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I find an exercice in a book which I am trying to solve: Find all functions f having these properties or show that they do not exist

  1. f holomorphic on a neighborhood of $0$ such that for all $n\in\Bbb{N},\quad2^{-n}<\vert f(1/n)\vert<2^{1-n}.$

I get that such a function doesn't exist using the fact that otherwise we have $f(0)=0$ and so $f(z)=z^pg(z)$ with $g(0)\ne 0$ but using the hypothesis I get $g(0)=0$ so such a function doesn't exist.

  1. f holomorphic on a neighborhood of $0$ such that for all $n\in\Bbb{N},\quad \vert f(\frac{1}n)-\frac{cos(\pi n)}{2n+1}\vert<\frac{1}{n^2}$

I tried using odd and even numbers, we get first $\vert f(\frac{1}n)-\frac{(-1)^n}{2n+1}\vert<\frac{1}{n^2}$ and $\vert f(\frac{1}n)+\frac{(-1)^n}{2n+1}\vert<\frac{1}{n^2}$ but I do not know how to continue.

  1. f holomorphic on $U:=\{z\in \Bbb{C}: 1/2<\vert z\vert<1\}$ such that for all $z\in U,\quad f(z)=f(e^{2i\pi \alpha}z)$ where $\alpha$ is irrational.

No idea for this one.

How can I continue ?

1 Answers 1

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For 2., you may write $f(z)=a_0+a_1 z + O(z^2)$ (and use even/odd values of $n$) to run into a contradiction. You may also show that $f'(0)$ does not exist (it is not a problem of holomorphicity, the function simply can't be differentiable at zero).

For 3. Iterating the condition, noting that $k\alpha\ {\rm mod} \ 1$, $k\in {\Bbb N}$ is dense in $S^1$, you get $f(z) = f( e^{i\phi} z)$ for all $\phi$ (and $z$ in the annulus). You may then conclude.

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    thanks, but still,writing $f(z)=a_0+a_1z+O(z^2)$ I get for off values of $n$ that $\vert a_0+\frac{a_1}{2p+1}+O(\frac{1}{(2p+1)^2})+\frac{(-1)^p}{4p+3}\vert<\frac{1}{(2p+1)}^2$(so that $a_0=0$ by taking the limit) and $\vert a_0+\frac{a_1}{2p}+O(\frac{1}{4p^2})+\frac{(-1)^p}{4p+1}\vert<\frac{1}{(4p^2)}^2$, how can I get a contradiction ? – 2017-02-13
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    After eliminating $a_0=0$ you may multiply by $p$ and conclude that this is not possible when $p\rightarrow \infty$ – 2017-02-13
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    I get that $\vert a_1\vert 1/2$ for odd and even values of $n.$ I don't see any contradiction, I presume that I miss something ? – 2017-02-13
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    Well, you get that $a_1=1/2$ and $a_1=-1/2$ which is difficult to satisfy simultaneously. – 2017-02-13
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    arf, from the beginning I have $(-1)^n$ after I took odd and even values, that's why I cannot get the result... thanks – 2017-02-13
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    How do you conclude for the 3. ? I don't get the argument – 2017-02-14
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    You may e.g. use that a non-constant holomorphic function has isolated zeroes. If $a\in U$ then $f(z)-f(a)=f(|z|)-f(|a|)$ vanishes on the circle of radius $|a|$. – 2017-02-14