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

As usual, we throughout fix a nonstandard natural and a complex polynomial of degree whose zeroes are all in . We assume that is a zero of whose standard part is , and assume that has no critical points in . Let be a random zero of and a random critical point. Under these circumstances, is uniformly distributed on and is almost surely zero. In particular,

and is infinitesimal in probability, hence infinitesimal in distribution. Let be the expected value of (thus also of ) and its variance. I think we won’t need the nonstandard-exponential bound 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 be a compact set. Then

uniformly for .

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

By underspill, it suffices to show that for every standard we have

We first give the proof for .

First suppose that . Since is infinitesimal in distribution,

here we need the and the since is not a bounded continuous function of . Since we have

but we know that

so, solving for , we get

we absorbed a into the . That gives

Since is a polynomial of degee and is monic (so the top coefficient of is ) this gives a bound

even for .

Now for , we use the bound

to transfer the above argument.

2. Uniform convergence of

Lemma 2 There is a standard compact set and a standard countable set such that

all elements of are isolated in , and is infinitesimal.

Tao claims

where 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 was assumed far from we have

We also have

so for every standard natural there is a standard natural such that

Multiplying both sides by we see that

where is the variety of critical points . Let be the set of standard parts of zeroes in ; then has cardinality and so is finite. For every zero , either

1. For every ,

   

   so the standard part of is , or

2. There is an such that is infinitesimal.

So we may set ; then is standard and countable, and does not converge to a point in , so is standard and is infinitesimal.

I was a little stumped on why is compact; Tao doesn’t prove this. It turns out it’s obvious, I was just too clueless to see it. The construction of forces that for any , there are only finitely many with , so if clusters anywhere, then it can only cluster on . This gives the desired compactness.

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 , 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 outside of

We break up the approximation lemma into multiple parts. Let be a standard compact set which does not meet . Given a curve we denote its arc length by ; 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 . Then there is a curve from to which misses and satisfies the uniform estimate

Proof: We use the decomposition of into the arc

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

Therefore we can instead consider a curve which goes all the way around , leaving . This curve has length for close to (and if are far from we can just perturb a line segment without generating too much error). Using our uniform max bound above we see that this choice of is valid.

Recall that the moments of are infinitesimal.

Since is infinitesimal, and is a positive distance from any infinitesimals (since it is standard compact), we have

uniformly in . Therefore has no critical points near and so is holomorphic on .

We first need a version of the fundamental theorem.

Lemma 4 Let be a contour in of length . Then

Proof: Our bounds on imply that we can take the Taylor expansion

of in terms of , which is uniform in . Taking expectations preserves the constant term (since it doesn’t depend on ), kills the linear term, and replaces the quadratic term with a , thus

At the start of this series we showed

Plugging in the Taylor expansion of we get

Simplifying the integral we get

whence the claim.

Lemma 5 Uniformly for one has

Proof: Applying the previous two lemmata we get

It remains to simplify

Taylor expanding and using the self-similarity of the Taylor expansion we get

which gives that bound.

Lemma 6 Let . Then

uniformly in .

Proof: We may assume that is small enough depending on , since the constant in the big- notation can depend on as well, and only appears next to implied constants. Now given we can find from to 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 to be a line segment which has been perturbed to miss the discrete set , and possibly arced to miss (say if is far from ). By compactness of we can choose the bounds on to be not just uniform in time but also in space (i.e. in ), and besides that is a curve through a compact set which misses . Indeed, one can take to be a closed ball containing , and then cut out small holes in around and , whose radii are bounded below since is compact. Since the moments of are infinitesimal one has

Here we used to enforce

By the previous lemma,

Integrating this result along we get

Applying our preliminary bound, the previous paragraph, and the fact that , thus

we get

We treat the first term first:

For the second term, while , so is bounded from below, whence

Thus we simplify

It will be convenient to instead write this as

Now we deal with the pesky integral. Since is moving towards at a speed which is bounded from below uniformly in “spacetime” (that is, ), there is a standard such that if then

since is going towards . (Tao’s argument puzzles me a bit here because he claims that the real inner product is uniformly bounded from below in spacetime, which seems impossible if . I agree with its conclusion though.) Exponentiating both sides we get

which bounds

Since is standard, it dominates the infinitesimal , so after shrinking a little we get a new bound

Since is exponentially small in , in particular it is smaller than . Plugging in everything we get the claim.

4. Control on zeroes away from

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

Lemma 7 Let be a standard compact set which misses and standard. Then for every zero of ,

Last time we showed that this implies

Thus all the zeroes of either live in or a neighborhood of a level set of . Proof: Plugging in in the approximation

we get

Several posts ago, we proved as a consequence of Grace’s theorem, so . In particular, if we solve for we get

Using

plugging in , and taking logarithms, we get

Now and misses the standard compact set , so since we have

(since and is infinitesimal). So we can Taylor expand in about :

Taking expectations and using ,

Plugging in we see the claim.

