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 1Let 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:

- Either or is infinitesimal, and
- 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:

- .
- .
- One has
and

- One has
and

- One has
and

- One has
and

Moreover,

- 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.

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

- Either is infinitesimal (case zero), or
- 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 3There 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 4One 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 5There 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 hasand

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 :

- Uniformly in ,
- 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 :

- One has
- For every compact set and ,
- For every standard smooth function ,
- 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.