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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!!

leo
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katherinebarry
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2 Answers2

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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: $f(z)=u(z)+iv(z)$) $v \geq 0$ for $|z| \leq 1$. Same reasoning with $h:z\mapsto e^{-if(z)}$ leads to $v \leq 0$ 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

Bach
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xounamoun
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  • Great, thank you so much :) – katherinebarry Nov 02 '12 at 10:20
  • Maybe I am overthinking, but I do not understand that why $v \geq 0$ for $\lvert z \rvert \leq 1$. Because in this case x might be between -1 and zero and so the norm of v is greater than one, the same for v less than zero happen. Also, I do not see why open mapping theorem shows that f should be constant because it says f maps opens to opens. Then how do you deduce that f is constant? @xounamoun – Tim Jul 08 '23 at 06:56
  • Inside the disk you have $|e^{if}| = e^{-v} \leq 1$ which entails $v \geq 0$ in this area. The open mapping theorem tells you that non constant holomorphic functions send opens to opens. Now if $f$ takes real values on a non-empty open portion this does mean that it is constant on this open (and thus everywhere by unique continuation argument). – xounamoun Jul 09 '23 at 14:27
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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.

N. S.
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    Is necessary the hypothesis of being entire? Or we can only require holomorphic on a neighborhood of the unit disk? – Daniel Nov 03 '12 at 23:27
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    @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 03 '12 at 23:34