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Problem of the Week #79 - December 2nd, 2013

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Chris L T521

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Jan 26, 2012
995
Here's this week's problem.

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Problem: If $f$ is a holomorphic function on the strip $-1<y<1$, $x\in\mathbb{R}$ with \[\left|f(z)\right|\leq A(1+|z|)^{\eta},\quad\eta\text{ a fixed real number}\]
for all $z$ in that strip, show that for each integer $n\geq 0$ there exists $A_n\geq 0$ so that
\[|f^{(n)}(x)|\leq A_n(1+|x|)^{\eta},\quad\text{for all }x\in\mathbb{R}.\]

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Hint:


Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Chris L T521

Well-known member
Staff member
Jan 26, 2012
995
No one answered this week's problem. You can find my solution below.

For $x\in\mathbb{R}$, applying Cauchy's inequalities to $f(z)$ on the open disk $D_r(x)$ centered at $x$ with radius $0<r<1$, we get
$$\begin{aligned}|f^{(n)}(x)| &\leq \frac{n!}{r^n} \sup\limits_{|z-x|=r}|f(z)|\\ & \leq\frac{n!}{r^n} A\sup\limits_{|z-x|=r} (1+|z|)^{\eta}\\ &\leq\frac{n!}{r^n} A(2+|x|)^{\eta}.\end{aligned}$$
Letting $r\rightarrow 1$, we get
$$|f^{(n)}(x)|\leq n! A(2+|x|)^{\eta}.$$
Thus, we can set
$$A_n=n! A\sup\limits_{x\in\mathbb{R}}\frac{(2+|x|)^{\eta}}{(1+|x|)^{\eta}}.$$
This completes the proof.
 
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