New answers tagged

2

Using only basic tools of complex analysis $f(1/z)$ is meromorphic at $0$: $\sum_{n=0}^N a_n(z)f(z)^n=0$ with $a_N\ne 0$, take $k$ such that $m=v_0(a_N(1/z)z^{-kN}) \le v_0(a_n(1/z)z^{-kn})$ (order of zero at $0$, negative if a pole) then $\sum_{n=0}^N b_n(z) (z^k f(1/z))^n = 0$ with the $b_n(z)=a_n(1/z)z^{-m-kn}$ holomorphic at $0$ and $b_N(0)\ne 0$. This ...


5

The only sufficient criterion I can think of is the case when $H$ is cyclic of prime power order. Then $Fix(H)$ is a smooth and complex (this holds for any compact group) but also connected by Smith theory. Edit I just checked that Smith theory actually works for arbitrary finite $p$-groups.


3

I doubt this bound can be improved much, at least for even $n$. Indeed: set $z_1 = z_2 = \ldots = z_k = k^{-1/2}$ and $z_{k+1} = z_{k+2} = \ldots = 0$, so that the $\ell^2$ norm of $(z_n)$ is $1$. (Intuitively, this is the worst-case scenario.) Then $$ g(\mu) = (E_1(k^{-1/2} \mu))^k = (1 - k^{-1/2} z)^k e^{z \sqrt k} .$$ As $k \to \infty$, the above ...


10

This function is (on the real line, at least) the product of $$ \exp( \mu^2 \sum_{i=1}^\infty |z_i|^2 - 2 \mu \Re(\sum_{i=1}^\infty z_i)) \quad (1)$$ and the Hadamard type product $$ \prod_{i=1}^\infty E_1( 2 \mu\Re z_i - \mu^2 |z_i|^2) \quad(2)$$ where $E_1$ is the first elementary factor $$ E_1(z) := (1-z) \exp(z).$$ The expression (1) is clearly entire in ...


12

The optimal exponent is $k$. Such examples are given by sparse power series. This is actually trivial in the case $k=0$ (which was not included in the OP). Then we can simply take $f(z)=\sum j^{-2} z^{N(j)}$, say. This is obviously bounded, and the coefficients $a_n$ will not satisfy $|a_n|\lesssim n^{-\epsilon}$ for any $\epsilon>0$ if $N(j)$ increases ...


6

To find $f(1)$ to high precision, we will expand $f$ as a Laurent series in $(z+1)^{-1}$, and solve for the coefficients. Setting $f(z)=g(z+1)$, we want to find $$g(z)=z+0+a_1z^{-1}+a_2z^{-2}+\cdots$$ satisfying $$g(z)=z-1+\frac{g(2z-1)}{g(2z)},$$ or $(g(z)-z+1)g(2z)=g(2z-1)$. Hence $$\begin{array}{rcrcrcrcrl} \bigg(1&+&a_1z^{-1}&+&a_2z^{-2}&...


13

Here is a more elementary proof. Suppose $F(z,f(z))=0$ where $F$ is a polynomial in two variables. How many solutions can the equation $f(z)=a$ for generic $a$ have? Pugging $f(z)=a$ we obtain $F(z,a)=0$ which has at most $d=\deg F$ solutions. So all equations $f(z)=a$ have at most $d$ solutions, therefore $f$ is rational. I was asked in the comment to stay ...


5

Rohrlich has conjectured that the multiplicative relations in $\mathbb{C}^\times / \overline{\mathbb{Q}}^\times$ between values of $\Gamma$ at rational numbers are generated by the multiplication formula and the reflection formula. In conceptual terms, Lang says that $\Gamma$ is an odd punctured distribution on $\mathbb{Q}/\mathbb{Z}$, and that conjecturally,...


16

A more abstract argument is also possible: $f$ satisfies $p(z,f(z))=0$, and let's for convenience assume that $p$ is irreducible (but the argument works in general). We have two meromorphic maps on the associated Riemann surface $R=\{ (z,w): p(z,w)=0\}$: the standard map $(z,w)\mapsto w$ and also $(z,w)\mapsto f(z)$, this being the composition of $(z,w)\...


