30
votes
Accepted
Are entire functions “essentially” determined by their maximum modulus function?
This is a classical problem, but only partial results are available:
MR3155684
Hayman, W. K.; Tyler, T. F.; White, D. J.
The Blumenthal conjecture, in Complex Analysis and Dynamical Systems V, 149–157....
16
votes
Accepted
Behavior of $|f'(z)|/(1+|f(z)|^2)$ as $|z| \rightarrow \infty$?
This is not true. The optimal estimate from below for transcendental entire functions is
$$\limsup_{z\to\infty}\frac{|z||f'(z)|}{\log|z|(1+|f(z)|^2)}=\infty,$$
and this is best possible,
J. Clunie and ...
15
votes
Behavior of $|f'(z)|/(1+|f(z)|^2)$ as $|z| \rightarrow \infty$?
A concrete example is given by $f(z)=\cos{\sqrt{z}}$. Then
$$
g_f(z) = \frac{|\sin\sqrt{z}|}{2|\sqrt{z}|(1+|\cos^2{\sqrt{z}}|)} .
$$
Obviously, this is small for large $|z|$ if $\sin{\sqrt{z}}$ is not ...
11
votes
Ways to prove the fundamental theorem of algebra
Here is a variant of d'Alembert's argument using the minimum of $|p(z)|$. It has the advantage that it proves more generally the Gelfand-Mazur theorem (usually proved by complex analysis): Any Banach ...
Community wiki
8
votes
Explicit triples of isomorphic Riemann surfaces
A particular example is that of the Lawson surface $\xi_{g,1}$ of genus $g$. As defined here, it is a compact minimal surface in the 3-sphere, obtained by reflecting the solution of the Plateau ...
Community wiki
7
votes
Explicit triples of isomorphic Riemann surfaces
There are indeed very few pairs (except spheres with 3 or 4 singularities, or tori, and what can be obtained from them by finite coverings, where correspondence 2)-3) is completely explicit. See:
H. P....
Community wiki
6
votes
Explicit triples of isomorphic Riemann surfaces
A classical, wonderful example in which is possible to explicitly see all the three descriptions is the Klein quartic.
Community wiki
5
votes
Accepted
For estimation on the integral $g(t)=\int_{-t}^{t}\left\vert\sum_{k=1}^Ne^{ikx}\right\rvert^2dx$ for small $t>0$
Using the formula for the sum of the first $n$ terms of a geometric series, we have
$$|f(x)|^2=\frac{\sin^2(Nx/2)}{\sin^2(x/2)}$$
and hence for $t\downarrow 0$
$$g(t)=2\int_0^t |f(x)|^2\,dx
=2\int_0^t ...
4
votes
Is the proof in "On Hilbert’s 8th Problem" published on Brazilian Journal of Probability and Statistics correct?
According to MR it has been retracted by the author, see Braz. J. Probab. Stat. 33 (2019). See also the MathReview for details on the proof.
4
votes
Construction of holomorphic function
Such a function does not exist, because the constant value $1$ of $|f(z)|^2$ on the circle $\{z\in\mathbb C\colon|z|^2=1/2\}$ is less than the value $e^{1/4}$ of $|f(z)|^2$ at the center $z=0$ of the ...
4
votes
Zeros of hypergeometric functions with complex variables
Much of the literature addresses the case that $a=-n$ is a negative integer, see Real zeros of $2F1$ hypergeometric polynomials (2013) and Zeros of the hypergeometric polynomial $F(-n,b;c;z)$ (2008). ...
3
votes
Zeros of hypergeometric functions with complex variables
The hypergeometric function is not a function in the usual meaning of this word: it is multivalued. For real parameters $a,b,c$, the question about
real zeros of real branches was investigated by ...
3
votes
Accepted
A question about complex Laplacian on compact Hermitian manifolds
For your first question, note that (let $\omega$ be the Hermitian form)
$$\int_M\Delta_c(f) \omega^n=\int_M(\mathrm{tr}_\omega \sqrt{-1}\partial
\overline{\partial}f)\omega^n=n\int_M \sqrt{-1}\partial
...
2
votes
Accepted
Estimate for an oscillatory integral of the first kind
Write $s=\eta t$ and note that $I(s,y)$ solves the 1D Schrödinger equation $iI_s-3I_{yy}=0$. Thus it satisfies the sharp estimate $|I(s,y)|\le c_0s^{-1/2}$ where $c_0$ is a multiple of $\int|I(0,y)|dy$...
2
votes
The monodromy in the proof of Little Picard via Klein's $J$
The usual proof of Picard's theorem along these lines used another
modular function which is called $\lambda$ and which is related to $J$
by
$$J=\frac{4}{27}\frac{(1-\lambda+\lambda^2)^3}{\lambda^2(1-\...
1
vote
Measure of preimage of Jordan disk under entire map
What you say is not correct, even for $e^z$. The preimage of the open disk
$\{z:|z-1|<1\}$ under $e^z$ does not satisfy your condition. However, if you consider a closed disk which does not contain ...
1
vote
Holomorphic vector fields acting on Dolbeault cohomology
Klemyatin proved that this action is trivial if the corresponding
${\Bbb C}$-flow is compatible with some metric (hence can be extended
to a compact torus action),
https://arxiv.org/abs/1909.04075,
(N....
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