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).


\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)


\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)


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


\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).


\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)


\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).


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


\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)


\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).


\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.


\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)


\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,


\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

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