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This is another question from a recent qualifying exam that really stumped me. I was thinking of using something with the Cauchy estimate for derivatives, but was clueless beyond that.

Let $f:[-1,1]\times [0,2]\rightarrow \mathbb{C}$ be real valued on the interval $[-1,1]$ in the $x$-axis. Show that if $|f|\leq 1$ on its domain, and is analytic on the interior of its domain, then $|f^{(8)}(\frac{i}{4})|\leq 120$.

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Does the fact that it's a map into the unit disc help at all? I'm not seeing a way to use the condition that it is real on the real axis... –  Potato Dec 20 '12 at 4:55
@Potato The condition of being real on the axis makes it possible to extend $f$ by the reflection principle to a larger rectangle. In the larger rectangle the point $i/4$ is not as close to the boundary, which leaves more room for experimenting with the Cauchy integral formula. –  user53153 Dec 20 '12 at 6:37
@PavelM Ah, so that's it. Thanks! –  Potato Dec 20 '12 at 6:40
Are you sure there's not a typo in the problem? $f^{(5)}$ instead of $f^{(8)}$ would make a lot more sense, especially as a qualifying exam problem. –  user53153 Jan 2 '13 at 4:44
Yes, I'm very sure. If you like you can view the exam here: kent.edu/math/graduate/resources/docs/upload/… –  Frank White Jan 2 '13 at 15:46
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2 Answers

up vote 4 down vote accepted

Based on Pavel M's answer, a simpler counter-example is $f_a(z)=e^{a(z-1)}$, where $a>0$ satisfies that $a^8>120e^a$. The range of $a$ is larger than $[2.5,18.5]$.

Write $z=x+iy$. On the one hand, $|f_a(z)|=f_a(x)=e^{a(x-1)}\le 1$ when $x\le 1$, so $f_a$ satisfies the conditions. On the other hand, since $f_a^{(8)}(z)=a^8f_a(z)$, according to the choice of $a$,


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Optimizing $a$ we get $a=8$, with $f^{(8)}(i/4)=(8/e)^8\approx 5628$. Very neat! –  user53153 Jan 4 '13 at 19:15
(I meant $|f^{(8)}(i/4)|$, of course.) –  user53153 Jan 4 '13 at 19:33
@PavelM: Of course. Thank you for your comment. –  Hu Zhengtang Jan 5 '13 at 1:08
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By the reflection principle, we can extend $f$ to the larger rectangle $R=[-1,1]\times [-2,2]$, and the upper bound $|f|\le 1$ holds for the extended function as well. The rectangle $R$ contains a disk of radius $1$ centered at $i/4$. Applying the Cauchy integral formula on this disk, we find that $|f^{(n)}(i/4)|\le n!$ for all $n$. In particular, $|f^{(5)}(i/4)|\le 120$, which I believe was the intent of the problem. For the 8th derivative we get $|f^{(8)}(i/4)|\le 40320$.

Indeed, I think I have a counterexample to the stated bound on $f^{(8)}(i/4)$. The idea is to map $R$ into the unit disk $\mathbb D$. Mapping rectangles is hard (elliptic functions, whatever). Instead I'm going to map a larger domain: the vertical strip $S=\{z:-1<\operatorname{Re}z<1\}$. This is a standard exercise with conformal maps: $$\psi(z)= i\,\frac{\exp(\pi i z/2)-1}{\exp(\pi i z/2)+1}$$ maps $S$ onto $\mathbb D$ and is real on the segment $[-1,1]$ (this segment is sent onto a half of the unit circle by the exponential map, and then onto horizontal diameter by the Möbius map).

No, $\psi$ is not my example. After all, $|\psi^{(8)}(i/4)|<24$ as you can find from your friendly computer algebra system, e.g., evalf(eval(diff(psi,z$8),z=Pi*I/4)) in Maple. When maximizing the 8th derivative, we should take the 8-fold cover: $f(z)=\psi(z)^8$. The same Maple thing now tells me that $|f^{(8)}(i/4)|>1800$.

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Executive summary of the 2nd paragraph: "map ... Mapping ... map ... maps ... maps ... map ... map". –  user53153 Jan 2 '13 at 8:27
So, is this a counterexample, or a solution? –  Frank White Jan 2 '13 at 15:50
@Frank I gave a solution to a different version of the problem, and a counterexample to the original one. I added a separator between them to improve readability. –  user53153 Jan 2 '13 at 15:56
This is helpful, as it stumped most of those who took the test, but I could really use something that doesn't require maple. –  Frank White Jan 2 '13 at 17:56
@Frank I don't have a counterexample that lends itself well to hand computations. We are talking about the 8th derivative, after all. I recommend you to talk to the professor who wrote the exam, and simply ask how s/he would solve this problem. –  user53153 Jan 2 '13 at 18:11
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