# entire function is constant

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.

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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 $a =0$.

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Nice solution N.S ! Thank you very much! ( You are too fast!) –  passenger Jul 5 '12 at 19:24
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 5 '12 at 19:28
You can give the value $f'(0)$ at $0$, and check that it's indeed a holomorphic function. –  Davide Giraudo Jul 5 '12 at 19:41
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 5 '12 at 20:21
O.K. Thank you! –  passenger Jul 5 '12 at 20:28
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.
@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 $n \in \mathbb{Z}^+$ –  user17762 Jul 5 '12 at 20:15