Random analytic functions: Ryll-Nardzewski's theorem

April 2021

In 1896, Émile Borel asked what can happen at the border of the disk of convergence of a Taylor series:

Étant donné une série de Taylor, il est intéressant de savoir si elle peut être prolongée en quelque manière au-delà de son cercle de convergence ou si ce cercle est une coupure.

This is a vast and difficult question, but surprisingly enough, it has an answer for random Taylor series, which are series whose coefficients are random.

At first sight, it might seem difficult to study the behaviour of a function which is itself random, and one might be tempted to say that random objects have random behaviours; but as Borel hinted in his 1896 paper, randomness comes with averaging, and averaging rules out eccentricity. That's the surprising result I would like to show in this note: this result essentially says that most random functions cannot be extended outside of their disk of convergence.

The precise statement is below; it is due to the Polish mathematician Czesław Ryll-Nardzewski in 1953.

Hadamard's formula

Let us remind a basic fact on the radius of convergence: if f(z)=anznf(z)=\sum a_n z^n is a power series, its radius of convergence is given by Hadamard's formula,

r=1lim supan1n. r = \frac{1}{\limsup |a_n|^\frac{1}{n}}.

Inside the disk D(0,r)D(0,r), the power series converges, but this says nothing on what happens on the border of the disk or outside. Does f(z)f(z) approach a limit as zz approaches the border, for instance if z1z\to 1? Does it erratically diverge? Are there some points on the boudary, such that ff can be extended analytically around this point?

Think about the series zn\sum z^n. Its radius of convergence is 1, but since the sum is equal to (1z)1(1-z)^{-1}, we can extend it at any point of the circle of radius 1, except at the point z=1z=1; in fact, we can extend this function at every point of C1\mathbb{C}\setminus 1, even if the series representation only holds in the disk D(0,1)D(0,1).

What happens at the border

We say that a complex number ww on the circle C1={z=1}C_1 = \{|z|=1\} is regular if there is a δ>0\delta > 0 such that ff can be extended analytically on D(w,δ)D(w, \delta). By the properties of analytic functions, this extension is unique. If a point is not regular, it is called singular.

The set R\mathscr{R} of regular points of ff is an open subset of C1C_1, and among functions with a radius of convergence equal to 1, it can have very different behaviours:

This last case where R\mathscr{R} is empty might seem pathological; it is actually not. If R\mathscr{R} is empty, we say that C1C_1 is a coupure for ff, following Borel's vocabulary.

Random analytic functions

In many of his works, Borel was interested in the behaviour of 'séries quelconques', as term he sometimes used to mean 'random'. A modern setting would be to take a sequence of (complex or real) random variables X0,X1,X2,...X_0, X_1, X_2, ..., and define

f(z)=n=0Xnzn. f(z) = \sum_{n=0}^\infty X_n z^n.

This is a random analytic function, and its radius of convergence r=1/lim supXn1/nr = 1/\limsup |X_n|^{1/n} is a random variable; however, classical theorems in probability such as Kolmogorov's zero-one law say that when the XnX_n are independent[1], then rr is almost surely a constant – possibly \infty.

Sometimes, this can be checked manually; this is the case for two common examples of random functions with a finite radius of convergence, the Rademacher and the Gaussian series.

Look at the behaviour of (two realizations of) these functions, inside their radius of convergence.

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 C1C_1, with denser zeros and windings. It is tempting to say that both functions have a coupure at C1C_1. That's actually the case and it's the content of the second part of this note.

Two theorems on random analytic functions

When ff is a random analytic function with a radius of convergence almost surely equal to 1, one can define R\mathscr{R} as the set of all points which are almost surely regular for ff, and this set is an open set[2]. If it is empty, then ff has a coupure at C1C_1.


If the XnX_n are symmetric random variables and if f(z)=Xnznf(z)=\sum X_n z^n has a radius of convergence of 11, then either every point in C1C_1 is regular, or C1C_1 is a coupure for ff.

By symmetric, we mean that the law of XnX_n and the law of Xn-X_n are equal.

Proof. Let us suppose that R\mathscr{R} is not empty. It contains a small arc interval of the circle, say {eit: t]a,b]}\{e^{it} : t \in ]a,b]\}, and there is an integer mm such that the arc length of this interval is bigger than 1/m1/m. Now, if the XnX_n are symmetric, then for any deterministic choice of signs sn=±1s_n = \pm 1, the law of the random analytic functions

snXnzn \sum s_n X_n z^n

are the same as the law of ff; in particular, if we set sns_n to be 1-1 if nkn \equiv k modulo mm and 11 otherwise, we obtain a new random series fkf_k with the same law as ff. But

f(z)fk(z)=2n=0Xk+mnzk+mn=2zkg(z) f(z) - f_k(z) = 2\sum_{n=0}^\infty X_{k + mn}z^{k + mn} = 2z^k g(z)

for some function gg, which clearly satisfies g(ze2iπ/m)=g(z)g(z e^{2i\pi/m}) = g(z). So, the left hand side of the preceding equation almost surely has R\mathscr{R} as a regular set, hence so does the right-hand side. But since gg is invariant by rotations of angle 2π/m2\pi/m, it means that the set of regular points of gg must also contain e2iπ/mRe^{2i\pi/m}\mathscr{R}, which contains the whole circle C1C_1; in other words, gg can be extended outside C1C_1, and so does 2zkg(z)=f(z)fk(z)2z^k g(z)=f(z)-f_k(z).