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: is a level set of , but one that a priori doesn’t tell us much about . We have just replaced it with a level set of , a function that is explicitly closely related to , but at the price of an error term.

5. Fine control

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

Let us first bound . Recall that so but in fact we can get a sharper bound. Since is discrete we can get arbitrarily close to whatever we want, say or . This will give us bounds on when we take the Taylor expansion

Lemma 8 Let be standard. Then

Proof: Let be a standard compact set which misses and a zero of . Since (since is close to ) and has positive standard part (since ) we can take Taylor expansions

and

in about . Taking expectations we have

and similarly for . Thus

since

Since

we have

Now so , whence

Now recall that is uniformly distributed on , so we can choose so that

Thus

which we can plug in to get the claim.

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

Lemma 9 One has

Proof: Let be standard reals such that . I don’t think the constants here actually matter; we just need or something. Anyways, summing up two copies of the inequality from the previous lemma with we have

since

That is,

Indeed,

so

If we square the tautology then we get

Taking expected values we get

or in other words

where we used the Taylor expansion

obtained by Taylor expanding about and applying . Using

we get

Thus

Dividing both sides by we have

In particular

Now we treat the imaginary part of . The previous lemma gave

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

Using the bounds

(Which follow from the previous paragraph and the bound ), we have

Since is discrete we can find arbitrarily close to which meets the hypotheses of the above equation. Therefore

Pkugging everything in, we get

Now since is infinitesimal; therefore we can discard that term.

Now we are ready to prove the second part. The point is that we are ready to dispose of the semi-infinitesimal . Doing so puts a lower bound on .

Lemma 10 Let be a standard compact set. Then for every ,

Proof: Since is uniformly distributed on , there is a zero of with . Since , we can find an infinitesimal such that

and . In the previous section we proved

Using and plugging in we have

Now

Taking expectations,

Taking a Taylor expansion,

so by Fubini’s theorem

using the previous lemma and we get

We also have

since is deterministic (and , and ; very easy to check!) I think Tao makes a typo here, referring to , which seems irrelevant. We do have

since . Plugging in

we get

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

so by the triangle inequality

By underspill, then, we can take .

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

Theorem 11 (Jensen’s formula) Let be a holomorphic function with zeroes and . Then

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

Let us finally prove the final part. In what follows, implied constants are allowed to depend on but not on .

Lemma 12 For any standard ,

Besides,

Proof: Let be the Haar measure on . We first prove this when . Since is discrete and is compact, for any standard (or semi-infinitesimal) , there is a standard compact set

such that

By the previous lemma, if then

and the same holds when we average in Haar measure:

We have

so, using the Cauchy-Schwarz inequality, one has

Meanwhile, if then the fact that

implies

and hence

We combine these into the unified estimate

valid for all , hence almost surely. Taking expected values we get

In the last lemma we bounded so we can absorb all the terms with in them to get

We also have

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

By underspill we can delete , thus

We now consider the specific case . Then

Now Tao claims and doesn’t prove

To see this, we expand as

using Fubini’s theorem. Now we use Jensen’s formula with , which has a zero exactly at . This seems problematic if , but we can condition on . Indeed, if then we have

which already gives us what we want. Anyways, if , then by Jensen’s formula,

So that’s how it is. Thus we have

Since , , so the same is true of its expected value . This gives the desired bound

We can use that bound to discard from the average

thus

Repeating the Jensen’s formula argument from above we see that we can replace with for any . So this holds even if is not necessarily nonnegative.

In this post we’re going to complete the proof of case zero and continue the proof of case one. In the last two posts we managed to prove:

Theorem 1 Let be a nonstandard natural, and let be a monic polynomial of degree on with all zeroes in . Suppose that is a zero of such that:

1. Either or is infinitesimal, and
2. has no critical points on .

Let be a random zero of and a random critical point of . Let be the expected value of . Let be a complex number outside some measure zero set and let be a contour that misses the zeroes of ,. Then:

1. .
2. .
3. One has

   

   and

   

4. One has

   

   and

   

5. One has

   

   and

   

6. One has

   

   and

   

Moreover,

1. If is infinitesimal (case zero), then are identically distributed and almost surely lie in

   

   Moreover, if is any compact set which misses , then

   

   so is infinitesimal in probability.

2. If is infinitesimal (case one), then is uniformly distributed on and is almost surely zero. Moreover,

   

We also saw that Sendov’s conjecture in high degree was equivalent to the following result, that we will now prove.

Lemma 2 Let be a nonstandard natural, and let be a monic polynomial of degree on with all zeroes in . Let be a zero of such that:

1. Either is infinitesimal (case zero), or
2. There is a standard such that

   

   (case one).

If there are no critical points of in , then .

1. Case zero

Now we prove case zero — the easy case — of Lemma 2.

Suppose that is infinitesimal. In this case, are identically distributed and almost surely are .

Lemma 3 There are such that for every ,

uniformly.

