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Consider the following problem:

If $f$ is a holomorphic function on the strip $S=\{z=x+iy:-1<y<1,x\in{\Bbb R}\}$ with $$ |f(z)|\leq A(1+|z|)^{\eta} \tag{1} $$ for all $z\in S$, where $\eta$ is a fixed real number, show that for each integer $n\geq 0$ there exists $A_n\geq 0$ so that $$ |f^{(n)}|\leq A_n(1+|x|)^{\eta}\tag{2} $$ for all $x\in{\Bbb R}$.

Let $C_R=\{z\in{\Bbb C}:|z|=R\}$. Then for every $0<R<1$, by Cauchy inequality, we have $$ |f^{(n)}(x)|\leq\frac{n!}{R^n}\|f\|_{x+C_R}\leq \frac{n!}{R^n}A(1+|x|+R)^\eta\tag{3} $$ where $x+C_R=\{x+z:z\in C_R\}$ and $\|f\|_{x+C_R}=\sup\{f(z):z\in x+C_R\}$. But I don't see how I can get rid of the $R$ here. Any help?

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I don't have an answer, but my hunch is you want to integrate over rectangles with height $2 - \epsilon$ and width $2R$ "centered" at $x$, and send $R \rightarrow \infty$ – uncookedfalcon Jan 9 '13 at 1:49
@uncookedfalcon A stretched-out rectangle does not offer any advantage over, say, a square of the same height. Indeed, about half of the integral over the square comes from its horizontal sides, and the rectangle contains those too. – user53153 Jan 9 '13 at 4:57
@Jack How is the second inequality in (3) justified? – Student G Oct 15 '14 at 13:01
@StudentG: it is from (1). – Jack Oct 16 '14 at 2:13
@Jack Oh … Thank you. I had managed to read “show that” where the exercise says “with”. – Student G Oct 16 '14 at 9:01

1 Answer 1

up vote 4 down vote accepted

Just fix $R=1/2$, estimate $3/2+|x| \le 3/2(1+|x|)$, and include $(3/2)^\eta$ in your $A_n$.

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A tighter bound would be to keep R a variable, include $(1+R)^\eta$ in your $A_n$, and then let $R\to 1$. Not that it matters for this problem in isolation – forgetfulfunctor Jul 13 at 12:27

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