U_i(e^{i\theta}(1 - \eta))}$, which seems irrelevant. We do have

$\displaystyle U_0(1 - \eta) = -\log|1 - \eta| \geq 0$

since ${|1 - \eta| \leq 0}$. Plugging in

$\displaystyle \text{Re } \eta \int_0^1 \frac{dt}{1 - t\eta} \geq 0$

we get

$\displaystyle U_\zeta(e^{i\theta} - e^{i\theta}\eta) - U_\zeta(e^{i\theta}) \geq -o(\sigma^2) - O((\varepsilon + o(1))^n).$

I think Tao makes another typo, dropping the Big O, but anyways,

$\displaystyle U_\zeta(a) - U_\zeta(e^{i\theta} - e^{i\theta}\eta) = o(\sigma^2) - O((\varepsilon + o(1))^n)$

so by the triangle inequality

$\displaystyle U_\zeta(a) - U_\zeta(e^{i\theta}) \geq -o(\sigma^2) - O((\varepsilon + o(1))^n).$

By underspill, then, we can take ${\varepsilon \rightarrow 0}$. $\Box$

We need a result from complex analysis called Jensen’s formula which I hadn’t heard of before.

Theorem 11 (Jensen’s formula) Let ${g: D(0, 1) \rightarrow \mathbf C}$ be a holomorphic function with zeroes ${a_1, \dots, a_n \in D(0, 1)}$ and ${g(0) \neq 0}$. Then

$\displaystyle \log |g(0)| = \sum_{j=1}^n \log |a_j| + \frac{1}{2\pi} \int_0^{2\pi} \log |g(e^{i\theta})| ~d\theta.$

In hindsight this is kinda trivial but I never realized it. In fact ${\log |g|}$ is subharmonic and in fact its Laplacian is exactly a linear combination of delta functions at each of the zeroes of ${g}$. If you subtract those away then this is just the mean-value property

$\displaystyle \log |g(0)| = \frac{1}{2\pi} \int_0^{2\pi} \log |g(e^{i\theta})| ~d\theta.$

Let us finally prove the final part. In what follows, implied constants are allowed to depend on ${\varphi}$ but not on ${\delta}$.

Lemma 12 For any standard ${\varphi \in C^\infty(\partial D(0, 1))}$,

$\displaystyle \int_0^{2\pi} \varphi(e^{i\theta}) U_\zeta(e^{i\theta}) ~d\theta = o(\sigma^2) + o(1)^n.$

Besides,

$\displaystyle U_\zeta(a) = o(\sigma^2) + o(1)^n.$

Proof: Let ${m}$ be the Haar measure on ${\partial D(0, 1)}$. We first prove this when ${\varphi \geq 0}$. Since ${T}$ is discrete and ${\partial D(0, 1)}$ is compact, for any standard (or semi-infinitesimal) ${\delta > 0}$, there is a standard compact set

$\displaystyle I \subseteq \partial D(0, 1) \setminus S$

such that

$\displaystyle m(\partial D(0, 1) \setminus I) < \delta.$

By the previous lemma, if ${e^{i\theta} \in I}$ then

$\displaystyle \varphi(e^{i\theta}) U_\zeta(a) - \varphi(e^{i\theta}) U_\zeta(e^{i\theta}) \geq -o(\sigma^2) - o(1)^n$

and the same holds when we average in Haar measure:

$\displaystyle U_\zeta(a)\int_I \varphi~dm - \int_I \varphi(e^{i\theta}) U_\zeta(e^{i\theta})~dm(e^{i\theta}) \geq -o(\sigma^2) - o(1)^n.$

We have

$\displaystyle |\log |e^{i\theta} - \zeta| + \text{Re } e^{-i\theta}\zeta| \leq |\log|3 - \zeta| + 3\text{Re } \zeta| \in L^2(dm(e^{i\theta}))$

so, using the Cauchy-Schwarz inequality, one has

$\displaystyle \int_{\partial D(0, 1) \setminus I} \varphi(e^{i\theta}) (\log |e^{i\theta} - \zeta| + \text{Re } e^{-i\theta}\zeta) ~dm(e^{i\theta}) = \sqrt{\int_I |\log |e^{i\theta} - \zeta| + \text{Re } e^{-i\theta}\zeta|} = O(\delta^{1/2}).$

Meanwhile, if ${|\zeta| \leq 1/2}$ then the fact that

$\displaystyle \log |e^{i\theta} - \zeta| = \text{Re }-\frac{\zeta}{e^{i\theta}} + O(|\zeta|^2)$

implies

$\displaystyle \log |e^{i\theta} - \zeta| + \text{Re } \frac{\zeta}{e^{i\theta}} = O(|\zeta|^2)$

and hence

$\displaystyle \int_{\partial D(0, 1) \setminus I} \varphi(e^{i\theta}) (\log |e^{i\theta} - \zeta| + \text{Re } e^{-i\theta}\zeta) ~dm(e^{i\theta}) = O(\delta|\zeta|^2).$

