Let $f$ be a holomorphic function on the open unit disc that satisfies $f({1\over 2})+f(-{1\over 2})=0$. Show $|f(0)|\le {1\over 4}$. There is a very similar question and I believe the solution should be quite parallel, but I have been wondering whether I can avoid using an additional map, which I really can't imagine myself considering during a stressful time. Is there another way to attack this?
$f$ holomorphic on $\Bbb{D}$ and $f({1\over 2})+f(-{1\over 2})=0$ then $|f(0)|\le {1\over 4}$, must I use a map?
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complex-analysis
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0There's something wrong with this problem. If you take any holomorphic function $g$ such that $g(1/2)+g(-1/2)=0$ and $g(0)\ne0$ you can construct $f(z)=g(z)/g(0)$, and get $|f(0)|=1$. – 2017-02-11
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0The question is whether or not such a theoretic construction is a certain way to represent $f$. If such $g$ is given, this is the function for which you have to show that $|g(0)|\le {1\over 4}$. Otherwise, $f(z)$ is not what is actually given, but is, to my limited perception, a modification of your own, Could you elaborate? – 2017-02-11
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1Just take $f(z)=\dfrac z{1-z}-\dfrac13$ as a counterexample. (This wasn't meant to be an answer.) – 2017-02-11
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0$f(1/2)+f(-1/2)=\dfrac{1/2}{1-1/2}-\dfrac13+\dfrac{-1/2}{1+1/2}-1/3=1-1/3-1/3-1/3=0$. – 2017-02-11
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0This could be an answer actually – 2017-02-11
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0May be you could fix the problem? – 2017-02-11
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0The problem was copied from a test word by word... – 2017-02-11
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0Maybe it is a typo, corrected during the test itself by giving it to the students attention, while the original paper remained the same. Thank you for ending my overthinking. – 2017-02-11