# Complex Inequality.

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• Jan 8th 2009, 12:58 PM
skamoni
Complex Inequality.
Q. Let f be a function which is differentiable and suppose that the function satisfies :

$|f(z)|\leq\sqrt{|z|}$

Then show f is identically equal to zero.

I know I probably need to use Liouville's Theorem here, but I don't know how to go about it. Any help would be appreciated. Thanks.
• Jan 8th 2009, 03:14 PM
NonCommAlg
Quote:

Originally Posted by skamoni
Q. Let f be a function which is differentiable and suppose that the function satisfies :

$|f(z)|\leq\sqrt{|z|}$

Then show f is identically equal to zero.

I know I probably need to use Liouville's Theorem here, but I don't know how to go about it. Any help would be appreciated. Thanks.

we have $f(0)=0.$ let $f_1(z)=(f(z))^2.$ then clearly $f_1$ is analytic, $f_1'(0)=2f(0)f'(0)=0$ and $|f_1(z)| \leq |z|.$ define: $g(z)=\begin{cases} \frac{f_1(z)}{z} & \text{if} \ z \neq 0 \\ 0 & \text{if} \ z=0 \end{cases}.$ then $g$ is bounded and analytic.

thus $g(z)=c$ (constant). hence $g \equiv 0,$ which will give us $f \equiv 0. \ \Box$
• Jan 8th 2009, 07:33 PM
ThePerfectHacker
Here is another way. Since $|f(z)| \leq \sqrt{|z|}$ it means $|f^2(z)| \leq |z|$.
Therefore, the function $f^2$ satisfies $f^2(z) = a+bz$ for $a,b\in \mathbb{C}$.
Thus, we have that $f(z) = \pm \sqrt{a+bz}$.
But to have this to be analytic we need that $b=0$.
Thus, $f(z) = c$ for some $c\in \mathbb{C}$.
However, $f(0) = 0 \implies f(z) = 0$.
• Jan 9th 2009, 03:06 PM
skamoni
Quote:

Originally Posted by ThePerfectHacker
Therefore, the function $f^2$ satisfies $f^2(z) = a+bz$ for $a,b\in \mathbb{C}$.

Could you explain this step to me please. Apologies if it's incredibly simple. Thank you.
• Jan 9th 2009, 03:31 PM
Jhevon
Quote:

Originally Posted by skamoni
Could you explain this step to me please. Apologies if it's incredibly simple. Thank you.

It follows from the Extended Liouville's theorem (you may know it under a different name (Thinking))

Basically, it says if $f$ is an entire function and

$|f(z)| \le A + B|z|^k$

then $f$ is a polynomial of degree at most $k$

here $k \ge 0$ is an integer and $A$ and $B$ are constants (positive constants if memory serves me correctly)

note that the case k = 0 is the original Liouville's theorem