Proof: Since is supported in , is holomorphic away from . Since is bounded, if is near then

which is nonzero near . So the variety is discrete, so there are such that whenever . To see this, suppose not; then for every we can find and with , so has infinitely many zeroes in the compact set . Since this is definitely not true, the claim follows by continuity of .

Let be the number of zeroes of in , so is a nonnegative standard natural since is standard and is compact. Let where ; then

by the argument principle.

We claim that in fact , which contradicts that is nonnegative. This will be proven in the rest of this section.

Here something really strange happens in Tao’s paper. He proves this:

Lemma 4 One has

in .

We now need to show that the convergence in above commutes with the use of the argument principle so that

this will be good because we have control on the zeroes and critical points of using our contradiction assumption. What’s curious to me is that Tao seems to substitute this with convergence in on an annulus. Indeed, convergence in does commute with use of the argument principle, but at no point of the proof does it seem like he uses the convergence in . So I include the proof of the latter in the next section as a curiosity item, but I think it can be omitted entirely. Tell me in the comments if I’ve made a mistake here.

If is a smooth cutoff supported on and identically one on (where ), one has

The term is easy to deal with, since for every , is a bounded continuous function of whenever (so ). By the definition of being infinitesimal in distribution we have

Therefore

is uniformly infinitesimal.

Now we treat the term. Interestingly, this is the main point of the argument where we use that is infinitesimal, and the rest of the argument seems to mainly go through with much weaker assumptions on .

Lemma 5 There is an such that if then

Proof: By the triangle inequality and its reverse, if then

Here is to be chosen.

Since we have

whenever is a compact set which misses , this in particular holds when . Since and it follows that

In particular,

We now claim

By splitting the integrand we first bound

since is nonstandard and the domain of integration has measure at most . On the other hand, the other term

since is standard while is nonstandard. This proves the claim.

Putting the above two paragraphs together and using Fubini’s theorem,

is infinitesimal. So outside of a set of infinitesimal measure, satisfies

If then there is a (deterministic) zero such that , thus lies in a set of measure . There are such sets since there are zeroes of , so their union has measure , which is infinitesimal. Therefore

which implies the claim.

Summing up, we have

in , where is as in the previous lemma. Pulling out the factor of , which is harmless, we can use the argument principle to deduce that is the number of zeroes minus poles of ; that is, the number of critical points minus zeroes of . Indeed, convergence in does commute with the argument principle, so we can throw out the infinitesimal .

But is infinitesimal, and we assumed that had no critical points in , which contains . So has no critical points, but has a zero ; therefore .

In a way this part of the proof was very easy: the only tricky bit was using the cutoff to get convergence in like we needed. The hint that we could use the argument principle was the fact that was infinitesimal, so we had control of the critical points near the origin.

2. Convergence in

Let be the distribution of and of . Since is infinitesimal in distribution, is infinitesimal in the weak topology of measures; that is, for every continuous function and compact set ,

Now

If is a compact set and is Lebesgue measure then

By Tonelli’s theorem

and the inner integral is finite since is Lebesgue integrable in codimension . So the outer integrand is a bounded continuous function, which implies that

which gives what we want when we recall

and we plug in .

3. Case one: Outlining the proof

The proof for case one is much longer, and is motivated by the pseudo-counterexample

Here is an th root of unity, and has no critical points on , but does have critical points at . Similar pseudo-counterexamples hold for

where is standard. We will seek to control these examples by controlling up to an error of size ; here is the variance of and is an infinitesimal raised to the power of , thus is very small, and forces us to balance out everything in terms of .

As discussed in the introduction of this post, is infinitesimal in probability (and, in particular, its expected value is infinitesimal); thus, with overwhelming probability, the critical points of are all infinitesimals. Combining this with the fact that is uniformly distributed on , it follows that sort of looks like .

We start with some nice bounds:

Lemma 6 (preliminary bounds) For any standard compact set , one has

and

uniformly in .

In other words, sort of grows like the translate of a homogeneous polynomial of degree .

It would be nice if was infinitesimal in , but this isn’t quite true; the following lemma is the best we can do.

Lemma 7 (uniform convergence of ) There is a standard compact set

where is countable, standard, does not meet , and consists of isolated points of , such that is infinitesimal in .

So we think of as some sort of generalization of . Away from we have good bounds on :

Lemma 8 (approximating outside ) For any standard compact set :

1. Uniformly in ,

   

2. For every standard and uniformly in ,

   

As a consequence, we can show that every zero of which is far from is close to the level set

This in particular holds for , since the standard part of is , and does not come close to (so neither does ). In fact the error term is infinitesimal:

Lemma 9 (zeroes away from ) For any standard compact set , any standard , and any zero ,

uniformly in .

