A problem of Erdős, Herzog, and Piranian [EHP58]. It is also listed as Problem 4.10 in [Ha74], where it is attributed to Erdős.

Let the maximal length of such a curve be denoted by $f(n)$.

• The length of the curve when $p(z)=z^n-1$ is $2n+O(1)$, and hence the conjecture implies in particular that $f(n)=2n+O(1)$.


• Dolzhenko [Do61] proved $f(n) \leq 4\pi n$, but few were aware of this work.


• Pommerenke [Po61] proved $f(n)\ll n^2$.


• Borwein [Bo95] proved $f(n)\ll n$ (Borwein was unaware of Dolzhenko's earlier work). The prize of \$250 is reported by Borwein [Bo95].


• Eremenko and Hayman [ErHa99] proved the full conjecture when $n=2$, and $f(n)\leq 9.173n$ for all $n$.


• Danchenko [Da07] proved $f(n)\leq 2\pi n$.


• Fryntov and Nazarov [FrNa09] proved that $z^n-1$ is a local maximiser, and solved this problem asymptotically, proving that\[f(n)\leq 2n+O(n^{7/8}).\]


• Tao [Ta25] has proved that $p(z)=z^n-1$ is the unique (up to rotation and translation) maximiser for all sufficiently large $n$.
  

Erdős, Herzog, and Piranian [EHP58] also ask whether the length is at least $2\pi$ if $\{ z: \lvert f(z)\rvert<1\}$ is connected (which $z^n$ shows is the best possible). This was proved by Pommerenke [Po59].

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The lower bound is easy: this is $\geq n$ and equality holds if and only if $p(z)=z^n$. The assumption that the set is connected is necessary, as witnessed for example by $p(z)=z^2+10z+1$.

The Chebyshev polynomials show that $n^2/2$ is best possible here. Erdős originally conjectured this without the $o(1)$ term but Szabados observed that was too strong. Pommerenke [Po59a] proved an upper bound of $\frac{e}{2}n^2$.

Eremenko and Lempert [ErLe94] have shown this is true, and in fact Chebyshev polynomials are the extreme examples.

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Conjectured by Erdős, Herzog, and Piranian [EHP58]. The lower bound $\gg n^{-4}$ follows from a result of Pommerenke [Po61]. The lower bound $\gg (\log n)^{-1}$ was proved by Krishnapur, Lundberg, and Ramachandran [KLR25].

Wagner [Wa88] proves, for $n\geq 3$, the existence of such polynomials with\[\lvert\{ z: \lvert p(z)\rvert <1\}\rvert \ll_\epsilon (\log\log n)^{-1/2+\epsilon}\]for all $\epsilon>0$. Krishnapur, Lundberg, and Ramachandran [KLR25] improved this upper bound to $\ll (\log\log n)^{-1}$.

In [EHP58] they also ask to determine the polynomials which achieve the minimum possible value of this measure.

Pólya [Po28] showed the upper bound\[\lvert\{ z: \lvert p(z)\rvert <1\}\rvert \leq \pi\]always holds, and this is achieved only when the $z_i$ are identical.

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This is Problem 4.1 in [Ha74] where it is attributed to Erdős.

The weaker conjecture that $\limsup M_n=\infty$ was proved by Wagner [Wa80], who show that there is some $c>0$ with $M_n>(\log n)^c$ infinitely often.

The second question was answered by Beck [Be91], who proved that there exists some $c>0$ such that\[\max_{n\leq N} M_n > N^c.\]Erdős (e.g. see [Ha74]) gave a construction of a sequence with $M_n\leq n+1$ for all $n$. Linden [Li77] improved this to give a sequence with $M_n\ll n^{1-c}$ for some $c>0$.

The third question seems to remain open.

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This is Problem 4.20 in [Ha74], where it is attributed to Erdős.

This was solved independently by Kristiansen [Kr74] (only in the case when $c_k\in\mathbb{R}$) and Saff and Sheil-Small [SSS73] (for general $c_k\in \mathbb{C}$).

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This is Problem 2.31 in [Ha74], where it is attributed to Erdős.

More generally, if $A,B\subseteq \mathbb{R}$ are two countable dense sets then is there an entire function such that $f(A)=B$?

Solved by Barth and Schneider [BaSc70], who proved that if $A,B\subset\mathbb{R}$ are countable dense sets then there exists a transcendental entire function $f$ such that $f(z)\in B$ if and only if $z\in A$. In [BaSc71] they proved the same result for countable dense subsets of $\mathbb{C}$.

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Clunie (unpublished) proved this if $a_n\geq 0$ for all $n$. This was disproved in general by Clunie and Hayman [ClHa64], who showed that the limit can take any value in $[0,1/2]$.

See also [513].

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Originally a conjecture of Littlewood. The answer is yes (for all $n\geq 2$), proved by Balister, Bollobás, Morris, Sahasrabudhe, and Tiba [BBMST20].

In Problem 4.31 of [Ha74] Erdős asks whether there exists a constant $c>0$ such that $\max_{\lvert z\rvert=1}\lvert P(z)\rvert > (1+c)\sqrt{n}$ for all large $n$.

See also [230].

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This is Problem 2.30 in [Ha74], where it is attributed to Erdős.

Solved in the affirmative by Barth and Schneider [BaSc72].

This problem has been formalised in Lean as part of the Google DeepMind Formal Conjectures project.

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This is Problem 4.31 in [Ha74], in which it is described as a conjecture of Erdős and Newman.

The lower bound of $\sqrt{n}$ is trivial from Parseval's theorem. The answer is no (contrary to Erdős' initial guess). Kahane [Ka80] constructed 'ultraflat' polynomials $P(z)=\sum a_kz^k$ with $\lvert a_k\rvert=1$ such that\[P(z)=(1+o(1))\sqrt{n}\]uniformly for all $z\in\mathbb{C}$ with $\lvert z\rvert=1$, where the $o(1)$ term $\to 0$ as $n\to \infty$.

