zeros on the circle of convergence

In this question some experiments were used to conjecture that the zeros of partial sums of a series converging to a function with natural boundary on the unit circle were (weakly) converging to the unit circle. Well, nothing is new under the Sun, and this is a Theorem of Jentsch, from his recent PhD thesis:

R. Jentsch, Untersuchungen zur Theorie der Folgen analytischer
Funktionen, Inauguraldissertation
Berlin, 1914,39 s.

Actually, Jentsch's theorem does not need the circle to be the natural boundary, but simply the circle of convergence, so the only thing it is lacking is the stronger statement that the empirical measure of the zeros converges to the Lebesgue measure on the unit circle.

EXACT STATEMENT Let $$f(z) = \sum_{i=0}^\infty a_n z^n,$$ with radius of convergence $1.$ Then every point of the unit circle is an accumulation point of zeros of partial sums of $f.$

END EXACT STATEMENT

The question is: does anyone here know of a more accessible place where the proof of this result could be found (or maybe there is a 10 line argument someone could just post)?

• I'm having some trouble parsing the first sentence of your post. – Anthony Quas Jul 12 '17 at 14:36
• @AnthonyQuas Fair point I fixed it (or tried to). – Igor Rivin Jul 12 '17 at 18:26
• @ChristianRemling See now. – Igor Rivin Jul 12 '17 at 19:22

Disclaimer: I learned the trickery below from N.K.Nikolskii, who was giving us a special topics course in complex analysis when I was a fourth year undegraduate student. I have no idea whether it can also be traced back 100 years but, certainly, it has been used by many people on many occasions and is worth teaching to our graduate students (IMHO).

Main Lemma: Let $r<1<R$. Assume that $P(z)=\sum_{k=0}^n a_kz^k$ is any polynomial of degree $n$ such that $|a_k|\le R^n$ for all $k=0,\dots,n$ and $|a_0|,|a_n|\ge r^n$. Then $\frac 1n\log|P(z)|-\log_+|z|$ is small in $L^1(\frac{dA(z)}{\max(1,|z|^4)})$ provided that $r$ and $R$ are close to $1$ and $n$ is large.

Proof: In the closed unit disk $\{|z|\le 1\}$, we have $|P(z)|\le(n+1)R^n$ whence $\frac 1n\log_+|P(z)|\le \frac{\log(n+1)}n+\log R$. Also, by Jensen, the average value of $\log|P(z)|$ over the unit disk is at least $n\log r$. Hence, we can bound the average value of $\frac1n|\log|P||$ over the disk by $\frac{2\log(n+1)}n+\log (R^2/r)$.

To treat the part outside the unit disk, just notice that the measure $\frac{dA(z)}{\max(1,|z|^4)}$ is invariant under the mapping $z\mapsto 1/z$ and apply the argument above to $z^nP(1/z)$ instead of $P$.

Corollary: If we have a sequence of polynomials like above with the parameters $n\to\infty$, $r=r_n\to 1$, and $R=R_n\to 1$, then the normalized counting measures of their zeroes tend weakly to the Haar measure on the unit circle.

Proof: Just take the Laplacians of both sides and observe that for probability measures the convergence in the sense of generalized functions is equivalent to weak convergence.

The applications to partial sums should be obvious now.

Recent? It's 103 years old! Try Jentzsch, Robert. Untersuchungen zur Theorie der Folgen analytischer Funktionen. Acta Math. 41 (1916), 219--251. doi:10.1007/BF02422945, which is available for free from Project Euclid.

• The proof given is actually quite simple once one has the right idea. If $f_n$ denotes the partial sums and $f_n\not=0$ near $z=1$ (say), then $f_n(z)^{1/n}$ is holomorphic there and (from the factorization as a polynomial) $|f_n(z)|\le (1+C|z|)^n$. So $f_n^{1/n}$ converges on a subsequence by normal families near $z=1$, and it then follows that in fact $f_n^{1/n}\to 1$. This gives $|f_n(z)|\le (1+\delta)^n$, but now one can look at $a_{n+1}=(f_{n+1}-f_n)/z^{n+1}$ to conclude that $R>1$. – Christian Remling Jul 12 '17 at 20:10
• Alternatively, you can look at the sequence of partial sums $f_n$ with $|a_n|^{1/n}\to 1$ and show that $\frac 1n\log|f_n|\to\log_+|z|$ in $L^1_{loc}$, so the Laplacians converge in the sense of distributions, i.e., the normalized counting measures of zeroes of $f_n$ converge weakly to the Haar measure on the unit circle. – fedja Jul 12 '17 at 22:03
• @fedja Is the log convergence basically Christian's argument? – Igor Rivin Jul 12 '17 at 23:46
• @IgorRivin He has it pointwise and under the assumption that there are no zeroes in the neighborhood and derives density. I have it in $L^1$ and under the assumption that $a_n$ is not too small and derive the equidistribution. But the basic underlying idea is, indeed, the same. – fedja Jul 13 '17 at 0:04
• @fedja Cool. If you want to make some version of this a separate answer, I would be happy to accept (and it will be easier for people to find). Note that your observations give a very quick proof on the distribution of zeros of "Kac polynomials". – Igor Rivin Jul 13 '17 at 0:07