Since satisfies the hypotheses of the above lemma,

is infinitesimal. This gives us some more bounds:

Lemma 10 (fine control) For every standard :

1. One has

   

2. For every compact set and ,

   

3. For every standard smooth function ,

   

4. One has

   

Here Tao claims

Apparently this follows from Fourier inversion but I don’t see it. In any case if we take the expected value of the left-hand side we get

by the fine control lemma, so

In particular this holds for the real part of . Since is infinitesimal, so are the first two moments of the real part of .

Since , one has

This is true since for any one has (which follows by Taylor expansion). In particular,

Let be the best approximation of on the arc , which exists since that arc is compact; then

Since , it has the useful property that

therefore

Plugging in the expansion for we have

We now use the inequality several times. First we bound

I had to think a bit about why this is legal; the point is that you can absorb the implied constant on into the implied constant on before applying the inequality. Now we bound

by similar reasoning.

Thus we conclude the bound

or in other words,

Applying the fine control lemma, or more precisely the result

as well as the fact that is infinitesimal, we have

for every standard , hence by underspill

By the fine control lemma,

Thus we bound

owing to the fact that so that .

Plugging in the above bounds,

By definition of variance we have

and is infinitesimal so we can spend the term as

But the fine control lemma said

So

In particular,

since is infinitesimal.

We used underspill to show

so

since was standard, which implies .

Next time, we’ll go back and fill in all the lemmata that we skipped in the proof for case one. This is a tricky bit — pages 25 through 34 of Tao’s paper. (For comparison, we covered pages 19 through 21, some of the exposition in pages 24 through 34, and pages 34 through 36 this time). Next time, then.

Before we begin, I want to fuss around with model theory again. Recall that if is a nonstandard complex number, then denotes the standard part of , if it exists. We previously defined what it meant for a nonstandard random variable to be infinitesimal in distribution. One can define something similar for any metrizable space with a notion of , where is infinitesimal provided that is. For example, a nonstandard random variable is infinitesimal in if for every compact set that can take values in, is infinitesimal, since is metrizable with

whenever is a compact exhaustion. If is nonstandard, is infinitesimal in some metrizable space, and is standard, then we call the standard part of in ; then the standard part is unique since metrizable spaces are Hausdorff.

If the metrizable space is compact, the case that we will mainly be interested in, then the standard part exists. This is a point that we will use again and again. Passing to the cheap perspective, this says that if is a compact metric space and is a sequence in , then there is a which is approximates infinitely often, but that’s just the Bolzano-Weierstrass theorem. Last time used Prokohov’s theorem to show that if is a nonstandard tight random variable, then has a standard part in distribution.

We now restate and prove Proposition 9 from the previous post.

Theorem 1 (distribution of random zeroes) Let be a nonstandard natural, a monic polynomial of degree with all zeroes in , and let be a zero of . Suppose that has no zeroes in . Let be a random zero of and a random zero of . Then:

1. If (case zero), then and are identically distributed and almost surely lie in the curve

   

   In particular, in probability. Moreover, for every compact set ,

   

2. If (case one), then is uniformly distributed on the unit circle and is almost surely zero. Moreover,

   

1. Moment-generating functions and balayage

We first show that and have equal moment-generating functions in a suitable sense.

To do this, we first show that they have the same logarithmic potential. Let be a random variable such that almost surely (that is, is almost surely bounded). Then the logarithmic potential

is defined almost everywhere as we discussed last time, and is harmonic outside of the essential range of .

Lemma 2 Let be a nonstandard, almost surely bounded, random complex number. Then the standard part of is according to the topology of under Lebesgue measure.

Proof: We pass to the cheap perspective. If we instead have a random sequence of and in distribution, then in , since up to a small error in we can replace with a test function ; one then has

where in the weak topology of measures, is the distribution of , is the distribution of , and is a compact set equipped with Lebesgue measure.

Lemma 3 For every , we have

In particular, .

Proof: By definition, , so . Now is a disc with diameter where is a rotation around the origin. Taking reciprocals preserves discs and preserves , so sits inside a disc with a diameter . Then is convex, so the expected value of is also . Therefore the Stieltjes transform

satisfies . In particular,

But we showed that

almost everywhere last time. This implies that for almost every ,

but all terms here are continuous so we can promote this to a statement that holds for every . In particular,

hence

Since while , is bounded from above and below by a constant times . Therefore the same holds of its logarithm , which is bounded from above and below by a constant times . This implies the first claim.

To derive the second claim from the first, we use the previous lemma, which implies that we must show that

in . But this follows since is integrable in two dimensions.

Lemma 4 Let be an almost surely bounded random variable. Then

Proof: One has the Taylor series

Indeed, by rescaling and using , we may assume . The summands expand as

and the imaginary parts all cancel by symmetry about . Using the symmetry about again we get

This equals the left-hand side as long as . Taking expectations and commuting the expectation with the sum using Fubini’s theorem (since is almost surely bounded), we see the claim.

Lemma 5 For all , one has

In particular, and have identical moments.