For more details see the paper [BoBo09] of Bombieri and Bourgain and where Kahane's construction is improved to yield such a polynomial with\[P(z)=\sqrt{n}+O(n^{\frac{7}{18}}(\log n)^{O(1)})\]for all $z\in\mathbb{C}$ with $\lvert z\rvert=1$.


See also [228].

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Erdős and Szekeres [ErSz59] proved that $\lim f(n)^{1/n}=1$ and $f(n)>\sqrt{2n}$. Erdős proved an upper bound of $\log f(n) \ll n^{1-c}$ for some constant $c>0$ with probabilistic methods. Atkinson [At61] showed that $\log f(n) \ll n^{1/2}\log n$.

This was improved to\[\log f(n) \ll n^{1/3}(\log n)^{4/3}\]by Odlyzko [Od82].

If we denote by $f^*(n)$ the analogous quantity with the assumption that $a_1<\cdots<a_n$ then Bourgain and Chang [BoCh18] prove that\[\log f^*(n)\ll (n\log n)^{1/2}\log\log n.\]Atkinson [At61] noted this is related to the Chowla cosine problem [510], in that if for any set of $n$ integers $A$ there exists $\theta$ such that $\sum_{n\in A}\cos(n\theta) < -M_n$ then\[\log f^*(n) \ll M_n \log n.\]The answer to the specific question asked is no - Belov and Konyagin [BeKo96] proved that\[\log f(n) \ll (\log n)^4.\]

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A reverse Littlewood-Offord problem. Erdős originally asked this with $\sqrt{2}$ replaced by $1$, but Carnielli and Carolino [CaCa11] observed that this is false, choosing $z_1=1$ and $z_k=i$ for $2\leq k\leq n$, where $n$ is even, since then the sum is at least $\sqrt{2}$ always.

Solved in the affirmative by He, Juškevičius, Narayanan, and Spiro [HJNS24]. The bound of $1/n$ is the best possible, as shown by taking $z_k=1$ for $1\leq k\leq n/2$ and $z_k=i$ otherwise.

See also [498].

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This is Problem 4.4 in [Ha74], where it is attributed to Erdős.

First investigated by Rényi and Rédei [Re47]. Erdős [Er49b] proved that $f(k)<k^{1-c}$ for some $c>0$. The conjecture that $f(k)\to \infty$ is due to Erdős and Rényi.

This was solved by Schinzel [Sc87], who proved that\[f(k) > \frac{\log\log k}{\log 2}.\]In fact Schinzel proves lower bounds for the corresponding problem with $P(x)^n$ for any integer $n\geq 1$, where the coefficients of the polynomial can be from any field with zero or sufficiently large positive characteristic.

Schinzel and Zannier [ScZa09] have improved this to\[f(k) \gg \log k.\]

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A problem of Selfridge and Straus [SeSt58], who prove that this is true if $k=2$ and $\lvert A\rvert \neq 2^l$ (for $l\geq 0$). On the other hand, there are examples with two distinct $A,B$ both of size $2^l$ such that $A_2=B_2$.

Selfridge and Straus [SeSt58] also show that the answer is yes if $k=3$ and $\lvert A\rvert>6$ or $k=4$ and $\lvert A\rvert>12$.

More generally, they prove that $A$ is uniquely determined by $A_k$ if $\lvert A\rvert$ is divisible by a prime greater than $k$.

Kruyt notes that this is trivially false if $\lvert A\rvert=k$ then rotating $A$ around an appropriate point produces a distinct set with the same sum. Presumably some condition like '$\lvert A\rvert$ sufficiently large' is intended. Tao also notes that this is false if $\lvert A\rvert=2k$.

Gordon, Fraenkel, and Straus [GFS62] proved that, for all $k$, the multiset $A_k$ uniquely determines $A$ provided $\lvert A\rvert$ is sufficiently large.

(In [Er61] Erdős states this problem incorrectly, replacing sums with products. This product formulation is easily seen to be false, as observed by Steinerberger: consider the case $k=3$ and subsets of the 6th roots of unity corresponding to $\{0,1,2,4\}$ and $\{0,2,3,4\}$ (as subsets of $\mathbb{Z}/6\mathbb{Z}$). The correct problem statement can be found in the paper of Selfridge and Straus that Erdős cites.)

This is mentioned in problem C5 of Guy's collection [Gu04].

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A strong form of the Littlewood-Offord problem. Erdős [Er45] proved this is true if $z_i\in\mathbb{R}$, and for general $z_i\in\mathbb{C}$ proved a weaker upper bound of\[\ll \frac{2^n}{\sqrt{n}}.\]This was solved in the affirmative by Kleitman [Kl65], who also later generalised this to arbitrary Hilbert spaces [Kl70].

See also [395].

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Cartan proved this is true with $2$ replaced by $2e$, which was improved to $2.59$ by Pommerenke [Po61]. Pommerenke [Po59] proved that $2$ is achievable if the set is connected (see [1046]).

The generalisation of this to higher dimensions was asked by Erdős as Problem 4.23 in [Ha74].

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Chowla's cosine problem. Ruzsa [Ru04] (improving on an earlier result of Bourgain [Bo86]), proved an upper bound of\[-\exp(O(\sqrt{\log N})).\]Polynomial bounds were proved independently by Bedert [Be25c] and Jin, Milojević, Tomon, and Zhang [JMTZ25]. The best bound follows from the method of Bedert [Be25c], which proved the existence of some $c>0$ such that, for all $A$ of size $N$,\[\sum_{n\in A}\cos(n\theta) < -cN^{1/7}.\]The example $A=B-B$, where $B$ is a Sidon set, shows that $N^{1/2}$ would be the best possible here.

