Let $ f $ a complex entire function such that:
$$ |f(z)| \leq \sqrt{2|z|} + \frac{1}{\sqrt{2|z|}} \quad \forall z \neq 0 $$
Prove that $ f$ is constant.
Thank's in advance!
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Let $g(z)=\frac{f(z)-f(0)}{z}$. Then $g(z)$ is entire and for all $z$ with $|z| \geq 1$ you have
$$ \left| g(z) \right| \leq \left|\frac{f(z)}{z} \right| + \left|\frac{f(0)}{z} \right| \leq \left|\frac{\sqrt{2|z|} + \frac{1}{\sqrt{2|z|}}}{z} \right| + \left|f(0) \right| $$
Now from here it is easily to prove that $g$ is bounded (note that by continuity $g$ is bounded on $|z| \leq 1$), thus, since entire is constant.
This proves that $f(z)=az+b $ for some $a,b$. Plug this in your equation and see that $a =0$.
N. S.
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Nice solution N.S ! Thank you very much! ( You are too fast!) – passenger Jul 05 '12 at 19:24
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I have one small question. $g$ is not defined in $ z=0$ so why is entire? Isn't in holomorphic in $ \mathbb C - {0} $ ? – passenger Jul 05 '12 at 19:28
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You can give the value $f'(0)$ at $0$, and check that it's indeed a holomorphic function. – Davide Giraudo Jul 05 '12 at 19:41
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Or alternately, you can use the Taylor series of $f$ at $z=0$ to show that $g$ has a Taylor series at $z=0$.... – N. S. Jul 05 '12 at 20:21
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O.K. Thank you! – passenger Jul 05 '12 at 20:28
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Since $f(z)$ is entire, from Cauchy integral formula, we have $$f^{(n)}(0) = \dfrac{n!}{2 \pi i}\oint_{C_r} \dfrac{f(z)}{z^{n+1}} dz$$ Now argue out why $f^{(n)}(0) = 0$ for $n > 0$ by letting $r \to \infty$ and looking at what happens to the integral.
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@passenger $$\left \vert f^{(n)}(0)\right \vert \leq \dfrac{n!}{2 \pi i} \oint_{C_r} \dfrac{\sqrt{2} \vert \sqrt{z} \vert + \dfrac1{\sqrt{2} \vert \sqrt{z} \vert}}{\vert z \vert^{n+1}}$$ and let $r \to \infty$ for $n \in \mathbb{Z}^+$ – Jul 05 '12 at 20:15
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