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Math Help - Complex Inequality.

  1. #1
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    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.
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  2. #2
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    Quote Originally Posted by skamoni View Post
    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
    Last edited by NonCommAlg; January 8th 2009 at 04:17 PM.
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  3. #3
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    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.
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  4. #4
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    Quote Originally Posted by ThePerfectHacker View Post
    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.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by skamoni View Post
    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 )

    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
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