This problem is Problem 81 on Green's open problems list.

This is related to [256].

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This is Problem 4.9 in [Ha74], where it is attributed to Erdős.

A problem of Erdős, Herzog, and Piranien [EHP58], who ask more generally whether\[\sum_C \mathrm{diam}(C) \leq n2^{1/n}\]and\[\sum_{C}\max(0, \mathrm{diam}(C)-1)\ll 1\]for all such $f$, where $C$ ranges over the connected components of the set in question. The example $f(z)=z^n-1$ has $\sum_C \mathrm{diam}(C)=(1+o(1))n2^{1/n}$.

They also asked whether, if the roots of $f$ are all in the disc $\{\lvert z\rvert\leq 1\}$, the total number of connected components with diameter $>1$ is absolutely bounded, but noted in an addendum that the answer is no: consider $z^n+1$ and move the zeros $e^{i\pi/n}$ and $e^{-i\pi/n}$ a short distance along the circle towards $1$. The set $\{z: \lvert f(z)\rvert \leq 1\}$ for $f(z)=z^n+1$ looks like $n$ 'leaves' joined at $0$, and this moving of two roots makes $\approx n/2$ of the leaves become disconnected at $0$. Each of the resulting components has diameter $>2^{1/n}-\epsilon$, and hence there are $\gg n$ components of diameter $>1$.

In [Er61] Erdős asks the weaker question given here, with the definition of the set altered so that $<1$ is replaced by $\leq 1$.

Pommerenke [Po61] proved that the answer is no to most of these questions, by showing that, for any $0<d<4$ and $k\geq 1$ there exist monic polynomials $f\in \mathbb{C}[x]$ such that $\{z: \lvert f(z)\rvert\leq 1\}$ has at least $k$ connected components of diameter $\geq d$.

This was independently proved by Huang [Hu25] (unaware of the previous work of Pommerenke). Pólya [Po28] showed that $4$ is the best possible here, in that no connected component can have diameter $>4$.

A picture of the set in question for $z^5-1$ is here.

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Littlewood's conjecture, proved independently by Konyagin [Ko81] and McGehee, Pigno, and Smith [MPS81].

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It is trivial that this value is in $[1/2,1)$. Kövári (unpublished) observed that it must be $>1/2$. Clunie and Hayman [ClHa64] showed that it is $\leq 2/\pi-c$ for some absolute constant $c>0$. Some other results on this quantity were established by Gray and Shah [GrSh63].

See also [227].

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Boas (unpublished) has proved the first part, that such a path must exist.

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Huber [Hu57] proved that for every $\lambda>0$ there is such a path $C_\lambda$ such that this integral is finite.

This is true. The case when $f$ has finite order was proved by Zhang [Zh77]. The general case was proved by Lewis, Rossi, and Weitsman [LRW84], who in fact proved this with $\lvert f\rvert$ replaced by $e^u$ where $u$ is any subharmonic function.

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A problem of Pólya [Po29]. Results of Wiman [Wi14] imply that if $(n_{k+1}-n_k)^2>n_k$ then $\limsup \frac{m(r)}{M(r)}=1$. Erdős and Macintyre [ErMa54] proved this under the assumption that\[\sum_{k\geq 2}\frac{1}{n_{k+1}-n_k}<\infty.\]This was solved in the affirmative by Fuchs [Fu63], who proved that in fact for any $\epsilon>0$\[\log m(r)> (1-\epsilon)\log M(r)\]holds outside a set of logarithmic density $0$.

Kovari [Ko65] has shown that the $\limsup$ is also $1$ for an arbitrary entire function given the stronger assumption that $n_k>k(\log k)^{2+c}$ for some $c>0$. It is conjectured that this condition can be replaced with $\sum \frac{1}{n_k}<\infty$. This would be best possible, as Macintyre [Ma52] has shown that, given any $n_k$ with $\sum \frac{1}{n_k}=\infty$, there is a corresponding entire function which tends to zero along the positive real axis.

In [Er61] this is asked with $m(r)=\max_n \lvert a_nr^n\rvert$, but with this definition the desired equality is a simple consequence of [Wi14] (see the comment by Quanyu Tang).

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A conjecture of Fejér and Pólya.

Fejér [Fe08] proved that if $\sum\frac{1}{n_k}<\infty$ then $f(z)$ assumes every value at least once, and Biernacki [Bi28] proved that if $\sum\frac{1}{n_k}<\infty$ then $f(z)$ assumes every value infinitely often.

Pólya [Po29] proved that if $f$ has finite order then $f(z)$ assumes every value infinitely often under the assumption that $\limsup (n_{k+1}-n_k)=\infty$.

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A problem of Turán, who proved that this maximum is $\gg 1/n$. This was solved by Atkinson [At61b], who showed that $c=1/6$ suffices. This has been improved by Biró, first to $c=1/2$ [Bi94], and later to an absolute constant $c>1/2$ [Bi00]. Based on computational evidence it is likely that the optimal value of $c$ is $\approx 0.7$.

See also [973].

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Erdős and Offord [EO56] showed that the number of real roots of a random degree $n$ polynomial with $\pm 1$ coefficients is $(\frac{2}{\pi}+o(1))\log n$.

It is ambiguous in [Er61] whether Erdős intended the coefficients to be uniformly chosen from $\{-1,1\}$ or $\{0,1\}$. In the latter case, the constant $\frac{2}{\pi}$ should be $\frac{1}{\pi}$ (see the discussion in the comments).