We combine these into the unified estimate

$\displaystyle \int_{\partial D(0, 1) \setminus I} \varphi(e^{i\theta}) (\log |e^{i\theta} - \zeta| + \text{Re } e^{-i\theta}\zeta) ~dm(e^{i\theta}) = O(\delta^{1/2}|\zeta|^2)$

valid for all ${|\zeta| \leq 1}$, hence almost surely. Taking expected values we get

$\displaystyle \int_{\partial D(0, 1) \setminus I} \varphi(e^{i\theta})U_\zeta(e^{i\theta}) + \varphi(e^{i\theta}) \text{Re }e^{-i\theta}\mu ~dm(e^{i\theta}) = O(\delta^{1/2}(|\mu|^2 + \sigma^2)) + o(\sigma^2) + o(1)^n.$

In the last lemma we bounded ${|\mu|}$ so we can absorb all the terms with ${\mu}$ in them to get

$\displaystyle \int_{\partial D(0, 1) \setminus I} \varphi(e^{i\theta})U_\zeta(e^{i\theta}) ~dm(e^{i\theta}) = O(\delta^{1/2}\sigma^2) + o(\sigma^2) + o(1)^n.$

We also have

$\displaystyle \int_{\partial D(0, 1) \setminus I} \varphi ~dm = O(\delta)$

(here Tao refers to a mysterious undefined measure ${\sigma}$ but I’m pretty sure he means ${m}$). Putting these integrals together with the integrals over ${I}$,

$\displaystyle \ U_\zeta(a)int_{\partial D(0, 1)} \varphi ~dm - \int_{\partial D(0, 1)} \varphi(e^{i\theta}) U_\zeta(e^{i\theta}) ~dm(e^{i\theta}) \geq -O(\delta^{1/2}\sigma^2) - o(\sigma^2) - o(1)^n.$

By underspill we can delete ${\delta}$, thus

$\displaystyle U_\zeta(a)\int_{\partial D(0, 1)} \varphi ~dm - \int_{\partial D(0, 1)} \varphi(e^{i\theta}) U_\zeta(e^{i\theta}) ~dm(e^{i\theta}) \geq - o(\sigma^2) - o(1)^n.$

We now consider the specific case ${\varphi = 1}$. Then

$\displaystyle U_\zeta(a) - \int_{\partial D(0, 1)} U_\zeta ~dm \geq -o(\sigma^2) - o(1)^n.$

Now Tao claims and doesn’t prove

$\displaystyle \int_{\partial D(0, 1)} U_\zeta ~dm = 0.$

To see this, we expand as

$\displaystyle \int_{\partial D(0, 1)} U_\zeta ~dm = -\mathbf E \frac{1}{2\pi} \int_0^{2\pi} \log|\zeta - e^{i\theta}| ~d\theta$

using Fubini’s theorem. Now we use Jensen’s formula with ${g(z) = \zeta - z}$, which has a zero exactly at ${\zeta}$. This seems problematic if ${\zeta = 0}$, but we can condition on ${|\zeta| > 0}$. Indeed, if ${\zeta = 0}$ then we have

$\displaystyle \int_0^{2\pi} \log|\zeta - e^{i\theta}| ~d\theta = \int_0^{2\pi} \log 1 ~d\theta = 0$

which already gives us what we want. Anyways, if ${|\zeta| > 0}$, then by Jensen’s formula,

$\displaystyle \frac{1}{2\pi} \int_0^{2\pi} \log|\zeta - e^{i\theta}| ~d\theta = \log |\zeta| - \log |\zeta| = 0.$

So that’s how it is. Thus we have

$\displaystyle -U_\zeta(a) \leq o(\sigma^2) + o(1)^n.$

Since ${|a - \zeta| \geq 1}$, ${\log |a - \zeta| \geq 0}$, so the same is true of its expected value ${-U_\zeta(a)}$. This gives the desired bound

$\displaystyle U_\zeta(a) = o(\sigma^2) + o(1)^n.$

We can use that bound to discard ${U_\zeta(a)}$ from the average

$\displaystyle U_\zeta(a)\int_{\partial D(0, 1)} \varphi ~dm - \int_{\partial D(0, 1)} \varphi(e^{i\theta}) U_\zeta(e^{i\theta}) ~dm(e^{i\theta}) \geq - o(\sigma^2) - o(1)^n,$

thus

$\displaystyle \int_{\partial D(0, 1)} \varphi(e^{i\theta}) U_\zeta(e^{i\theta}) ~dm(e^{i\theta})= o(\sigma^2) + o(1)^n.$

Repeating the Jensen’s formula argument from above we see that we can replace ${\varphi}$ with ${\varphi - k}$ for any ${k \geq 0}$. So this holds even if ${\varphi}$ is not necessarily nonnegative. $\Box$