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Source. Theorem 5 and the Lemma after it, pp. 1174-1175, of P. Erdős, "Some remarks on polynomials," Bull. Amer. Math. Soc. 53 (1947), 1169-1176. Pages are the journal's own, as on the source card.

Setting

Page 1174. For a closed set $M$ in the plane and a point $z_0$, let $\omega_n(M,z_0)$ be the maximum of $\lvert f_n'(z_0)\rvert$ over all polynomials $f_n$ of degree $n$ with $\lvert f_n(z)\rvert\le1$ for all $z$ in $M$. The paper quotes Szegő (Math. Z. 23 (1925), 45-61): if the transfinite diameter of $M$ is positive, then $\lim\omega_n(M,z_0)^{1/n}<\infty$. It quotes Fekete (Math. Z. 26 (1927), 324-344): if $z_0$ is not in $M$, then $\lim\omega_n(M,z_0)^{1/n}$ exists, and it is finite if the transfinite diameter of $M$ is positive and infinite if that diameter is $0$.

For $z_0$ in $M$ the paper lists two open questions: (1) does $\lim\omega_n(M,z_0)^{1/n}$ exist; (2) if the transfinite diameter of $M$ is $0$, is $\lim\omega_n(M,z_0)^{1/n}=\infty$? It answers both in the negative, the second by Theorem 5 and the first by Theorem 6.

Statement

Theorem 5 (p. 1174). Let $M$ be the set consisting of $0$ and the points $1/2^k$, $k=0,1,2,\ldots$. Then

$$\omega_n(M,0)<c^n .$$

The set $M$ is closed and countable, so its transfinite diameter is $0$.

Lemma (pp. 1174-1175). Let $a,b,d$ be real numbers with $d-b=b-a$. If $\lvert f_n(z)\rvert<1$ for $a<z<b$, then $f_n'(d)<c_1^n/(b-a)$. The paper notes that the case $a=0$, $b=1$ follows from a result of Szegő and the general case by a linear transformation.

Read depth. Claims checked: the statement, the lemma and the short proof were read on the print.

Proof pointer

Page 1175. The equation $f_n^2(z)=1$ has at most $2n$ real roots, and $\lvert f_n(1/2^k)\rvert<1$ for every $k$, so for some $k$ the bound $\lvert f_n(z)\rvert<1$ holds on the whole interval $1/2^{k+1}<z<1/2^k$. The lemma, applied to that interval and the point $0$, gives $\lvert f_n'(0)\rvert<2^{n+1}c_1^n$, which is below $c^n$ for a suitable $c$. The print says "for some $k>n+1$" [sic]; the bound $2^{n+1}$ needs $k\le n$, which the count of roots gives, since each of the intervals $k=0,1,\ldots,n$ on which the bound fails holds at least two of the roots.