Proof: If we take then we conclude that

The left-hand side is a Fourier series, and by uniqueness of Fourier series it holds that for every ,

This gives a bound on the difference of moments

which is only possible if the moments of and are identical. The left-hand side doesn’t depend on , but if , , then and so the claim holds. On the other hand, if then this claim still holds, since we showed last time that

and obviously .

Here I was puzzled for a bit. Surely if two random variables have the same moment-generating function then they are identically distributed! But, while we can define the moment-generating function of a random variable as a formal power series , it is not true that has to have a positive radius of convergence, in which case the inverse Laplace transform of is ill-defined. Worse, the circle is not simply connected, and in case one, we have to look at a uniform distribution on the circle, whose moments therefore aren’t going to points on the circle, so the moment-generating function doesn’t tell us much.

2. Balayage

We recall the definition of the Poisson kernel :

whenever is a radius. Convolving the Poisson kernel against a continuous function on solves the Dirichlet problem of with boundary data .

Definition 6 Let be a random variable. The balayage of is

Balayage is a puzzling notion. First, the name refers to a hair-care technique, which is kind of unhelpful. According to Tao, we’re supposed to interpret balayage as follows.

If is an initial datum for Brownian motion , then is the probability density of the first location where passes through . Tao asserts this without proof, but conveniently, this was a problem in my PDE class last semester. The idea is to approximate by the lattice , which we view as a graph where each vertex has degree , with one edge to each of the vertices directly above, below, left, and right of it. Then the Laplacian on is approximated by the graph Laplacian on , and Brownian motion is approximated by the discrete-time stochastic process wherein a particle starts at the vertex that best approximates and at each stage has a chance of moving to each of the vertices adjacent to its current position.

So suppose that and are actually vertices of . The probability density is harmonic in with respect to the graph Laplacian since it is the mean of as ranges over the adjacent vertices to ; therefore it remains harmonic as we take . The boundary conditions follow similarly.

Now if is a random initial datum for Brownian motion which starts in , the balayage of is again a probability density on that records where one expects the Brownian motion to escape, but this time the initial datum is also random.

I guess the point is that balayage serves as a substitute for the moment-generating function in the event that the latter is just a formal power series. We want to be able to use analytic techniques on the moment-generating function, but we can’t, so we just use balayage instead.

Let be the balayage of . Since is bounded, we can use Fubini’s theorem to commute the expectation with the sum and see that

provided that . It will be convenient to rewrite this in the form

so is uniquely determined by the moment-generating function of . In particular, and have identical balayage, and one has a bound

We claim that

which implies the bound

To see this, we discard the term since , which implies that

Up to a constant factor we may assume that the logarithms are base in which case we get a bound

The constant is absolute since .

By the integral test, we get a bound

Using the bound

for any and the change of variable (thus ), we get a bound

since the error in the exponent can’t affect the exponential decay of the integral in . Since we certainly have

this is a suitable tail bound.

To complete the proof of the claim we need to bound the main term. To this end we bound

Here denotes exponentiation. Now if is small enough (say ), this supremum will be attained when , thus . Therefore

Luckily is easy to differentiate: its critical point is . This gives

so

which was the bound we needed, and proves the claim. Maybe there’s an easier way to do this, because Tao says the claim is a trivial consequence of dyadic decomposition.

Let’s interpret the bound that we just proved. Well, if the balayage of is supposed to describe the point on the circle at which a Brownian motion with random initial datum escapes, a bound on a difference of two balyages should describe how the trajectories diverge after escaping. In this case, the divergence is infinitesimal, but at different speeds depending on . As , our infinitesimal divergence gains a positive standard part, while if stays close to , the divergence remains infinitesimal. This makes sense, since if we take a bigger circle we forget more and more about the fact that are not the same random variable, since Brownian motion has more time to “forget more stuff” as it just wanders around aimlessly. So in the regime where is close to , it is reasonable to take standard parts and pass to and , while in the regime where is close to this costs us dearly.

3. Case zero

Suppose that is infinitesimal.

We showed last time that , so is infinitesimal. Therefore almost surely.

I think there’s a typo here, because Tao lets range over and considers points , which don’t exist since while every point in has . I think this can be fixed by taking closures, which is what I do in the next lemma.

Tao proves a “qualitative” claim and then says that by repeating the argument and looking out for constants you can get a “quantitative” version which is what he actually needs. I’m just going to prove the quantitative argument straight-up. The idea is that if is a compact set which misses and then a Brownian motion with initial datum will probably escape through an arc which is close to , but is not close to so a Brownian motion which starts at will probably not escape through . Therefore have very different balayage, even though the difference in their balayage was already shown to be infinitesimal.

I guess this shows the true power of balayage: even though the moment-generating function is “just” a formal power series, we know that the essential supports of must “look like each other” up to rescaling in radius. This still holds in case one, where one of them is a circle and the other is the center of the circle. Either way, you get the same balayage, since whether you start at some point on a circle or you start in the center of the circle, if you’re a Brownian motion you will exhibit the same long-term behavior.

