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Complex analysis, holomorphic functions, automorphic group actions and forms, pseudoconvexity, complex geometry, analytic spaces, analytic sheaves.

1 vote

The integral inequality

No, such an inequality need not hold: one can construct $f$ of exponential type and a sequence $\{a_n\}$ of real numbers such that $$ \frac1{f(a_n)} \int_{a_n - \frac12}^{a_n - \frac12} \left|\phant …
Noam D. Elkies's user avatar
21 votes

Integral of $\log|e^{it}-1|$

Here's another elementary proof, again assuming one has checked that the integral converges absolutely despite the singularity at $t=0$. Let $I$ be the average of the $2\pi$-periodic function $\log \l …
Noam D. Elkies's user avatar
3 votes

What is $\sum (x+\mathbb{Z})^{-2}$?

For the record, one can prove the product formula for the sine without complex analysis (and without the Gamma function), from which (as David Speyer noted) one can recover $\sum_{i \in \bf Z} (x+i)^{ …
Noam D. Elkies's user avatar
4 votes

on completeness of R_mn, the set of all rational functions of type (m,n)

At the risk that I'm doing somebody's homework... Yes, $R_{m,n}[a,b]$ is complete with respect to the uniform metric $\left\| \cdot \right\|$. Let $f \in {\cal C}([a,b])$ be a uniform limit of ratio …
Noam D. Elkies's user avatar
11 votes
Accepted

Abscissa of convergence of Dirichlet series

That's not true even for power series with real coefficients. Let $$ D(s) = \sqrt{1-2^{-s}} = 1 - \frac12 2^{-s} - \frac18 4^{-s} - \frac1{16} 8^{-s} - \frac5{128} 16^{-s} - \cdots . $$ Then $P(s) = …
Noam D. Elkies's user avatar
1 vote

Complex Zeroes of Stirling functions of the second kind

Alexandre Eremenko already showed that for each $n>1$ the function $$ S_n(x) := \frac1{n!} \sum_{k=1}^n {n \choose k} (-1)^{n-k} k^x $$ has infinitely many zeros $x \in {\bf C}$. One can still say mor …
Noam D. Elkies's user avatar
4 votes
Accepted

Sums of entire surjective functions

One expects there to be no such $a_n$ in general, because the "typical" entire functions is surjective (those that aren't are of the special form $z \mapsto c + \exp g(z)$). An explicit example is $ …
Noam D. Elkies's user avatar
4 votes
Accepted

A generalization of polynomial algebra on a Riemann surface

Counterexample to the first question ("is $A$ an algebra of functions?"): Let $M$ be a vertical strip such as {$x + iy : 0 < x < 1$}, and define $f_1,f_2$ as the restriction to $M$ of the entire func …
Noam D. Elkies's user avatar
65 votes

Absolute value inequality for complex numbers

In general, once you've proven an inequality like this in ${\bf R}$ it holds automatically in any Euclidean space (including ${\bf C}$) by averaging over projections. ("Inequality like this" = inequa …
Noam D. Elkies's user avatar
9 votes

Question on a Basel-like sum

[Edited to fill in a couple of steps, correct some typos, standardize the $L$-function notation, and add comments about the relation with the "Basel sum"] Here's a proof via transformation of the sum …
Noam D. Elkies's user avatar
8 votes
Accepted

Modular form not meromorphic at $\infty$

[expanding on my comment to convert it to an answer] An example is $e^j \varphi$ where $j$ is the $j$-invariant and $\varphi$ is any nonzero form of weight $k$. In general a holomorphic (or meromorphi …
Noam D. Elkies's user avatar
16 votes
Accepted

A statement on complex polynomials

The conjecture is easily seen to be true for $n<3$. We give a counterexample for $n=3$. Let $p_i = z^2 - \omega_i$ where the $\omega_i$ are the cube roots of unity. These are linearly dependent becaus …
Noam D. Elkies's user avatar
32 votes
Accepted

A 14th and 26th-power Dedekind eta function identity?

This question asks in effect to show that $\eta^n$ is a $\pm p^{n/2}$ eigenfunction for the Hecke operator $T_p$. The claim holds because each of these $\eta^n$ happens to be a CM form of weight $n/2 …
Noam D. Elkies's user avatar
24 votes

Rational functions with a common iterate

Over ${\bf C}$, An easy counterexample to question 3 is $f(x) = x^2$, $g(x) = cx^2$ where $c$ is a nontrivial cube root of unity. Then $f(f(x)) = g(g(x)) = x^4$ but $f$ and $g$ do not commute. There a …
Noam D. Elkies's user avatar
15 votes
Accepted

Conformal mappings that preserve angles and areas but not perimeters?

No. Either condition implies $|\,f'(z)| = 1$ for all $z$ in the domain of $\,f$, which in turn implies that $\,f'(z)$ is locally constant (for instance using the open mapping theorem) and thus that $ …
Noam D. Elkies's user avatar

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