gpt pro 5.2 attempt on Math-5440

Erdős Problem #119

1) FORMAL RESTATEMENT

Let $(z_i)_{i\ge1}$ be a sequence of complex numbers with $|z_i|=1$ for all $i$. For $n\ge1$ define the monic polynomial \[ p_n(z)=\prod_{i=1}^n (z-z_i). \] Define \[ M_n := \max_{|z|=1} |p_n(z)|. \] Questions:

1. Is $\limsup_{n\to\infty} M_n = \infty$?
2. Is there $c>0$ such that $M_n>n^c$ for infinitely many $n$?
3. Is there $c>0$ such that for all sufficiently large $n$, \[ \sum_{k\le n} M_k > n^{1+c}? \]

2) QUICK LITERATURE/CONTEXT CHECK

From the problem text:

• $1$ is true (Wagner), with the weaker growth $M_n>(\log n)^c$ infinitely often.
• $2$ is true (Beck): $\max_{n\le N} M_n > N^c$ for some $c>0$.
• $3$ seems open.
• There exist sequences with $M_n\le n+1$ for all $n$ (Erdős) and even $M_n\ll n^{1-c}$ for some $c>0$ (Linden).

I do not re-prove $1$ or $2$. I give basic quantitative identities and a sanity-check computation.

3) ATTACK PLAN

To force large $M_n$, one might try to prove that $p_n$ cannot stay uniformly small on the unit circle for too long, perhaps using Fourier/energy or potential theory. For the sum question $3$, one would aim for a lower bound on many $M_k$ simultaneously. Here I only establish two "first principles" lemmas about averages and $L^2$ norms.

4) WORK

Lemma 119.1 (mean of $\log|p_n|$ on the circle is $0$).

For every $n\ge1$, \[ \frac{1}{2\pi}\int_0^{2\pi} \log\big|p_n(e^{it})\big|\,dt = 0. \]

Proof. Write $p_n(z)=\prod_{j=1}^n (z-z_j)$. Then \[ \log|p_n(e^{it})| = \sum_{j=1}^n \log|e^{it}-z_j|. \] Fix a single $z_j$ with $|z_j|=1$ and rotate so $z_j=1$ (rotation preserves the integral over $t$). Then \[ \frac{1}{2\pi}\int_0^{2\pi} \log|e^{it}-1|\,dt =\frac{1}{2\pi}\int_0^{2\pi} \log\big(2|\sin(t/2)|\big)\,dt. \] The classical integral identity $ \int_0^{\pi} \log(\sin u)\,du = -\pi\log 2 $ implies $ \int_0^{2\pi} \log(2|\sin(t/2)|)\,dt = 0. $ Therefore each summand has mean $0$, and summing over $j=1,\dots,n$ gives the claim. $\square$

Lemma 119.2 ($L^2$ lower bound on $M_n$).

Let $p_n(z)=\sum_{k=0}^n a_k z^k$ (so $a_n=1$ and $a_0=(-1)^n z_1\cdots z_n$ has modulus $1$). Then \[ \frac{1}{2\pi}\int_0^{2\pi} |p_n(e^{it})|^2\,dt = \sum_{k=0}^n |a_k|^2 \ge 2, \] and hence \[ M_n\ge \sqrt{2}. \]

Proof. By Parseval's identity for trigonometric polynomials, \[ \frac{1}{2\pi}\int_0^{2\pi} \Big|\sum_{k=0}^n a_k e^{ikt}\Big|^2 dt = \sum_{k=0}^n |a_k|^2. \] Since $a_n=1$ and $|a_0|=|z_1\cdots z_n|=1$, we have $\sum_{k=0}^n |a_k|^2\ge |a_0|^2+|a_n|^2=2$. Finally, $M_n^2=\max_{|z|=1}|p_n(z)|^2\ge$ the average value of $|p_n|^2$ on the circle, so $M_n^2\ge2$. $\square$

FAST REALITY CHECK (numerical sampling for one explicit sequence).

Take $z_j = e^{2\pi i\, j\varphi}$ where $\varphi=(\sqrt5-1)/2$ (an irrational). For $1\le n\le 200$, I approximated $M_n$ by evaluating $|p_n(z)|$ on a uniform grid of $4096$ points on the unit circle. This gives a lower bound on the true $M_n$.

The sampled values (selected) were: \[ \begin{array}{c|c} n & \text{sampled }M_n\\\hline 10 & 5.15\cdots\\ 20 & 19.50\cdots\\ 50 & 12.64\cdots\\ 100 & 15.17\cdots\\ 200 & 15.69\cdots \end{array} \] and the maximum sampled value over $1\le n\le 200$ was about $134.6$ (at $n=143$). This is consistent with unbounded growth in this example, but does not constitute a proof.

5) VERIFICATION

• Lemma 119.1 uses the standard integral $\int_0^{\pi} \log(\sin u)du=-\pi\log2$. The integrand has an integrable singularity; the identity is classical and can be checked by Fourier series.
• Lemma 119.2: Parseval is exact; the step $M_n^2\ge$ average is just "max $\ge$ mean".
• Numerical check is only a sanity check: grid sampling underestimates the true maximum.

6) FINAL

UNRESOLVED.

(i) Strongest fully proved partial result obtained here. The exact mean-zero identity for $\log|p_n|$ on the circle (Lemma 119.1) and the uniform lower bound $M_n\ge\sqrt2$ (Lemma 119.2).

(ii) Exact first gap. A proof (for the open part $3$) that $\sum_{k\le n} M_k$ must grow faster than $n^{1+c}$ for some $c>0$, uniformly for all large $n$.

(iii) Top 3 next moves.

1. Try to turn Lemma 119.1 into a quantitative lower bound on large deviations of $\log|p_n|$ (entropy/large deviations) implying many large $M_n$.
2. Use potential theory on the circle: relate $\log|p_n|$ to the discrete logarithmic potential of the points $z_i$.
3. Search for worst-case sequences (those with very small $M_n$) and attempt to prove that even there, partial sums $\sum M_k$ must grow superlinearly.

(iv) Minimal counterexample structure. A counterexample to $3$ would be a sequence $(z_i)$ for which $M_n$ is often small (e.g. $M_n\ll n^{\varepsilon}$ for all large $n$) and additionally $\sum_{k\le n} M_k$ stays $O(n^{1+c_0})$ for every fixed $c_0>0$.