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 1Let 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 2There 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

- For every ,
so the standard part of is , or

- 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 3Let . 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 4Let 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 5Uniformly 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 6Let . 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 7Let 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 8Let 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 9One 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 10Let 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 12For 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.