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I recently came across the following problem

Let $h : \Bbb C → \Bbb C$ be an analytic function such that $h(0) = 0, h(1/2) = 5$, and $|h(z)| < 10$ for $|z| < 1$. Then pick out the correct statement(s):

  • the set $\{z : |h(z)| = 5\}$ is unbounded by the Maximum Principle
  • the set $\{z : |h'(z)| = 5\}$ is a circle of strictly positive radius
  • $h(1) = 10$
  • regardless of what $h'$ is, $h'' \equiv 0$.

My try

Consider $f(z)=\frac{h(z)}{10}$ and $f$ satisfies the hypothesis of Schwarz lemma, so $$\vert f(z) \vert \leq \vert z \vert$$ and so $$\vert h(z) \vert \leq 10 \vert z \vert ^1$$ so by extended Lioville's theorem, $h$ is a polynomial of degree atmost $1$. So $h$ is either constant or linear. But given hypothesis implies $h$ is not constant and $h$ is exactly $10z$. That is, $$(\forall z \in \Bbb C): h(z)\equiv 10z$$

So the set in the first bullet is a circle $\vert z \vert =1/2$ ,which is bounded and so it is false

The second set is $\varnothing$ and so the second one is false

Third and fourth are true. Since $h$ is linear , its second derivative vanishes!

Is my reasoning correct ? Any thoughts?

Edit: I agree that my reasoning in wrong as Mr. Kavi Rama Murthy says in the comment. So this $h(z)=10z$ helps to eliminate the first two options. But how to conclude the last two. Any help ?

Community
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Chinnapparaj R
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  • By Schwarz Lemma you only get $|h(z)| \leq 10|z|$ for $|z| <1$ . This does not tell you that $h$ is a ploynomial. – Kavi Rama Murthy Feb 14 '19 at 12:05
  • @KaviRamaMurthy: So how to proceed further ? kindly explain ! – Chinnapparaj R Feb 14 '19 at 12:08
  • Consider the function defined on the disk $f(z)=h(z)/(10z)$. By Schwarz lemma, $|f|$ is $1$ at $1/2$ and $1$ is the supremum of $|f|$ on the unit disk. So the right argument is not Liouville but something else. – Aphelli Feb 14 '19 at 12:29

3 Answers3

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Since $h$ is entire with $h(0)=0$, the function $f(z):=h(z)/10z$ is entire. Since $|h(z)|<10$ on the open unit disk, by continuity, $|h(z)|\le 10$ and thus $|f(z)|\le 1$ on the unit circle. By maximum modulus principle, unless $f$ is a constant, the maximum of $|f(z)|$ over closed unit disk can be only achieved on the boundary, that is, the unit circle.

Suppose to a contrary that $f$ is not a constant. Then $|f|\le1$ in the closed unit disk. But we have that $|f(\frac12)|=1$, which asserts that the maximum of $|f|$ over closed unit disk is achieved by an interior point. This gives us a contradiction.

Therefore $f$ must be a constant. Since $f(\frac12)=1$, $f(z)=1$ for all $z$. So $h(z)=10z$.

Eric Yau
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Let $g(z)= h(z)/10.$ Then $g$ restricted to $\mathbb D,$ let's call it $\tilde g,$ maps $\mathbb D$ into $\mathbb D,$ sends $0$ to $0,$ and takes $1/2$ to $1/2.$ By the Schwarz Lemma, $\tilde g(z)= z.$ That is the same as saying $g(z)=z$ in $\mathbb D.$ By the identity principle, $g(z) = z$ on all of $\mathbb C.$ Therefore $h(z) = 10z$ on $\mathbb C.$ In order then, we see the given statements are false, false, true, and true.

zhw.
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A partial solution : Schwartz Lemma does not work completely( on boundary ) . To discard some options take counterexample as $f(z)=10z$.

neelkanth
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