In the following lemmata, let be a compact set. The set is compact since it is the preimage of a compact set, so contained a compact interval .

Lemma 7 One has

Proof: Since is compact the minimum is attained. Let be the minimum. Since is a real-valued harmonic function in , thus

the maximum principle implies that the worst case is when meets and , say . Then

Of course this is just a formal power series and doesn’t make much sense. But if instead where is very small depending on a given , then, after discarding quadratic terms in ,

This follows since in general

Now

since the integrand is maximized when , in which case the integrand evaluates to the measure of , which is since and has positive measure. Therefore

On the other hand, for any one has

so this implies gives a lower bound on the integral over .

Lemma 8 If then

Proof: Let in the previous lemma, conditioning on the event , to see that

where . Taking expectations and dividing by the probability that , we can use Fubini’s theorem to deduce

where . Applying the bound on from the section on balayage, we deduce

We already showed that . So in order to show

which was the bound that we wanted, it suffices to show that for every such that ,

Tao says that “one can show” this claim, but I wasn’t able to do it. I think the point is that under those cirumstances one has and even as , so we have some control on . In fact I was able to compute

which suggests that this is the right direction, but the bounds I got never seemed to go anywhere. Someone bug me in the comments if there’s an easy way to do this that I somehow missed.

Now we take to complete the proof.

4. Case one

Suppose that is infinitesimal. Let be the expected value of (hence also of ). Let be a standard real.

We first need to go on an excursion to a paper of Dégot, who proves the following theorem:

Lemma 9 One has

Moreover,

I will omit the proof since it takes some complex analysis I’m pretty unfamiliar with. It seems to need Grace’s theorem, which I guess is a variant of one of the many theorems in complex analysis that says that the polynomial image of a disk is kind of like a disk. It also uses some theorem called the Walsh contraction principle that involves polynomials on the projective plane. Curious.

In what follows we will say that an event is standard-possible if the probability that happens has positive standard part.

Lemma 10 For every , is standard-possible. Besides, .

Proof: Since almost surely and

but

we have

Combining this with the lemma we see that the standard part of is , so

On the other hand,

and since is nonstandard, is infinitesimal, so the constant in gets eaten. In particular,

which implies that

and hence

Since this is true for arbitrary standard , underspill implies that there is an infinitesimal such that

But almost surely, and we just showed

So the claim holds.

We now allow to take the value , thus .

Lemma 11 One has

and

Moreover, if , so has no zeroes in that disk.

Proof: Since

one has

Now and .

Here I drew two unit circles in , one entered at the origin and one at (since is infinitesimal); is (up to infinitesimal error) in the first circle and out of the second. The rightmost points of intersection between the two circles are on a vertical line which by the Pythagorean theorem is to the left of the vertical line , which in turn is to the left of the perpendicular bisector . Thus , and if then the real part of is . In particular, if the standard real part of is then , so has positive standard part.

By the previous lemma, it is standard-possible that the standard real part of is , so the standard real part of is standard-possibly positive and is almost surely nonnegative. Plugging into the above we deduce the existence of a standard absolute constant such that

In particular,

Keeping in mind that is nonstandard, this doesn’t necessarily mean that has nonpositive standard part, but it does give a pretty tight bound. Taking a first-order Taylor approximation we get

But one has

from the Dégot lemma. Clearly this term dominates so we have

Since one has a lower bound this implies is controlled from below by an absolute constant.

We also claim . In fact, we showed last time that

we want to show that , so it suffices to show that , or in other words that

Since by assumption on , this is trivial. We deduce that

and hence

Now Tao claims that the proof that is similar, if . Since was a valid choice of we have . Since , if then where is an absolute constant. Applying the fact that is standard-possible and is almost surely nonnegative we get

so we indeed have the claim.

We now prove the desired bound

Actually,

as we proved last time, so the bound guarantees the claim.4

In particular

by Fatou’s lemma. So almost surely. Therefore is harmonic on , and we already showed that if was small enough, thus

if was small enough. That implies on an open set and hence everywhere. Since

we can plug in and conclude that all moments of except the zeroth moment are zero. So is uniformly distributed on the unit circle.

By overspill, I think one can intuit that if is a random polynomial of high degree which has a zero close to , all zeroes in , and no critical point close to , then sort of looks like

where is a primitive root of unity of the same degree as . Therefore looks like a cyclotomic polynomial, and therefore should have lots of zeroes close to the unit sphere, in particular close to , a contradiction. This isn’t rigorous but gives some hint as to why this case might be bad.

Now one has

and in particular by Fatou’s lemma

But it was almost surely true that , thus that . So this enforces almost surely. In particular, almost surely,

Since is a contractible curve, its complement is connected. We recall that near infinity, and since we already know the distribution of , we can use it to compute near infinity. Tao says the computation of is a straightforward application of the Newtonian shell theorem; he’s not wrong but I figured I should write out the details.

