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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 $$ |f^{(n)}(x)|\leq A_n(1+|x|)^{\eta}\tag{2} $$ 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 09 '13 at 01:49
  • @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. –  Jan 09 '13 at 04:57
  • Are you tacitly assuming that $\eta$ is non-negative? – CuriousKid7 Sep 23 '19 at 15:54
  • @CuriousKid7: no, $\eta$ is some fixed real number. –  Sep 23 '19 at 19:22
  • But does the second inequality in (3) hold when $\eta <0$? – CuriousKid7 Sep 23 '19 at 19:57
  • @Jack what is $||f||_{x+C_R}$ ? Is it a circle $C_R$ centered at x? I din't understand why |x| is added to the right of the equation in $A(1+|x|+R)^\eta$? – u_any_45 Sep 22 '20 at 01:16
  • @CuriousKid7 In case that $\eta<0$, just fix $R=1/2$ and bound it from above by $n!2^n(|x|+1/2)^\eta\leq n!4^n(|x|+1)^\eta$ when $|x|>1$. Also, note that $f^{(n)}(x)$ is continuous on $[-1,1]$, so it is naturally bounded. Putting these together we show that such an $A_n$ exists. – dromniscience Feb 09 '22 at 07:28

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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 –  Jul 13 '15 at 12:27