In the case of $\{-1,1\}$ Do [Do24] proved that, if $R_n[-1,1]$ counts the number of roots in $[-1,1]$, then, almost surely,\[\lim_{n\to \infty}\frac{R_n[-1,1]}{\log n}=\frac{1}{\pi}.\]See also [522].

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Random polynomials with independently identically distributed coefficients are sometimes called Kac polynomials - this problem considers the case of Rademacher coefficients, i.e. independent uniform $\pm 1$ values. Erdős and Offord [EO56] showed that the number of real roots of a random degree $n$ polynomial with $\pm 1$ coefficients is $(\frac{2}{\pi}+o(1))\log n$.

There is some ambiguity whether Erdős intended the coefficients to be in $\{-1,1\}$ or $\{0,1\}$ - see the comments section.

A weaker version of this was solved by Yakir [Ya21], who proved that\[\frac{R_n}{n/2}\to 1\]in probability. (This weaker claim was also asked by Erdős, and also appears in a book of Hayman [Ha67].) More precisely,\[\lim_{n\to \infty} \mathbb{P}(\lvert R_n-n/2\rvert \geq n^{9/10}) =0.\]See also [521].

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Salem and Zygmund [SZ54] proved that $\sqrt{n\log n}$ is the right order of magnitude, but not an asymptotic.

This was settled by Halász [Ha73], who proved this is true with $C=1$.

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A problem of Salem and Zygmund [SaZy54]. Chung showed that, for almost all $t$, there exist infinitely many $n$ such that\[M_n(t) \ll \left(\frac{n}{\log\log n}\right)^{1/2}.\]Erdős (unpublished) showed that for almost all $t$ and every $\epsilon>0$ we have $\lim_{n\to \infty}M_n(t)/n^{1/2-\epsilon}=\infty$.

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Random polynomials with independently identically distributed coefficients are sometimes called Kac polynomials - this problem considers the case of Rademacher coefficients, i.e. independent uniform $\pm 1$ values. The first problem asks whether $m(f)<1$ almost surely. Littlewood [Li66] conjectured that the stronger $m(f)=o(1)$ holds almost surely.

The answer to both questions is yes: Littlewood's conjecture was solved by Kashin [Ka87], and Konyagin [Ko94] improved this to show that $m(f)\leq n^{-1/2+o(1)}$ almost surely. This is essentially best possible, since Konyagin and Schlag [KoSc99] proved that for any $\epsilon>0$\[\limsup_{n\to \infty} \mathbb
(m(f) \leq \epsilon n^{-1/2})\ll \epsilon.\]Cook and Nguyen [CoNg21] have identified the limiting distribution, proving that for any $\epsilon>0$\[\lim_{n\to \infty} \mathbb
(m(f) > \epsilon n^{-1/2}) = e^{-\epsilon \lambda}\]where $\lambda$ is an explicit constant.

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It is unclear to me whether Erdős also intended to assume that $\lvert a_{n+1}\rvert\leq \lvert a_n\rvert$.

It is 'well known' that, for almost all $\epsilon_n=\pm 1$, the series diverges for almost all $\lvert z\rvert=1$ (assuming only $\sum \lvert a_n\rvert^2=\infty$).

Dvoretzky and Erdős [DE59] showed that if $\lvert a_n\rvert >c/\sqrt{n}$ then, for almost all $\epsilon_n=\pm 1$, the series diverges for all $\lvert z\rvert=1$.

This is true, and was proved by Michelen and Sawhney [MiSa25], who in fact proved that the set of such $z$ has Hausdorff dimension $1$.

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Bernstein [Be31] proved that for any choice of $a_i^n$ there exists $x_0\in [-1,1]$ such that\[\limsup_{n\to \infty} \sum_{1\leq i\leq n}\lvert p_{i}^n(x)\rvert=\infty.\]Erdős and Vértesi [ErVe80] proved that for any choice of $a_i^n$ there exists a continuous $f:[-1,1]\to \mathbb{R}$ such that\[\limsup_{n\to \infty} \lvert \mathcal{L}^nf(x)\rvert=\infty\]for almost all $x\in [-1,1]$.

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Erdős [Er82e] writes that this was solved in the affirmative 'more than ten years ago', but gives no reference or indication who solved it. From context he seems to attribute this to Barth and Schneider [BaSc72], but this paper contains no such result.

Tang points out that the problem is trivial if we take $f$ to be a polynomial, so presumably it is intended the function $f$ is transcendental.

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A conjecture of Erdős from the early 1950s. Answered in the affirmative by de Bruijn [dB51].

See also [908].

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A conjecture of de Bruijn and Erdős. Answered in the affirmative by Laczkovich [La80].

See also [907].

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The space of rational points in Hilbert space has this property for $n=1$. This was proved for general $n$ by Anderson and Keisler [AnKe67].

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A question of Erdős and Ingham [ErIn64]. The simplest case they could not decide this question for was the finite sequence $\{2,3,5\}$.

Their interest in this problem arose from their proof that the statement that there are no such zeros is equivalent to the fact that, for any non-decreasing $f:\mathbb{R}\to \mathbb{R}_{\geq 0}$ which is bounded on every bounded interval and is $=0$ for $x<1$, the relationship\[f(x)+\sum_k f(x/a_k)=\left(1+\sum_k \frac{1}{a_k}+o(1)\right)x\]implies $f(x)=(1+o(1))x$.

A related question on MathOverflow concerns the vanishing of $2^{-1-it}+3^{-1-it}+5^{-1-it}$, and GH shows that the Four Exponentials conjecture implies this never vanishes.

Yip [Yi25] has proved that this is not always true - in fact, for any real $t\neq 0$, there exists a sequence of integers $1<a_1<\cdots$ such that $\sum \frac{1}{a_i}<\infty$ and $1+\sum_{k}\frac{1}{a_k^{1+it}}=0$.