For one has

where the denotes that this is a line integral in rather than in . Translating we get

which is the integral of the fundamental solution of the Laplace equation over . If (reasonable since is close to infinity), this implies the integrand is harmonic, so by the mean-value formula one has

and so this holds for both and near infinity. But then is harmonic away from , so that implies that

Since the distribution of is the Laplacian of one has

Therefore almost surely. In particular, is infinitesimal almost surely. This completes the proof in case one.

By the way, I now wonder if when one first learns PDE it would be instructive to think of the fundamental solution of the Laplace equation and the mean-value formulae as essentially a consequence of the classical laws of gravity. Of course the arrow of causation actually points the other way, but we are humans living in a physical world and so have a pretty intuitive understanding of what gravity does, while stuff like convolution kernels seem quite abstract.

Next time we’ll prove a contradiction for case zero, and maybe start on the proof for case one. The proof for case one looks really goddamn long, so I’ll probably skip or blackbox some of it, maybe some of the earlier lemmata, in the interest of my own time.

Having some more free time than usual, I figured I would read a recent paper that looked interesting. However, I’m something of a noob at math, so I figure it’s worth it to take it slowly and painstakingly think through the details. This will be a sort of stream-of-conciousness post where I do just that.

The paper I’ll be reading will be Terry Tao’s “Sendov’s conjecture for sufficiently high degree polynomials.” This paper looks interesting to me because it applies “cheap nonstandard” methods to prove a result in complex analysis. In addition it uses probability-theoretic methods, which I’m learning a lot of right now. Sendov’s conjecture is the following:

Sendov’s conjecture Let be a polynomial of degree that has all zeroes . If is a zero then has a zero in .

Without loss of generality, we may assume that is monic and that . Indeed, if is a polynomial in the variable we divide by the argument of and then rescale by the top coefficient.

Tao notes that by Tarski’s theorem, in principle it suffices to get an upper bound on the degree of a counterexample and then use a computer-assisted proof to complete the argument. I think that for every you’d get a formula in the theory of real closed fields that could be decided in time, where is an absolute constant (which, unfortunately, is exponential in the number of variables of the formula, and so is probably quite large). Worse, Tao is going to use a compactness argument and so is going to get an astronomical bound . Still, something to keep in mind — computer-assisted proofs seem like the future in analysis.

More precisely, Tao proves the following:

Proposition 1 For every in a monotone sequence in , let be a monic polynomial of degree with all zeroes in and let satisfy . If for every , has no zeroes in , then .

It’s now pretty natural to see how “cheap nonstandard” methods apply. One can pass to a subsequence countably many times and still preserve the hypotheses of the proposition, so by diagonalization and compactness, we may assume good (but ineffective) convergence properties. For example, we can assume that , where does not depend on and is with respect to .

Using overspill, one can view the proposition model-theoretically: it says that if is a nonstandard natural number, a monic polynomial of degree with a zero , and there are no zeroes of in , then . Tao never fully takes this POV, but frequently appeals to results like the following:

Proposition 2 Let be a first-order predicate. Then:

1. (Overspill) If for every sufficiently small standard , , then there is an infinitesimal such that .
2. (Underspill) If for every infinitesimal , , then there are arbitrarily small such that .
3. (-saturation) If is where is compact, then the implied constant in the statement of is independent of .

Henceforth we will use asymptotic notation in the nonstandard sense; for example, a quantity is if it is infinitesimal. This is equivalent to the cheap nonstandard perspective where a quantity is iff it is with respect to , where is ranging over some monotone sequence of naturals. I think the model-theoretic perspective is helpful here because we are going to pass to subsequences a lot, and at least in the presence of the boolean prime ideal theorem, letting be a fixed nonstandard natural number is equivalent to choosing a nonprincipal ultrafilter on that picks out the subsequences we are going to pass to, in the perspective where ranges over a monotone sequence of standard naturals.

This follows because the Stone-Cech compactification is exactly the space of ultrafilters on . Indeed, if ranges over a monotone sequence of standard naturals, then in , converges to an element of , which then is a nonprincipal ultrafilter. If denotes the ultrapower of with respect to , then I think the equivalence class of the sequence in is exactly the limit of . Conversely, once a nonprincipal ultrafilter has been fixed, we have a canonical way to pass to subsequences: only pass to a subsequence which converges to . This is possible since is compact.

I think it will be at times convenient to go back to the “monotone sequence of standard naturals” perspective, especially when we’re doing computations, so I reserve the right to go between the two. We’ll call the monotone sequence perspective the “cheap perspective” and the model-theoretic perspective the “expensive perspective”.