We finally note that f0++fm1=(m2)ff_0+ \dotsb + f_{m-1} = (m-2)f, so that (ff0)++(ffm1)=2f(f - f_0) + \dotsb + (f - f_{m-1}) = 2f. But this would mean that the set R\mathscr{R} of regular points of ff contains CC.

In other words, either R\mathscr{R} is empty, or it is equal to C1C_1.

We now turn to the general case, where the distribution of the XnX_n is not necessarily symmetric; with a clever argument, we can actually symmetrize them and use the preceding result to get Ryll-Nardzewski's theorem.

Ryll-Nardzewski's theorem

If the XnX_n are random complex numbers such that f(z)=Xnznf(z) = \sum X_n z^n almost surely has radius of convergence 1, then only two things can happen.

(1) Either the circle C1C_1 is a coupure of ff.

(2) Or, there is a deterministic analytic function f~\tilde{f} with radius of convergence equal to 1, and such that:

  • the radius of convergence of ff~f-\tilde{f} is s>1s>1

  • the circle CsC_s is a coupure for ff~f-\tilde{f}.

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 XnX_n's.

Proof. Let us suppose that we are not in the first case, that is, the set of almost sure regular points R\mathscr{R} is not empty. We introduce another sequence of random variables X~n\tilde{X}_n, which are independent of the XnX_n's and have the same law. We note F~(z)=X~nzn\tilde{F}(z) = \sum \tilde{X}_n z^n their Taylor series: it has the same distribution as ff.

Now, the coefficients of the random analytic function g(z)=f(z)F~(z)g(z) = f(z) - \tilde{F}(z) are XnX~nX_n - \tilde{X}_n, so they are symmetric and the preceding result applies: almost surely, gg has a coupure at its radius of convergence ss. Since both f,F~f,\tilde{F} have radius of convergence 1, we must have s1s\geqslant 1: but can we have s=1s=1? Well, since ff and F~\tilde{F} both have a nonenmpty set of regular points on the circle of radius 1, it must mean that these points are also regular for gg: if gg has a coupure, it cannot be on the circle of radius 1, and we obtain s>1s>1.

Now comes the end: since ff and F~\tilde{F} are independent, we can essentially consider that F~\tilde{F} is not random... More precisely, conditionnaly on (almost every) realization f~\tilde{f} of F~\tilde{F}, the random function fF~=ff~f - \tilde{F} = f - \tilde{f} has a coupure at s>1s>1.

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 F~\tilde{F}...

The L1L^1 case

The last argument of the proof will be less shady if we allow an extra assumption: we can suppose that the XnX_n are L1L^1. Instead of considering one particular realization of F~\tilde{F}, we can simply average with respect to F~\tilde{F}, which has the same distribution as ff, so

E[F~(z)]=E[X~n]zn=E[Xn]zn.\mathbf{E}[\tilde{F}(z)] = \sum \mathbf{E}[\tilde{X}_n]z^n = \sum \mathbf{E}[X_n]z^n.

The second case of the theorem now reads:

f(z)n=1E[Xn]znf(z) - \sum_{n=1}^\infty \mathbf{E}[X_n]z^n has a coupure at CsC_s for some s>1s>1.

A Poisson example

I recently stumbled across an example illustrating this: consider the case where XnX_n are independent Poisson random variables with parameter dnd^n for some common dd. Routine arguments show that the radius of convergence of ff is 1/d1/d, but in fact there is no coupure at 1/d1/d. To see why, note that

E[Xn]zn=dnzn=11zd\sum \mathbf{E}[X^n]z^n = \sum d^n z^n = \frac{1}{1-zd}

so that

f(z)11zd=(Xndn)zn.f(z) - \frac{1}{1 - zd}= \sum (X_n - d^n)z^n.

By elementary concentration arguments, it is possible to prove that lim supXndn1/n=d\limsup|X_n - d^n|^{1/n} = \sqrt{d}, and consequently f(z)(1zd)1f(z) - (1-zd)^{-1} has a radius of convergence equal to 1/d1/\sqrt{d}, and the singularity at 1/d1/d was really isolated. It is also possible to show that f(z)(1zd)1f(z) - (1-zd)^{-1} has a coupure at 1/d1/\sqrt{d}, 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)


[1] Note that we only asked the XnX_n to be independent, not necessarily identically distributed.
[2] The right way to define R\mathscr{R} is as the union of all arc intervals {e2iπt:t]a,b[}\{e^{2i\pi t} : t\in ]a,b[\} with a,ba,b rationals, and such that almost surely every point in this interval is regular for ff. That way, R\mathscr{R} is measurable (as a countable union of intervals) and every point in R\mathscr{R} is almost surely regular.