It remains open whether this is true for every finite sequence of integers.

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This is Problem 7.3 in [Ha74], where it is attributed to Erdős.

Erdős proved (as described on p.35 of [Tu84b]) that such a sequence does exist with $\lvert z_i\rvert\leq 1$. Indeed, Erdős' construction gives a value of $C\approx 1.32$.

In [Er92f] (a different) Erdős refines this analysis, proving that if\[M_2=\min_{z_j} \max_{2\leq k\leq n+1} \left\lvert \sum_{1\leq j\leq n}z_j^k\right\rvert,\]where the minimum is take over all $z_j\in \mathbb{C}$ with $\max \lvert z_j\rvert=1$, then\[(1.746)^{-n} < M_2 < (1.745)^{-n}.\]Tang notes in the comments that Theorem 6.1 of [Tu84b] implies that, if $\lvert z_i\rvert \geq 1$ for all $i$, then\[\max_{2\leq k\leq n+1}\left\lvert \sum_{1\leq i\leq n}z_i^k\right\rvert \geq (2e)^{-(1+o(1))n}.\]See also [519].

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A conjecture of Turán. Erdős speculates that this may be true if there are two distinct $(n-1)$-tuples of consecutive values of $s_k$ which are $0$. He does not elaborate on what the 'essentially' may mean precisely.

This is true (in the stronger form with only two such tuples) - in fact if $n$ is odd then the $z_i$ must be exactly the $n$th roots of unity, and if $n$ is even they must be the vertices of two regular $(n/2)$-gons with the same circumscribed circle centred at the origin. This was first proved by Tijdeman [Ti66]. An independent proof of this was given in the comments section by Hu, Tang, and Zhang.

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This is Problem 7.21 in [Ha74], where it is attributed to Erdős.

Erdős [Er64b] remarks it is 'easy to see' that\[\limsup_{k\to \infty}\left(\sup_n\left\lvert \sum_{j\leq n} e(kx_j)\right\rvert\right)=\infty.\]Erdős [Er65b] later found a 'very easy' proof that $A_k\gg \log k$ for infinitely many $k$. Clunie [Cl67] proved that $A_k\gg k^{1/2}$ infinitely often, and that there exist sequences with $A_k\leq k$ for all $k$. Tao has independently found a proof that $A_k\gg k^{1/2}$ infinitely often (see the comment section).

Liu [Li69] showed that, for any $\epsilon>0$, $A_k\gg k^{1-\epsilon}$ infinitely often, under the additional assumption that there are only a finite number of distinct points. Clunie observed in the Mathscinet review of [Li69], however, that under this assumption in fact $A_k=\infty$ infinitely often.

The question of whether $A_k=o(k)$ is possible (repeated in [Er65b] and [Ha74]) seems to still be open.

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Erdős and Turán [ErTu50] proved such an upper bound with $n$ replaced by $d$.

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A conjecture of Khintchine [Kh23].

In fact this is false, and was disproved by Marstrand [Ma70].

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Erdős [Er49d] constructed a lacunary sequence and $f\in L^2([0,1])$ such that, for every $\epsilon>0$, for almost all $\alpha$\[\limsup_{N\to \infty}\frac{1}{N(\log\log N)^{\frac{1}{2}-\epsilon}}\sum_{1\leq k\leq N}f(\{\alpha n_k\})=\infty.\]Erdős also proved that, for every lacunary sequence and $f\in L^2$, for every $\epsilon>0$, for almost all $\alpha$,\[\sum_{1\leq k\leq N}\sum_{1\leq k\leq N}f(\{\alpha n_k\})=o( N(\log N)^{\frac{1}{2}+\epsilon}).\]Erdős [Er64b] thought that his lower bound was closer to the truth.

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Raikov proved the conclusion always holds (for every $f\in L^2([0,1])$, with no assumption on $\| f-f_n\|_2$) if $n_k=a^k$ for some integer $a\geq 2$. Erdős [Er64b] also asked whether this is true for $n_k=\lfloor a^k\rfloor$ for some $a>1$.

Kac, Salem, and Zygmund [KSZ48] proved that the conclusion holds if\[\| f-f_n\|_2 \ll \frac{1}{(\log n)^{c}}\]for some constant $c>1$. Erdős [Er49d] proved that the conclusion holds if\[\| f-f_n\|_2 \ll \frac{1}{(\log\log n)^{c}}\]for some constant $c>1$. Matsuyama [Ma66] improved this to $c>1/2$.

In [Er64b] Erdős asked whether the conclusion holds for all bounded functions $f$ and lacunary sequences $n_k$.

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The notion of a well-distributed sequence was introduced by Hlawka and Petersen [Hl55].

Erdős proved that, if $n_k$ is a lacunary sequence, then the sequence $\{ \alpha n_k\}$ is not well-distributed for almost all $\alpha$.

He also claimed in [Er64b] to have proved that there exists an irrational $\alpha$ for which $\{\alpha p_n\}$ is not well-distributed. He later retracted this claim in [Er85e], saying "The theorem is no doubt correct and perhaps will not be difficult to prove but I never was able to reconstruct my 'proof' which perhaps never existed ."

The existence of such an $\alpha$ was established by Champagne, Le, Liu, and Wooley [CLLW24].

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A problem of Erdős and Szüsz. Hecke [He22] and Ostrowski ([Os27] and [Os30]) proved the converse.

This is true, and was proved by Kesten [Ke66].

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Kesten [Ke60] proved that if\[f(\alpha,\beta,n)=\frac{1}{\log n}\sum_{1\leq k\leq n}(\tfrac{1}{2}-\{\beta+\alpha k\})\]then $f(\alpha,\beta,n)$ has asymptotic distribution function\[g(c)=\frac{1}{\pi}\int_{-\infty}^{\rho c}\frac{1}{1+t^2}\mathrm{d}t,\]where $\rho>0$ is an explicit constant.

