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Complex analysis, holomorphic functions, automorphic group actions and forms, pseudoconvexity, complex geometry, analytic spaces, analytic sheaves.
0
votes
A question that arises in trying to make mathematically precise a well known informal statem...
To say Robert Bryant's explanation in another way: assuming (as in the OP) that the $f(z_i)$ are already the values of a function $f$ analytic in a nbd $U$ of $0$, there is a unique power series $P$ s …
4
votes
Accepted
A New Analytic Inequality
Yes, indeed
$$\frac{1}{2\pi}\int_0^{2\pi} f(e^{it})\overline{f(e^{-it})}dt= |f(0)|^2,$$
as you can check applying the Cauchy formula to the holomorphic function $f(z)\overline{f(\bar z)}$.
16
votes
Accepted
Analytic functions with integer coefficients
We may assume w.l.o.g $0<\beta<1$. Write $\beta^{-1}$ in the $ \beta^{-2}$ expansion as:
$$\beta^{-1}=\sum_{k=0}^\infty d_k\beta^{2k}$$
with integer digits $0\le d_k<\beta^{-2}$, and define
$$f(x):=-1 …
1
vote
Roots of a polynomial inside the unit circle
I think it should be true in the stronger assumption $q>p+2$, but I suspect it could be false for integer $p\ge1$ and $q=p+2$, and an even $k$. Let me explain why -it could be possibly useful to make …
5
votes
Smooth functions with vanishing normal derivatives
To make it a bit quantitative, let $\omega$ be a modulus of continuity of $u$. Extend $u\in C^0(B_1)$ to $u_0(x):=u\big(\frac x{ \|x\|\vee1 }\big)$, which incidentally has modulus of continuity $ \om …
1
vote
A functional equation concerning analytic functions
An elementary argument based on the identity principle for power series.
If $\text{deg}(P)>1$, it has a fixed point $z_0$. We can assume $z_0=0$ (we can replace $f$ with $F(z):=f(z_0+z)$ and $p$ wit …
7
votes
Real integral which cannot be evaluated without complex analysis?
I'm quite sure one could camuflage any complex analysis technique into a real calculus computation, leaving no appeal even to the notion of complex numbers. For instance, the Cauchy's formula on a cir …
4
votes
Accepted
A sequence that tell us if a holomorphic function of several variables is identically zero
Start from a countable dense subset $S$ of the unit ball, and take a sequence $Z_\nu$ such that for all $s\in S$, one has $\nu Z_\nu=s$ infinitely often. Then, any holomorphic function in $\mathbb{C}^ …
8
votes
Complete metric on the space of Jordan curves?
I fear this is possibly a bit too abstract for your purposes, but tells that there is such a complete distance topologically equivalent to the uniform distance, and can be made more concrete (check th …
11
votes
Roots of $x^n-x^{n-1}-\cdots-x-1$
Let's prove the equivalent claim:
The only complex solutions to $2z=z^{n+1} +1$ with $|z|\le 1$ are $z={1\over\alpha}$ and $z=1$.
Remark: The number $\beta:={1\over\alpha}$ is the minimum fixed …
25
votes
Asymptotic approximation of $x^\alpha$ by entire functions
As a matter of fact, real entire functions (that is, entire functions that map the real line into itself, or equivalently, functions represented by a power series centered in 0, with real coefficient …
2
votes
What is the angle between two complex vectors?
Actually, the case of complex vector spaces is rather a particular case than an extension, with respect to the case of real vector spaces. Recall that, as a vector space over $\mathbb{R}$, your $\math …
11
votes
Why are lacunary series so badly behaved?
"Objection, the question assumes facts not in evidence!"
Talking about the general question as in the title, I wonder in what measure can we say that lacunary series are particularly badly behaved. M …
3
votes
Accepted
Asymptotics of the derivatives of analytic functions
I think the answer to both is negative. By a result by Carleman, entire real functions are dense in $C(\mathbb{R},\mathbb{R})$ with the Withney topology, so there is an entire $f$ such that $$1/x + …
7
votes
Accepted
complex polynomials and inequalities
For $n\ge 3$, let's take $z_j$, for $1\le j\le n$, be the $n$-th roots of unity, and $w_j:=z_j+4\bar{z_j}$, so that the assumption are satisfied by $$f(z):=z+4z^{n-1}$$ $$g(z):=4z+z^{n-1}.$$ Howev …