GPT 6 Astra Ultra attempt on Math-5706

FORMAL RESTATEMENT

Erdős #515; one path for all positive exponents, 2026-09-24. Given a transcendental entire function $f$, construct one locally rectifiable path $\Gamma$ tending to infinity such that, for every $\lambda>0$, \[ \int_\Gamma |f(z)|^{-\lambda}\,|dz|<\infty. \] The integral is with respect to arclength, and the path must be independent of $\lambda$.

QUICK LITERATURE AND CONTEXT CHECK

Lewis--Rossi--Weitsman proved a stronger subharmonic theorem. We reconstruct its level-set argument for $u=\log|f|$, using their short-path lemma as a stated external input. The growth hypothesis for a general subharmonic function is essential and will be verified from transcendence here.

OBJECTIVE AND ATTACK PLAN

Join successively higher level sets by paths whose lengths grow subexponentially in the current level. The integrand then decays exponentially faster than the length, simultaneously for every exponent.

WORK

The geometric input. Lemma 1 of Lewis--Rossi--Weitsman supplies absolute constants $C>0$ and $0<\delta_0<1$ with the following property. If $D$ is a proper simply connected plane domain, $v$ is subharmonic on $D$, $0\leq v\leq1$, and $v(a)\geq1-\delta_0$, there is a rectifiable path from $a$ to some $b\in\partial D$, staying in $D$ before its endpoint, on which $v>0$ before the endpoint, of length at most $C\operatorname{dist}(a,\partial D)$. Its proof uses Hall's lemma and conformal mapping; it is not reproved here.

Growth and the successive levels. Write $u=\log|f|$ and $M_u(r)=\log\max_{|z|=r}|f(z)|$. Cauchy's coefficient estimate gives $M_u(r)\geq d\log r+\log|a_d|$ whenever the Taylor coefficient $a_d$ is nonzero. Such degrees $d$ are unbounded, so $M_u(r)/\log r\to\infty$. Choose $a_1$ with $u_1=u(a_1)>0$, and choose $0<\delta<1$ with $(1-\delta)^2\geq1-\delta_0$. Put $q=(1-\delta)^{-1}$. Given $a_j$ with $u_j=u(a_j)>0$, let $D_j$ be the component of $\{u<q u_j\}$ containing $a_j$. This is a proper simply connected domain. To see simple connectivity, every Jordan curve inside $D_j$ bounds a region on which the maximum-modulus principle keeps $|f|$ below its boundary maximum, itself strictly below $e^{q u_j}$; the filled region remains in the same component. On $D_j$ define \[ v_j=\frac{1-\delta}{u_j}\max(u-\delta u_j,0). \] The maximum removes the singularities of $\log|f|$ at its zeros; $v_j$ is subharmonic, $0\leq v_j<1$, and $v_j(a_j)=(1-\delta)^2$. The lemma gives a path $\gamma_j$ ending at $a_{j+1}\in\partial D_j$ with \[ L_j\leq C\operatorname{dist}(a_j,\partial D_j),\qquad u\geq\delta u_j\ \hbox{on }\gamma_j,\qquad u_{j+1}=q u_j. \tag{1} \] For the last equality, continuity of $|f|$ forces $|f(a_{j+1})|=e^{q u_j}$: a boundary point with a strictly smaller value would be an interior point of the same component. In particular $u_j=q^{j-1}u_1$.

Controlling the lengths without choosing $\lambda$. Fix any $\varepsilon>0$. For all large $j$, the growth just proved implies $M_u(e^{\varepsilon u_j})>q u_j$. Some point on this circle lies outside $D_j$, so \[ L_j\leq C(|a_j|+e^{\varepsilon u_j}),\qquad |a_{j+1}|\leq(1+C)|a_j|+Ce^{\varepsilon u_j}. \tag{2} \] Iterating $2$ gives $|a_j|\leq K(1+C)^j(1+j e^{\varepsilon u_j})$, with a constant absorbing the finitely many initial terms. Therefore $\limsup_j\log(1+|a_j|)/u_j\leq\varepsilon$, since $j/u_j\to0$. As $\varepsilon$ was arbitrary, this ratio tends to zero, and $2$ also gives $\log(1+L_j)=o(u_j)$.

Concatenate the paths $\gamma_j$. Equation $1$ shows that their tail leaves every compact set, because $u$ is bounded above on each compact set and $\delta u_j\to\infty$. Thus the concatenation tends to infinity and is locally rectifiable. For each fixed $\lambda>0$, \[ \int_{\gamma_j}e^{-\lambda u}|dz| \leq L_j e^{-\lambda\delta u_j} \leq e^{-\lambda\delta u_j/2} \] for all sufficiently large $j$. The last series converges because $u_j$ grows geometrically. The finitely many initial integrals are finite by $1$. This single construction therefore works for every $\lambda>0$.

RESULT AND LIMITATIONS

Attempt status: solved. The deduction for every transcendental entire function, the arclength estimate, and independence of the exponent are proved. The only substantive external geometric input is the precise short-path lemma stated above. This reconstructs the known solution and does not claim a new theorem.

SOURCES

https://www.erdosproblems.com/515; J. Lewis, J. Rossi and A. Weitsman, On the growth of subharmonic functions along paths, Lemma 1 and Theorem 1, https://archive.ymsc.tsinghua.edu.cn/pacm_download/116/7409-11512_2006_Article_BF02384375.pdf.

Completion rate estimate: 100%.