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A problem of Erdős, Herzog, and Piranian [EHP58], who proved that the measure of the set in question is always at most $2\sqrt{2}$ under the assumption that all the roots are in $\{-1,1\}$, and conjecture this is the best possible upper bound.

They also note that the infimum of the set in question is less than $2$, as witnessed by $f(x)=(x+1)(x-1)^m$ for $m\geq 3$. They further note that if the roots are restricted to $[-2,2]$ then the infimum is zero, as witnessed by a small perturbation of the Chebyshev polynomials.

They further conjectured that, if the roots are restricted to $[-2,2]$, then\[\lvert \{ x\in \mathbb{R} : \lvert f(x)\rvert < 1\}\rvert\geq n^{-c}\]for an absolute constant $c>0$. This was proved by Pommerenke [Po61], who in fact showed that this set must contain an interval of width $\gg n^{-4}$.

The current best known bounds (see the discussion in the comments) are\[1.519\approx 2^{4/3}-1\leq \inf \leq 1.835\cdots\]and\[\sup = 2\sqrt{2}\approx 2.828.\]

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A problem of Erdős, Herzog, and Piranian, who note that $f(z)=z^n-1$ has $\rho(f) \leq \frac{\pi/2}{n}$.

Pommerenke [Po61] proved that\[\rho(f) \geq \frac{1}{2en^2}.\]Krishnapur, Lundberg, and Ramachandran [KLR25] proved\[\rho(f) \gg \frac{1}{n\sqrt{\log n}}.\]

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A problem of Erdős, Herzog, and Piranian [EHP58], who show that the answer is yes if $F$ is a line segment or disc, and that if the transfinite diameter is $<1$ then $\{ z: \lvert f(z)\rvert < 1\}$ always contains a disc of radius $\gg_F 1$.

Erdős and Netanyahu [ErNe73] proved that if $F$ is also bounded and connected, with transfinite diameter $0<c<1$, then $\{ z: \lvert f(z)\rvert < 1\}$ always contains a disc of radius $\gg_c 1$.

The transfinite diameter of $F$, also known as the logarithmic capacity, is defined by\[\rho(F)=\lim_{n\to \infty}\sup_{z_1,\ldots,z_n\in F}\left(\prod_{i<j}\lvert z_i-z_j\rvert\right)^{1/\binom{n}{2}}.\]

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A problem of Erdős, Herzog, and Piranian [EHP58], who proved that this set always has a connected component containing at least two of the roots.

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A problem of Erdős, Herzog, and Piranien [EHP58], who proved that if $F$ is the disc of radius $1$ then this set can have $n$ connected components (for example $f(z)=z^n+1$).

The transfinite diameter of $F$, also known as the logarithmic capacity, is defined by\[\rho(F)=\lim_{n\to \infty}\sup_{z_1,\ldots,z_n\in F}\left(\prod_{i<j}\lvert z_i-z_j\rvert\right)^{1/\binom{n}{2}}.\]This was solved by Ghosh and Ramachandran [GhRa24], who proved that, if $d$ is the transfinite diameter of $F$, then

• if $0<d<1$ then the set has at most $(1-c)n$ connected components for some $c>0$ depending on $F$;


• if $d\leq 1/4$ and $F$ is connected then the set has only one connected component;


• there are examples with $d=1$ such that, for infinitely many $n$, the set can have $n$ connected components.

They also note that the answer cannot depend only on the transfinite diameter of $F$ - for example, both $F_1=\{ z: \lvert z\rvert\leq 1/2\}$ and $F_2=[-1,1]$ have transfinite diameter $1/2$, but the former always has one connected component, and the latter can have $\gg n$ many connected components.

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A problem of Erdős, Herzog, and Piranian [EHP58].

Pommerenke [Po61] (using his previous work [Po59]) proved that the answer is no, and there exists such an $f$ for which the projection of this set onto every line has measure at least $2.386$. On the other hand, Pommerenke also proved there always exists a line such that the projection has measure at most $3.3$.

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A problem of Erdős, Herzog, and Piranian [EHP58].

This has been resolved by Tang, who proved that the infimum of $\Lambda(f)$ over all such $f$ is $2$. Tang also suggests that, if the degree $n$ is fixed, then the the infimum over all such $f$ of degree $n$ is attained by $f_n(z)=z^n-1$ (and proves this for $n=1$ and $n=2$).

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A problem of Erdős, Herzog, and Piranian [EHP58], who proved that, for any monic polynomial $f$, if $\{ z: \lvert f(z)\rvert <1\}$ is connected and $f$ has roots $z_1,\ldots,z_n$ then $\prod_{i\neq j}\lvert z_i-z_j\rvert <n^n$.

The value of $\Delta$ when the $z_i$ are the vertices of a regular polygon is $n^n$ when $n$ is even and\[\cos(\pi/2n)^{-n(n-1)}n^n \sim e^{\pi^2/8}n^n\]when $n$ is odd.

Pommerenke [Po61] proved that $\Delta \leq 2^{O(n)}n^n$ for all $z_i$ with $\lvert z_i-z_j\rvert \leq 2$.

Hu and Tang (see the comments) found examples when $n=4$ and $n=6$ that show that vertices of a regular polygon do not maximise $\Delta$. Cambie (also in the comments) showed that, in general, the vertices of a regular polygon are not a maximiser for all even $n\geq 4$.

There is a lot of discussion of this problem in the comments. It is now known that, for even $n$,\[\liminf \frac{\max \Delta}{n^n}\geq C\]for some $C>0$. This was proved with $C\approx 1.0378$ by Sothanaphan [So25]. An alternative construction by Cambie, Dong, and Tang (see the comments by Stijn Cambie) achieves $C\approx 1.304457$ for $6\mid n$, and $C\approx 1.26853$ for all even $n$.


