Exceptional radii cannot be removed from the logarithmic-derivative estimate

Set $r_n:=2^{n-1}$ and define integers ${\lambda }_1:=2$, ${\lambda }_{n+1}:=4n{\lambda }_n{2}^{n{\lambda }_n}+n+1$ for $n\ge 1$. Then ${\lambda }_n$ is strictly increasing with ${\lambda }_n>n$ and ${\lambda }_n\ge 2^n$, and ${\lambda }_{n+1}\ge 4n{\lambda }_n{2}^{n{\lambda }_n}\ge 16n{\lambda }_n$ for every $n$.

Consequences for $\nu \ge 2$: $\log {\lambda }_{\nu }\ge \log (4(\nu -1){\lambda }_{\nu -1})+(\nu -1){\lambda }_{\nu -1}\log 2\ge (\nu -1){\lambda }_{\nu -1}\log 2$, and ${\lambda }_{\nu -1}\to \mathrm{\infty }$, so $\log {\lambda }_{\nu }\mathrm{/}\nu \to \mathrm{\infty }$ and $\log {\lambda }_{\nu }\mathrm{/}{\lambda }_{\nu -1}\to \mathrm{\infty }$. Hence $\nu =o(\log {\lambda }_{\nu })$, $\log \nu =o(\log {\lambda }_{\nu })$ and ${\lambda }_{\nu -1}=o(\log {\lambda }_{\nu })$.

Define coefficients $a_j:=r_n^{-{\lambda }_n}$ when $j={\lambda }_n$ for some $n$, and $a_j:=0$ otherwise; strict increase of ${\lambda }_n$ makes this unambiguous. On $\mathrm{\mid }z\mathrm{\mid }\le R$, all but finitely many nonzero terms of ${\sum }_{j\ge 0}{a}_j{z}^j$ are bounded by $2^{-{\lambda }_n}$, and ${\sum }_n{2}^{-{\lambda }_n}<\mathrm{\infty }$. Thus [F2] gives absolute uniform convergence on every such disc; the power series has infinite radius and its nonzero terms, in increasing degree order, are precisely $f(z):={\sum }_{n\ge 1}(z\mathrm{/}{r}_n{)}^{{\lambda }_n}$. By [F3], $f$ is entire and $f^{\mathrm{\prime }}(z)={\sum }_{n\ge 1}\frac{{\lambda }_n}{r_n}(z\mathrm{/}{r}_n{)}^{{\lambda }_n-1}$; this is the differentiated power series with zero coefficients omitted. Its coefficient of $z^2$ is $r_1^{-2}=1$, so $f$ is nonconstant.

(Upper bound on lacunary circles) For $\nu \ge 2$ and $\mathrm{\mid }z\mathrm{\mid }=r_{\nu }$, the $n$-th term of $f$ has modulus $2^{(\nu -n){\lambda }_n}$ for $n<\nu$, modulus $1$ for $n=\nu$, and modulus $2^{-(n-\nu ){\lambda }_n}$ for $n>\nu$. In the first block each modulus is at most $2^{{\lambda }_{\nu -1}}$: for $n\le \nu -2$ one has $(\nu -n){\lambda }_n\le (\nu -1){\lambda }_{\nu -2}\le {\lambda }_{\nu -1}$ by step 1.1, while $n=\nu -1$ gives exponent ${\lambda }_{\nu -1}$ exactly. Hence the first block is at most $(\nu -1)2^{{\lambda }_{\nu -1}}$, and the tail is at most ${\sum }_{m\ge 1}{2}^{-m}=1$. So $\mathrm{\mid }f(z)\mathrm{\mid }\le (\nu -1)2^{{\lambda }_{\nu -1}}+2$ and $m_0(r_{\nu },f)\le \log ((\nu -1)2^{{\lambda }_{\nu -1}}+2)\le {\lambda }_{\nu -1}\log 2+\log (2(\nu -1))$.

