Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $f(z)$ be entire function. Consider the functions $e^{if(z)}$ and $e^{−if(z)}$ and applying the Maximum Modulus Theorem, show that if $f(z)$ is real when $|z| = 1$, then $f(z)$ must be a constant function.

(We take $f(z)=u(z)+iv(z)$)

I am confused as so far I have $|g(z)|=|e^{if(z)}|=|e^{-v(z)}|$ and then since $f(z)$ is real, $f(z)=u(z)$ and $v(z)=0$ so I assumed it would follow that $|g(z)|=|e^{v(z)}|=1$.

Similarly, $|g(z)|=|e^{-if(z)}|=|e^{v(z)}|=1$.

Using Liouville I assumed one could say that both $g(z)$ and $h(z)$ are bounded entire functions, they are constant and so it follows that $v(z)$ is constant, meaning that both its partial derivatives are equal to 0 and, due to Cauchy Riemann, both of the partial derivatives of $u(z)$ are equal to zero. It would then follow that $f(z)$ is constant.

I don't know how to go about the question using the Maximum Modulus Theorem, also I feel I am overlooking the importance of $|z|=1$ perhaps?

Any help would be much appreciated!!

share|improve this question

2 Answers 2

up vote 10 down vote accepted

The function $g:z\mapsto e^{if(z)}$ is entire. Since $f$ is real on the unit circle $\mathbb{S}^1$, it turns out that $|g|=1$ on this set. But since $g$ is entire, using the Maximum Modulus Theorem, we know that $|g(z)| \leq 1$ for all $|z| \leq 1$. This means that (using your notations) $v \leq 0$ for $|z| \leq 1$. Same reasoning with $h:z\mapsto e^{-if(z)}$ leads to $v \geq 0$ on the unit disk and hence $v(z)=0$ on the unit disk, that is $f$ takes only real values on the whole unit disk which happens only if $f$ is constant (open mapping theorem).

Ayman

share|improve this answer
    
Great, thank you so much :) –  katherinebarry Nov 2 '12 at 10:20

The maximum modulus principle says that $|g(z)|=|e^{if(z)}|$ attains its maximum for $D=\{ |z| \leq 1 \}$ on the boundary.

Thus

$$|e^{if(z)}| \leq 1 \,;\, \forall |z| \leq 1 \,.$$

Applying it to $h(z)=|e^{-if(z)}|$ you get again

$$|e^{-if(z)}| \leq 1 \,;\, \forall |z| \leq 1 \,.$$

Now,

$$e^{-if(z)}=\frac{1}{e^{if(z)}} \,.$$

Plug this in the second identity and you are done.

share|improve this answer
    
Is necessary the hypothesis of being entire? Or we can only require holomorphic on a neighborhood of the unit disk? –  Daniel Nov 3 '12 at 23:27
    
@Daniel That is enough. Actually MMP says that if $g,h $ are holomorphic inside $|z| <1$, then they cannot have a max/min inside the domain. And if $g,h$ are continuous on the closed unit disk, then they attain their max/min by compactness. So I think that holomorphic on the open disk and continuous on the closed disk is all you need. –  N. S. Nov 3 '12 at 23:34

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.