I’m not familiar with the literature on Sendov’s conjecture, so I’m going to blackbox the reduction that Tao carries out. The reduction says that, due to the Gauss-Lucas theorem and previously existing partial results on Sendov’s conjecture, to prove Proposition 1, it suffices to show:

Proposition 3 Let be a nonstandard natural, let be a monic polynomial of degree with all zeroes in and let satisfy . Suppose that has no zeroes in and

1. (Theorem 3.1 in Tao) either , or
2. (Theorem 5.1 in Tao) there is a standard such that

   

Then .

In the former case we have and in the latter we have . We’ll call the former “case zero” and the latter “case one.”

Tao gives a probabilistic proof of Proposition 3, and that’s going to be the bulk of this post and its sequels. Let be a random zero of , drawn uniformly from the finite set of zeroes. Le denote a random zero of chosen independently of .

In the cheap perspective, depends on , we are going to study properties of the convergence of as , by using our chosen ultrafilter to repeatedly pass to subsequences to make converge in some suitable topology. The probability spaces that lives in depend on , but as long as we are interested in a deterministic limit that does not depend on , this is no problem. Indeed will converge to uniformly (resp. in probability) provided that for every , almost surely (resp. ), and this makes sense even though the probability space we are studying depends on . The usual definition of convergence in distribution still makes sense even for random variables converges to a random variable that deos not depend on provided that their distributions converge vaguely to .

Okay, it’s pretty obvious what being infinitesimally close to a deterministic standard real is in the uniform or probabilistic sense. Expanding out the definition of the vague topology of measures, a nonstandard measure on a locally compact Hausdorff space is infinitesimally close to a standard measure provided that for every continuous function with compact support,

This induces a definition of being infinitesimally close in distribution.

Okay, no more model-theoretic games, it’s time to start the actual proof.

Definition 4 Let be a bounded complex random variable. The logarithmic potential of is

Here denotes expected value. Tao claims but does not prove that this definition makes sense for almost every . To check this, let be a compact set equipped with Lebesgue measure and let be the distribution of . Then

and the integrand is singular along the set , which has real codimension in . The double integral of a logarithm makes sense almost surely provided that the logarithm blows up with real codimension (to see this, check the double integral of log on ) so this looks good.

Definition 5 Let be a bounded complex random variable. The Stieltjes transform of is

Then is “less singular” than , so this definition is inoffensive almost everywhere.

Henceforth Tao lets denote the distribution of , where is any bounded complex random variable. Then and where are the complex derivative and Cauchy-Riemann operators respectively. Since we have

This just follows straight from the definitions. Of course might not be an absolutely continuous measure, so this only makes sense if we use the calculus of distributions.

Does this make sense if is deterministic, say almost surely? In that case is a Dirac measure at and . Everything looks good, since .

For the next claims I need the Gauss-Lucas theorem:

Theorem 6 (Gauss-Lucas) If is a polynomial on , then all zeroes of belong to the convex hull of the variety .

Lemma 7 (Lemma 1.6i in Tao) surely lies in and surely lies in .

Proof: The first claim is just a tautology. For the second, by assumption on all zeroes of lie in the convex set , so so does their convex hull. In particular lies in almost surely by the Gauss-Lucas theorem. Our contradiction hypothesis says that .

Lemma 8 (Lemma 1.6ii-iv in Tao) One has . For almost every ,

and

Moreover,

and

Moreover,

and

Proof: We pass to the cheap perspective, so is a large standard natural. Since is large, in particular , if is the coefficient of in then the roots of sum to . The roots of sum to by calculus. So .

We write

and

Taking of both sides and then dividing by we immediately get and . Then we take the complex derivative of both sides of the and formulae to get the formulae for and .

Now the formula for follows by subtracting the above formulae, as does the formula for .

Since the distributions of and have bounded support (it’s contained in ) by Prokhorov’s theorem we can find standard random variables and such that is infinitesimal in distribution and similarly for . The point is that and give, up to an infinitesimal error, information about the behavior of , , and by the above lemma and the following proposition.

Proposition 9 (Theorem 1.10 in Tao) One has:

1. In case zero, and are identically distributed and almost surely lie in , so is infinitesimal in probability. Moreover, for every compact set ,

   

2. In case one, is uniformly distributed on and is almost surely zero. Moreover,

   

The proposition gives quantitative bounds that force the zeroes to all be in certain locations. Looking ahead in the paper, it follows that:

1. In case zero, the Stieltjes transform of is infinitesimally close to , so by a stability-of-zeroes argument to show that has no zeroes near the origin, even though .
2. In case one, if is the standard deviation of , then we have control on the zeroes of up to an error of size , which we can then use to deduce a contradiction.

Next time I’ll start the proof of this proposition. Its proof apparently follows from the theory of Newtonian potentials, which is not too surprising since if is the Laplacian of . It needs the following lemma:

Lemma 10 (Lemma 1.6vi in Tao) If is a curve in that misses the zeroes of and then

and

Proof: One has

by the previous lemma. Breaking up into finitely many parts one can assume that is a contractible curve in a simply connected set, in which case we have a branch of the logarithm along . Now apply the fundamental theorem. The case for is the same.