14

The following argument is based on Christian Remling's proof (given in a comment), but is more elementary. Let us examine the behavior of $f(1/z)$ as $z\to 0$. The function $f(1/z)$ is algebraic over $\mathbb{C}(z)$, hence there are complex polynomials $p_n(z)$ such that $$\sum_{n=0}^N p_n(z)f(1/z)^n=0.$$ Here $N$ is a positive integer. Without loss of ...


3

It is easy to prove that for every region $D$ there exists a function $f$ analytic in $D$ such that $\partial D$ is the "natural boundary" that is $f$ does not have an analytic continuation into any larger region. So whatever you mean by ``$\partial D$ is $2$-dimensional'', you can always have such an example. Such function is easy to construct: ...


4

The integral operator $$Ph(z)=-\frac{1}{\pi}\int\int h(\zeta)\left(\frac{1}{\zeta-z}-\frac{1}{z}\right)dxdy$$ acts on $L^p$, $p>2$, and the result satiafies $(Ph)_{\overline{z}}=h$ in the sense of distributions. For continuous $h$, this equation may not have a classical $C^1$ solution. Edit. The following example was suggested by user @Fedja. It is known ...


4

Existence. (Maybe not a "nice example" as requested, though.) Take any simple closed curve $S$ with Hausdorff dimension $2$. (I am assuming you mean Hausdorff dimension when you say "dimension".) Take any function $F(z)$ on the unit disk with the unit circle $T$ as its natural boundary. By the Riemann mapping theorem, we get a ...


1

For a small $\alpha > 0$, write $$\beta = \frac{\sin^2 \alpha}{\sin(2 \alpha)} = \frac{\tan \alpha}{2} ,$$ and define $$g(z) = f(z) \exp(i \beta z^2).$$ Then $$|g(z)| \leqslant |f(z)| \leqslant C \exp(|z|^2)$$ for $z \in A$, $$|g(i r)| = |f(i r)| \leqslant M$$ for $r > 0$, and $$\begin{aligned}|g(r e^{i \alpha})| & = |f(r e^{i \alpha}))| \exp(-\...


4

In general, holomorphic maps $f: \mathbb{C} \rightarrow \mathbb{C}$ have no fixed points; but using Picard theorem we can show that $f\circ f$ always have fixed point. Theorem (Fixed-point theorem) Let $f: \mathbb{C} \rightarrow \mathbb{C}$ be holomorphic. Then $f \circ f: \mathbb{C} \rightarrow \mathbb{C}$ always has a fixed point unless $f$ is a ...


7

The set of all polynomials associated with a given one is described as follows: Let the given polynomial be $$P(z)=c(z-z_1)\ldots(z-z_n).$$ Then any associated polynomial is of the form $$Q(z)=\lambda c(z-\sigma_1(z_1))\ldots(z-\sigma_n(z_n)),$$ where each $\sigma_j(z)=z$ or $-\overline{z}$, and $|\lambda|=1$. So, besides the continuous parameter $\lambda$ ...


8

Yes, there is a classification. An isolated branch point can be algebraic or logarithmic. If the branch point is at 0, algebraic means that $f(z^n)$ has a pole or removable singularity at 0. It can also have an essential singularity, but this does not have an accepted name. In the case of a logarithmic point $f(e^z)$ is an (arbitrary) meromorphic function in ...


8

The answer is no in general: for example, if $\phi$ is a cuspidal automorphic form in a cuspidal automorphic representation $\pi$ of $\mathrm{GL}_n(\mathbb{A}_F)$ and $\Phi$ is an Eisenstein series in the noncuspidal automorphic representation $\Pi = \widetilde{\pi} \boxplus \omega$ of $\mathrm{GL}_n(\mathbb{A}_F)$, where $\omega$ is a Hecke character, then ...


12

A differential equation for ${\cal A} (x) $ can be obtained as follows, $$ \frac{d^3}{dx^3 } {\cal A} (x) = \int_{-\infty }^{\infty } dk\, (-ik^3 ) e^{ikx} e^{-k^4 } = \frac{x}{4} \int_{-\infty }^{\infty } dk\, e^{ikx} e^{-k^4 } = \frac{x}{4} {\cal A} (x) $$ where integration by parts has been used in the second equality. The leading asymptotic behavior of ...


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