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.
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 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 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
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
Several posts ago, we proved as a consequence of Grace’s theorem, so . In particular, if we solve for we get
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
in about . Taking expectations we have
and similarly for . Thus
Now so , whence
Now recall that is uniformly distributed on , so we can choose so that
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
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
Dividing both sides by we have
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
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
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 ,
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
By the previous lemma, if then
and the same holds when we average in Haar measure:
so, using the Cauchy-Schwarz inequality, one has
Meanwhile, if then the fact that
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
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.