The Gaussian function is on top, the Rademacher function at the bottom. The small inset on the left is the color scheme for these domain colourings.
We can see a more and more erratic behaviour closer to the border of , with denser zeros and windings. It is tempting to say that both functions have a coupure at . That's actually the case and it's the content of the second part of this note.
When is a random analytic function with a radius of convergence almost surely equal to 1, one can define as the set of all points which are almost surely regular for , and this set is an open set. If it is empty, then has a coupure at .
If the are symmetric random variables and if has a radius of convergence of , then either every point in is regular, or is a coupure for .
By symmetric, we mean that the law of and the law of are equal.
Proof. Let us suppose that is not empty. It contains a small arc interval of the circle, say , and there is an integer such that the arc length of this interval is bigger than . Now, if the are symmetric, then for any deterministic choice of signs , the law of the random analytic functions
are the same as the law of ; in particular, if we set to be if modulo and otherwise, we obtain a new random series with the same law as . But
for some function , which clearly satisfies . So, the left hand side of the preceding equation almost surely has as a regular set, hence so does the right-hand side. But since is invariant by rotations of angle , it means that the set of regular points of must also contain , which contains the whole circle ; in other words, can be extended outside , and so does .
We finally note that , so that . But this would mean that the set of regular points of contains .
In other words, either is empty, or it is equal to .
We now turn to the general case, where the distribution of the is not necessarily symmetric; with a clever argument, we can actually symmetrize them and use the preceding result to get Ryll-Nardzewski's theorem.
If the are random complex numbers such that almost surely has radius of convergence 1, then only two things can happen.
(1) Either the circle is a coupure of .
(2) Or, there is a deterministic analytic function with radius of convergence equal to 1, and such that:
the radius of convergence of is
the circle is a coupure for .
This was beforehand a conjecture, and it was proved by the Polish mathematician C. Ryll-Nardzewski; a few simpler proofs appeared afterwise. This one is notably simple and is drawn from J.-P. Kahane's book, Some random series of functions. The same statement actually holds on assumptions weaker than independence of the 's.
Proof. Let us suppose that we are not in the first case, that is, the set of almost sure regular points is not empty. We introduce another sequence of random variables , which are independent of the 's and have the same law. We note their Taylor series: it has the same distribution as .
Now, the coefficients of the random analytic function are , so they are symmetric and the preceding result applies: almost surely, has a coupure at its radius of convergence . Since both have radius of convergence 1, we must have : but can we have ? Well, since and both have a nonenmpty set of regular points on the circle of radius 1, it must mean that these points are also regular for : if has a coupure, it cannot be on the circle of radius 1, and we obtain .
Now comes the end: since and are independent, we can essentially consider that is not random... More precisely, conditionnaly on (almost every) realization of , the random function has a coupure at .
This last statement is actually much stronger than the statement of the theorem, since it means that there are many deterministic functions such that the conclusion holds: almost every realization of ...
The last argument of the proof will be less shady if we allow an extra assumption: we can suppose that the are . Instead of considering one particular realization of , we can simply average with respect to , which has the same distribution as , so
The second case of the theorem now reads:
I recently stumbled across an example illustrating this: consider the case where are independent Poisson random variables with parameter for some common . Routine arguments show that the radius of convergence of is , but in fact there is no coupure at . To see why, note that
By elementary concentration arguments, it is possible to prove that , and consequently has a radius of convergence equal to , and the singularity at was really isolated. It is also possible to show that has a coupure at , thus giving a full illustration of Ryll-Nardzewski's theorem.
Jean-Pierre Kahane, Some random series of functions
Czesław Ryll-Nardzewski, D. Blackwell's conjecture on power series with random coefficients (Studia Math., 1953)
Émile Borel, Sur les séries de Taylor (C.R. de l'Acad., 1896)
|||Note that we only asked the to be independent, not necessarily identically distributed.|
|||The right way to define is as the union of all arc intervals with rationals, and such that almost surely every point in this interval is regular for . That way, is measurable (as a countable union of intervals) and every point in is almost surely regular.|