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Thread: cauchy inequalities [complex analysis]

  1. #1
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    cauchy inequalities [complex analysis]

    hello all,..

    i need help with this problem

    If $\displaystyle f$ is a holomorphic function on the strip $\displaystyle -1 < y < 1, x \in \mathbb{R}$ with $\displaystyle |f(z)| \leq A (1 + |z|)^{\lambda}$, where $\displaystyle \lambda$ is a fixed real number, for all z in that strip.

    show that for each integer $\displaystyle n \geq 0$ there exist $\displaystyle A_n \geq 0$ s.t

    $\displaystyle |f^{(n)}(x)| \geq A_n(1 + |x|)^{\lambda}$ for all $\displaystyle x \in \mathbb{R}$.

    my uncomplete solution :

    fixed n. for any $\displaystyle x_0 \in \mathbb{R}$, we can make a circle $\displaystyle C$centered at $\displaystyle x_0$ with radius $\displaystyle R < 1$.

    by cauchy inequalities, we have

    $\displaystyle |f^{(n)}(x_0)| \leq \frac{n!||f||_C}{R^n}$

    with $\displaystyle ||f||_C = \sup_{z \in C}|f(z)|$.

    let $\displaystyle \sup_{z \in C}|f(z)| = |f(z_0)|$

    because $\displaystyle |f(z_0)| \leq A (1 + |z_0|)^{\lambda}$, then

    $\displaystyle |f^{(n)}(x_0)| \leq \frac{n!A (1 + |z_0|)^{\lambda}}{R^n}$

    what should I do next?

    thx for any comment
    Last edited by dedust; Feb 15th 2010 at 02:27 PM.
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  2. #2
    Senior Member
    Joined
    Nov 2009
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    Quote Originally Posted by dedust View Post
    hello all,..

    i need help with this problem

    If $\displaystyle f$ is a holomorphic function on the strip $\displaystyle -1 < y < 1, x \in \mathbb{R}$ with $\displaystyle |f(z)| \leq A (1 + |z|)^{\lambda}$, where $\displaystyle \lambda$ is a fixed real number, for all z in that strip.

    show that for each integer $\displaystyle n \geq 0$ there exist $\displaystyle A_n \geq 0$ s.t

    $\displaystyle |f^{(n)}(x)| \geq A_n(1 + |x|)^{\lambda}$ for all $\displaystyle x \in \mathbb{R}$.

    my uncomplete solution :

    fixed n. for any $\displaystyle x_0 \in \mathbb{R}$, we can make a circle $\displaystyle C$centered at $\displaystyle x_0$ with radius $\displaystyle R < 1$.

    by cauchy inequalities, we have

    $\displaystyle |f^{(n)}(x_0)| \leq \frac{n!||f||_C}{R^n}$

    with $\displaystyle ||f||_C = \sup_{z \in C}|f(z)|$.

    let $\displaystyle \sup_{z \in C}|f(z)| = |f(z_0)|$

    because $\displaystyle |f(z_0)| \leq A (1 + |z_0|)^{\lambda}$, then

    $\displaystyle |f^{(n)}(x_0)| \leq \frac{n!A (1 + |z_0|)^{\lambda}}{R^n}$

    what should I do next?

    thx for any comment

    Cauchy inequalities

    If $\displaystyle f$ is holomorphic in an open set that contains the closure of a disc $\displaystyle D$ centered at $\displaystyle z_0$ and of radius $\displaystyle R$, then

    $\displaystyle |f(z_0)| \leq \frac{n!||f||_C}{R^n}$,

    where $\displaystyle ||f||_C = \sup_{z \in C} |f(z)|$ denotes the supremum of $\displaystyle |f|$ on the boundary circle $\displaystyle C$
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