(Lower bound for the derivative) For $\nu \ge 2$ and $\mathrm{\mid }z\mathrm{\mid }=r_{\nu }$: the $\nu$-th term of $f^{\mathrm{\prime }}$ has modulus ${\lambda }_{\nu }\mathrm{/}{r}_{\nu }$; the earlier terms satisfy ${\sum }_{n<\nu }\frac{{\lambda }_n}{r_n}(\frac{r_{\nu }}{r_n}{)}^{{\lambda }_n-1}=\frac{1}{r_{\nu }}{\sum }_{n<\nu }{\lambda }_n{2}^{(\nu -n){\lambda }_n}\le \frac{(\nu -1){\lambda }_{\nu -1}{2}^{(\nu -1){\lambda }_{\nu -1}}}{r_{\nu }}\le \frac{{\lambda }_{\nu }}{4r_{\nu }}$ by step 1.1; and the later terms satisfy ${\sum }_{n>\nu }\frac{{\lambda }_n}{r_n}(\frac{r_{\nu }}{r_n}{)}^{{\lambda }_n-1}=\frac{1}{r_{\nu }}{\sum }_{m\ge 1}{\lambda }_{\nu +m}{2}^{-m{\lambda }_{\nu +m}}\le \frac{2}{r_{\nu }}$ because ${\lambda }_{\nu +m}\ge 2$ makes each summand at most $2^{-m}$. Hence $\mathrm{\mid }{f}^{\mathrm{\prime }}(z)\mathrm{\mid }\ge \frac{1}{r_{\nu }}({\lambda }_{\nu }-\frac{{\lambda }_{\nu }}{4}-2)\ge \frac{{\lambda }_{\nu }}{2r_{\nu }}$ for all $\nu$, so $m_0(r_{\nu },f^{\mathrm{\prime }})\ge \log {\lambda }_{\nu }-\log 2-(\nu -1)\log 2\ge \frac{1}{2}\log {\lambda }_{\nu }$ for all large $\nu$ by step 2.1.

(Standard proximity of the quotient) For finite complex $u,v\mathrm{\ne }0$ one has ${\log }^{+}\mathrm{\mid }u\mathrm{/}v\mathrm{\mid }\ge {\log }^{+}\mathrm{\mid }u\mathrm{\mid }-{\log }^{+}\mathrm{\mid }v\mathrm{\mid }$; taking angular means gives $m_0(r_{\nu },f^{\mathrm{\prime }}\mathrm{/}f)\ge m_0(r_{\nu },f^{\mathrm{\prime }})-m_0(r_{\nu },f)\ge \frac{1}{2}\log {\lambda }_{\nu }-{\lambda }_{\nu -1}\log 2-\log (2(\nu -1))\ge \frac{1}{3}\log {\lambda }_{\nu }$ for all large $\nu$, by step 2.1.

(Comparison scale) By step 3.1 and [F1], ${\log }^{+}T(r_{\nu },f)+\log r_{\nu }=O({\lambda }_{\nu -1}+\nu +\log \nu )=o(\log {\lambda }_{\nu })$ by step 2.1.

(Infinite order) At $\mathrm{\mid }z\mathrm{\mid }=r_{\nu +1}=2r_{\nu }$ the $\nu$-th term of $f$ equals $2^{{\lambda }_{\nu }}$, the terms with $n<\nu$ sum to at most $2^{{\lambda }_{\nu }-1}$ by step 1.1 (each of the $\nu -1$ terms has exponent $(\nu +1-n){\lambda }_n\le 2{\lambda }_{\nu -1}$, and $(\nu -1)2^{2{\lambda }_{\nu -1}}\le 2^{{\lambda }_{\nu }-1}$ because ${\lambda }_{\nu }-1\ge 2{\lambda }_{\nu -1}+{\log }_2(\nu -1)$ whenever ${\lambda }_{\nu }\ge 16(\nu -1){\lambda }_{\nu -1}$ and ${\lambda }_{\nu -1}\ge 2^{\nu -1}$), and the terms with $n>\nu$ sum to at most $2$ as in step 3.1. Hence $\mathrm{\mid }f\mathrm{\mid }\ge 2^{{\lambda }_{\nu }-1}-2\ge 2^{{\lambda }_{\nu }-2}$ on that circle, so $m_0(r_{\nu +1},f)\ge ({\lambda }_{\nu }-2)\log 2$ and [F1] gives $T(r_{\nu +1},f)\ge m_0(r_{\nu +1},f)$. By step 2.1, $\log {\lambda }_{\nu }\mathrm{/}\nu \to \mathrm{\infty }$; since $\log r_{\nu +1}=\nu \log 2$, this gives $\log T(r_{\nu +1},f)\mathrm{/}\log r_{\nu +1}\to \mathrm{\infty }$, so $f$ has infinite order.

Combining steps 4.1 and 4.2, $\frac{m_0(r_{\nu },f^{\mathrm{\prime }}\mathrm{/}f)}{{\log }^{+}T(r_{\nu },f)+\log r_{\nu }}\ge \frac{\log {\lambda }_{\nu }\mathrm{/}3}{o(\log {\lambda }_{\nu })}\to \mathrm{\infty }$ along $r_{\nu }\to \mathrm{\infty }$; hence $m_0(r_{\nu },f^{\mathrm{\prime }}\mathrm{/}f)$ is not $O({\log }^{+}T(r_{\nu },f)+\log r_{\nu })$.

The construction is explicit: the exponents are given by a closed recursion, the radii are $r_{\nu }=2^{\nu -1}$, and no selection beyond the displayed formulas occurs. Countable Choice is carried only as in the statement.