It remains possible that the regular polygon is a maximiser for odd $n$.

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A problem of Erdős, Herzog, and Piranian [EHP58], who also ask, if $\{ z: \lvert f(z)\rvert\leq 1\}$ is connected, then what are the least possible diameter and greatest possible width of this set, and conjecture the answer is $2$ in both cases.

Their guess that the width is always at most $2$ is false, as Pommerenke [Po59] gave an example with width $>\sqrt{3}2^{1/3}\approx 2.18$.

The condition that $E$ is connected is equivalent to $E$ containing all zeros of $f'$.

The answer is yes, and in fact the centre of this disc can be taken to be $\frac{z_1+\cdots+z_n}{n}$, where the $z_i$ are the roots of $f$, as shown by Pommerenke [Po59].

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A question of Grunsky, which was reported by Erdős, Herzog, and Piranian [EHP58].

The answer is no, as shown by Pommerenke [Po61], who showed that, if $k$ is sufficiently large, and $f(z)=z^k(z-a)$ where $a$ is sufficiently close to $(1+\frac{1}{k})k^{\frac{1}{k+1}}$, then $\{ z: \lvert f(z)\rvert\leq 1\}$ has two components, and the component which contains $0$ is not convex.

Goodman [Go66] proved that one of the three components of\[\{ z: \lvert (z^2+1)(z-2)^2\rvert < 5^{3/2}/4\}\]is not convex (depicted here, and constructed an example with simple roots, of degree $4$. The referee of the paper also gave the example of $\lvert z(z^5-1)\rvert<5.6^{-6/5}$ (depicted here).

Goodman raises the question of the maximum number of non-convex components that are possible as a function of the degree of $f$.

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A problem of Erdős, Herzog, and Piranian [EHP58].

Pommerenke [Po61] proved the answer is no for $r>1$, showing that if $f(z)=z^n-r^n$ then $\{ z: \lvert f(z)\rvert \leq 1\}$ has $n$ connected components, all with diameter $\to 0$ as $n\to \infty$.

On the other hand, if $0<r\leq 1$, then the answer is yes, as also shown by Pommerenke [Po61]. Indeed, Pommerenke showed that:

• If $0\leq r\leq 1/2$ then the component which contains $0$ must have diameter $\geq 2$, which $f(z)=z^n$ shows is best possible,


• If $1/2<r\leq \frac{\sqrt{5}-1}{2}$ then the component which contains $0$ must have diameter $>1/r$, and


• If $\frac{\sqrt{5}-1}{2}\leq r\leq 1$ then the component which contains $0$ must have diameter $>2-r^2$.

The example\[f(z)=(z^n+1)(z-1)^2(z-\omega)^{-1}(z-\overline{\omega})^{-1},\]where $\omega=e^{i\pi/n}$, shows one can have the maximum diameter $<1+o(1)$ when $r=1$.

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All the zeros of $f'(x)$ are all distinct real numbers in $(a_0,a_m)$ by Rolle's theorem.

This was proved by Balint [Ba60b]. Balint gives no source for the conjecture - presumably it was from Erdős in personal communication. Some generalisations of this were given by Lorch [Lo76].

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A problem originally due to Hayman [Ha60], according to [GoEr79], although (confusingly) in the book [Ha74] by Hayman it is attributed to Erdős, as Problem 2.41.

Hayman [Ha60b] proved that if $\log M(r) \ll (\log r)^2$ then there exists a path $\Gamma$ on which $f(z)\to \infty$ and $\ell(r)=r$.

Disproved by Gol'dberg and Eremenko [GoEr79] who proved that for any function $\phi(r)$ which $\to \infty$ as $r\to \infty$ there is an entire function $f$ such that\[\log M(r) \ll \phi(r)(\log r)^2\]and there is no path $\Gamma$ on which $f(z)\to \infty$ and $\ell(r) \ll r$. They also construct such functions of any prescribed finite order in $[0,\infty)$.

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This is Problem 1.25 in [Ha74], where it is attributed to Erdős.

Gol'dberg [Go78] and Toppila [To76] have constructed entire functions with this property.

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This is Problem 2.16 in [Ha74], where it is attributed to Erdős.

The answer to the first question is yes, as shown by Herzog and Piranian [HePi68]. The second question is still open, although an 'approximate' affirmative answer is given by Glücksam and Pardo-Simón [GlPa24].

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This is Problem 2.40 in [Ha74] where it is attributed to Erdős. Hayman conjectured that\[\int_0^\infty \frac{r}{\log\log M(r)}\mathrm{d}r<\infty\]is true, and best possible, where $M(r)=\max_{\lvert z\rvert=r}\lvert f(z)\rvert$.

Hayman's strong conjecture was proved independently by Camera [Ca77] and Gol'dberg [Go79b].

The second question was answered in the negative by Gol'dberg [Go79b], who proved that if $T=\{ c>0 : \lvert E(c)\rvert <\infty\}$ then for any $m>0$ there exist entire functions $f$ such that $T=[m,\infty)$ or $T=(m,\infty)$. (It is clear that $T=\emptyset$ and $T=(0,\infty)$ are also possible.)

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This is Problem 2.46 in [Ha74], where it is attributed to Erdős.

Erdős [Er64g] proved (answering a question of Wetzel) that if there are only countably many distinct values for each $f_\alpha(z_0)$ then, if $\mathfrak{c}>\aleph_1$, the family $\{f_\alpha\}$ is itself countable, and also showed this is false if $\mathfrak{c}=\aleph_1$.

