Erdős Problem #119

FORMAL RESTATEMENT

Erdős #119; reconstruction of Korsky's determinant argument, 2026-09-23. Let $|z_j|=1$, $p_n(z)=\prod_{j=1}^n(z-z_j)$, and $M_n=\max_{|z|=1}|p_n(z)|$. The three questions ask whether $\limsup M_n=\infty$, whether $M_n>n^c$ infinitely often for some absolute $c>0$, and whether $\sum_{n\leq N}M_n>N^{1+c}$ for every sufficiently large $N$, for some absolute $c>0$.

QUICK LITERATURE AND CONTEXT CHECK

The current source records affirmative answers and describes the recent cumulative-bound proof of Samuel Korsky. The proof below reconstructs its determinant argument, which yields all three conclusions. The author and source are credited; this is not a claim of an original discovery or a claim that this file has been formally verified.

OBJECTIVE AND ATTACK PLAN

For each fixed $N$, construct $N+1$ polynomials of degree at most $N$ whose coefficient determinant is exactly one. Their $L^2$ norms are bounded by successive $M_j$ divided by square roots of consecutive integers. Hadamard's inequality then forces a factorial lower bound on the product of the $M_j$.

WORK

Give the unit circle normalized arclength measure and write $\|\cdot\|_2$ for its $L^2$ norm. Set $p_0=1$ and $M_0=1$. Fix $N\geq1$. For $1\leq k\leq N$, define \[ r_k=N-k+1,\qquad q_k(z)=\frac1{r_k+1}\sum_{\nu=0}^{r_k}(\overline{z_k}z)^\nu, \qquad f_k=p_{k-1}q_k. \tag{1} \] The conjugate in $1$ is essential. It ensures $q_k(z_k)=1$, including when $z_k$ is not real. Orthogonality of the monomials and $|z_k|=1$ give \[ \|q_k\|_2^2=\frac1{r_k+1}=\frac1{N-k+2}. \tag{2} \] Since $q_k(z_k)-1=0$, polynomial division gives $q_k=1+(z-z_k)s_k$ with $\deg s_k\leq N-k$. Therefore \[ f_k=p_{k-1}+p_ks_k. \tag{3} \] The polynomials $p_k,p_{k+1},\ldots,p_N$ span all multiples of $p_k$ of degree at most $N$: division by $p_k$ leaves monic polynomials of successive degrees $0,1,\ldots,N-k$. Thus $3$ says that replacing $p_{k-1}$ by $f_k$ adds only a linear combination of the later basis vectors. It follows that the change from $p_0,p_1,\ldots,p_N$ to $f_1,\ldots,f_N,p_N$ is triangular with diagonal entries one. The former basis has coefficient determinant one in the monomial basis because its polynomials are monic of successive degrees. The latter basis consequently has the same determinant.

Parseval identifies a polynomial's coefficient-vector norm with its $L^2$ norm on the circle. Hadamard's determinant inequality now implies \[ 1\leq \|p_N\|_2\prod_{k=1}^N\|f_k\|_2 \leq M_N\prod_{k=1}^N\frac{M_{k-1}}{\sqrt{N-k+2}} =\frac{\prod_{k=1}^NM_k}{\sqrt{(N+1)!}}. \tag{4} \] In the middle step we used the pointwise bound $|p_{k-1}|\leq M_{k-1}$, followed by $2$. No separation or distinctness assumption on the roots was used. Squaring $4$ proves \[ \prod_{k=1}^NM_k^2\geq(N+1)!. \tag{5} \] The arithmetic--geometric mean inequality and Stirling's formula then give \[ \sum_{k=1}^NM_k\geq N((N+1)!)^{1/(2N)} =(e^{-1/2}+o(1))N^{3/2}. \tag{6} \] The error term in this lower bound depends only on $N$, so the estimate is uniform over all root sequences. In particular $6$ exceeds $N^{1+c}$ for every sufficiently large $N$ whenever $0<c<1/2$, answering the third question.

There must also be infinitely many $n$ for which $M_n>\sqrt n$. Otherwise choose $K$ after the last exception. The finitely many initial factors yield a constant $A>0$ such that $\prod_{k=1}^NM_k^2\leq A N!$ for every $N\geq K$. This contradicts $5$ as soon as $N+1>A$. Thus the second question holds with $c=1/2$, and this same infinite subsequence implies $\limsup M_n=\infty$.

VERIFICATION AND SCOPE

The proof is complete using only polynomial division, Parseval, Hadamard, arithmetic--geometric mean, and Stirling. Repeated roots cause no problem: the nested monic-polynomial basis remains a basis. A large value among the first $N$ indices alone would not imply infinitely many growing-index witnesses; the product contradiction above checks that quantifier explicitly. No matching upper bound or optimality of the constant in $6$ is asserted.

SOURCES

• Current problem and attribution: https://www.erdosproblems.com/119.
• S. Korsky, Cumulative bounds for polynomials with zeros on the unit circle, Theorem 1.1 and its determinant proof, manuscript linked in the problem discussion: https://drive.google.com/file/d/1ibiaGcLV64Kpi7YP5icaadbe1cvernnD/view.

CONCLUSION

Attempt status: solved by reconstruction of the cited proof. Equations $5$--$6$ answer all three questions affirmatively. No novelty is claimed. Completion rate estimate: 100%.

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 $c_0>0$.