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Let $f(z)$ be a holomorphic function that maps the unit disk to the unit disk. Prove that $$|f^{(5)}(0)| \leq 120.$$ I use some concrete example it seems that this statement work out but i do not know how to come up with a formal proof help any one

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  • $\begingroup$ Is that $5$ a power or the order of the derivative? $\endgroup$ – ajotatxe Apr 28 '15 at 17:44
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Perhaps you should use Cauchy's integral formula $$f^{(5)}(0)=\frac{5!}{2\pi i}\int_\gamma\frac{f(z)}{z^6}dz$$

where $\gamma$ is the unit circle.

Then, use the Estimation Lemma: $$|f^{(5)}(0)|\le \frac{120}{2\pi}|\gamma|M$$ where $|\gamma|$ is the length of $\gamma$ and $M$ is an upper bound for $f(z)/z^6$ on $\gamma$.

Can you finish?

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  • $\begingroup$ i am still not following though. especially the part where you say to use cauchy integral $\endgroup$ – user146269 Apr 28 '15 at 17:54
  • $\begingroup$ i am really having a hard time with this problem $\endgroup$ – user146269 Apr 28 '15 at 17:55
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Assuming $f^5$ means really $f^{(5)}$. Then the hypothesis says that $f$ is holomorphic in the open unit disk $D(0,1)$, and that $|f(z)|\leq 1$ for all $z \in D(0,1)$. Then by the general Cauchy´s Integral Formula $$ |f^{(5)}(0)| = \left| \frac{5!}{2\pi i} \int_{|z|=1} \frac{f(z)}{z^6}dz\right| \leq \frac{|5!|}{|2\pi i|} \int_{|z|=1} \frac{|f(z)|}{|z|^6}|dz| \leq \frac{5!}{2\pi} \int_{|z|=1}|dz| = 5! = 120 $$ since $|f(z)|/|z|^6 \leq 1$ and $\int_{|z|=1}|dz|=2\pi$.

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  • $\begingroup$ i have one question shouldn't it $\frac {5!}{2 \pi i}$ $\endgroup$ – user146269 Apr 28 '15 at 18:04
  • $\begingroup$ where is the i that is the part i do not understand i feel like something is missing $\endgroup$ – user146269 Apr 28 '15 at 18:04
  • $\begingroup$ You are right, am editing that, however, since $|2\pi i|= 2\pi$, after the first inequality the $i$ becomes $|i|=1$. $\endgroup$ – Alonso Delfín Apr 28 '15 at 18:05
  • $\begingroup$ @user146269 yes, indeed the $i$ was missing, now i have made an edit? is everything clear now ? $\endgroup$ – Alonso Delfín Apr 28 '15 at 18:07
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    $\begingroup$ @ᴇʏᴇs , Well since $f$ sends the unit disk to the unit disk thats exactly what it says. By Integrating on the contour $|z|=1$ the region bounded is the unit disk, then $|f(z)|\leq 1$ and $|z|=1$, thus $|f(z)|/|z| \leq 1$ $\endgroup$ – Alonso Delfín Apr 28 '15 at 18:14
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Maybe overly pedantic, but we can't really integrate over $|z| = 1$ without some justification, since $f$ is only assumed to be holomorphic on $|z| < 1$. It's not that big of a deal though, since we can integrate over the circle $|z| = r$ for any $r < 1$ to get: $$ |f^{(5)}(0)| = \bigg| \frac{5!}{2\pi i}\int_{|z|=r} \frac{f(z)}{z^6}\,dz \bigg| \le \frac{120}{2\pi} \cdot 2\pi r \cdot \frac{1}{r^6} = \frac{120}{r^5} $$ since $|f(z)| < 1$ on $|z|=r$ by assumption. The above inequality holds for all $r$, so by letting $r \to 1^-$, it follows that $|f^{5}(0)| \le 120$.

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