In [Ha74] it is written that it is 'easy to see' the answer is yes if $\mathfrak{m}^+<\mathfrak{c}$, and also that it is possible that the question is undecidable.

Indeed, it has been shown that this is undecidable if $\mathfrak{m}^+=\mathfrak{c}$: Kumar and Shelah [KuSh17] have shown that there is a model in which $\mathfrak{c}=\aleph_2$ such that the answer is yes (with $\mathfrak{m}=\aleph_1$), while Schilhan and Weinert [ScWe24] have shown the answer can be no, in a different model with $\mathfrak{c}=\aleph_2$.

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This is Problem 4.22 in [Ha74], where it is attributed to Erdős. In [Ha74] it is reported that Clunie and Netanyahu (personal communication) showed that a path always exists which joins $z=0$ to $\lvert z\rvert=1$ in $A$.

Erdős wrote 'presumably this tends to infinity with $n$, but not too fast'.

The trivial lower bound for the length of this path is $1$, which is achieved for $f(z)=z^n$. The interesting side of this question is what the worst case behaviour is (as a function of $n$).

See also [1041].

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A problem of Kemperman [Ke69], who proved it is true if $f$ is measurable. Erdős [Er81b] wrote 'if it were my problem I would offer \$500 for it'.

This was solved by Laczkovich [La84].

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Proved independently by de Bruijn [dB66] and Jurkat [Ju65].

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The functions $l_k(x)$ are sometimes called the fundamental functions of Lagrange interpolation, and $\Lambda$ is sometimes called the Lebesgue constant.

Faber [Fa14] proved\[\Lambda(x_1,\ldots,x_n)\gg \log n\]for all choices of $x_i$, and Bernstein [Be31] proved it is $>(\frac{2}{\pi}-o(1))\log n$. Erdős [Er61c] improved this to\[\Lambda(x_1,\ldots,x_n)> \frac{2}{\pi}\log n-O(1).\]This is best possible, since taking the $x_i$ as the roots of the $n$th Chebyshev polynomial yields\[\Lambda(x_1,\ldots,x_n)< \frac{2}{\pi}\log n+O(1).\]Erdős thought that the minimising choice is characterised by the property that the sums\[\max_{x\in [x_i,x_{i+1}]}\sum_k \lvert l_k(x)\rvert\]are all equal for $0\leq i\leq n$ (where $x_0=-1$ and $x_{n+1}=1$).

If $x_1=-1$ and $x_n=1$ then there is a unique minimising set of $x_i$, which are symmetric around $0$. (Such a choice is called canonical.)

The minimising canonical choice is known only for $n\leq 4$. For $n=2$ the points are $-1,1$ (with $\Lambda=1$). For $n=3$ the points are $-1,0,1$ (with $\Lambda=1.25$), as shown by Bernstein [Be31]. Rack and Vajda [RaVa15] have shown that for $n=4$ the points are $-1,-t,t,1$ where $t\approx 0.4177$ is an explicit algebraic constant (with $\Lambda \approx 1.4229$).

In [Er67] Erdős suggests that an easier variant might be to have the $x_i\in \mathbb{C}$ with $\lvert x_i\rvert=1$, and seek to minimise $\max_{\lvert z\rvert=1}\sum_{k}\lvert l_k(z)\rvert$, adding it 'seems certain' that the minimising $x_i$ are the $n$th roots of unity. This was proved by Brutman [Br80] for odd $n$ and by Brutman and Pinkus [BrPi80] for even $n$.

See also [1130] and [1132].

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The functions $l_k(x)$ are sometimes called the fundamental functions of Lagrange interpolation.

Erdős [Er47] could prove\[\Upsilon(x_1,\ldots,x_n)< \sqrt{n}.\]Erdős thought that the maximising choice is characterised by the property that the sums\[\max_{x\in [x_i,x_{i+1}]}\sum_k \lvert l_k(x)\rvert\]are all equal for $0\leq i\leq n$ (where $x_0=-1$ and $x_{n+1}=1$), which would be the same (conjectured) characterisation as [1129].

See also [1129].

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Erdős first conjectured this minimum was achieved by taking the $x_i$ to be the roots of the integral of the Legendre polynomial, since Fejer [Fe32] had earlier shown these to be minimisers of\[\max_{x\in [-1,1]}\sum_k \lvert l_k(x)\rvert^2.\]This was disproved by Szabados [Sz66] for every $n>3$.

Erdős, Szabados, Varma, and Vértesi [ESVV94] proved that\[2-O\left(\frac{(\log n)^2}{n}\right)\leq \min I\leq 2-\frac{2}{2n-1}\]where the upper bound is witnessed by the roots of the integral of the Legendre polynomial as above.

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A result of Bernstein [Be31] implies that the set of $x\in(-1,1)$ for which\[\limsup_{n\to \infty}\frac{L_n(x)}{\log n}\geq \frac{2}{\pi}\]is everywhere dense.

Erdős [Er61c] proved that, for any fixed $x_1,\ldots,x_n\in [-1,1]$,\[\max_{x\in [-1,1]}\sum_{1\leq k\leq n}\lvert l_k(x)\rvert>\frac{2}{\pi}\log n-O(1).\]See also [1129] for more on $L_n(x)$.

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Erdős proved that, for any $C>0$, there exists $\epsilon>0$ such that if $n$ is sufficiently large and $m=\lfloor (1+\epsilon)n\rfloor$ then for any $x_1,\ldots,x_m\in [-1,1]$ there is a polynomial $P$ of degree $n$ such that $\lvert P(x_i)\rvert\leq 1$ for $1\leq i\leq m$ and\[\max_{x\in [-1,1]}\lvert P(x)\rvert>C.\]The conjectured statement would also imply this, but Erdős in [Er67] says he could not even prove it for $m=n$.

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