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Assume that f is entire, $f(0)=1$, and $\int_{0}^{2\pi} \lvert f(e^{i\theta}) \rvert d\theta = 2 \pi$. Prove that f is constant.

Now, I tried to bound the integral by $2\pi$ times the max of $\lvert f(e^{i\theta}) \rvert$ where $0 \leq \theta \leq 2\pi$ and then using the max-modulus principle tired to get a contradiction but we get an obvious inequality. Next, I tried to make use of the power series of f to evaluate the integral and use the properties of integral, but then we again get obvious inequalities. Any hint would be helpful.

Tim
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    Hint: $|\frac 1 {2\pi}\int_0^{2\pi} f(e^{i\theta})d\theta|=\frac 1 {2\pi}\int_0^{2\pi} |f(e^{i\theta}|d\theta$. What can you conclude from this ? – Kavi Rama Murthy Jul 07 '23 at 05:36
  • Is this true because we can make use of power series of f and except the first integral the rest are zero? Are you going to use the mean value theorem? @geetha290krm – Tim Jul 07 '23 at 05:50
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    I am using the Mean Value Theorem. – Kavi Rama Murthy Jul 07 '23 at 05:54
  • I see. Your first line is by the MVT. Then it would imply that Im(f) is zero a.e on the boundary of D. So, by the max-modulus it is less than or equal to zero on D a.e. If we have Im(f) is zero then C.R.E. will finish the proof. How can we get that? @geetha290krm – Tim Jul 07 '23 at 06:29
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    Partial dup https://math.stackexchange.com/q/226786/27978 – copper.hat Jul 07 '23 at 07:18

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Since $1=f(0) = {1 \over 2 \pi i}\int_C {f(z) \over z} dz = {1 \over 2 \pi }\int_0^{2 \pi}f(e^{i \theta}) d \theta $ we see that the integral is real and ${1 \over 2 \pi }\int_0^{2 \pi}f(e^{i \theta}) d \theta = {1 \over 2 \pi }\int_0^{2 \pi}|f(e^{i \theta})| d \theta $. In particular, $\operatorname{re} f(e^{i \theta}) = | f(e^{i \theta}) | $ everywhere and so $f(e^{i \theta}) = | f(e^{i \theta}) | $ is real everywhere.

Consider $h(z) = e^{ i f(z)}$ which is entire, and $|h(e^{i \theta})| = 1$, hence $|h(z)| \le 1$ on the unit disk. Repeating with $g(z) = e^{ -i f(z)}$ shows that $| e^{ i f(z)}| = 1$ on the unit disk and hence $f$ is constant.

Addendum:

Using the curve $C(t) = e^{it}$ we have

$f(0) = {1 \over 2 \pi i}\int_C {f(z) \over z} dz = {1 \over 2 \pi i}\int_0^{2 \pi} {f(e^{i \theta}) \over e^{i \theta}} i e^{i \theta} d \theta = {1 \over 2 \pi}\int_0^{2 \pi} f(e^{i \theta})d \theta $.

copper.hat
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  • I understand the second part of the proof, but in the first part I do not understand how you conclude that f is real. For instance, in the first line the second integral should be $\vert f(e^{i\theta}) \rvert$ but there is no absolute value there. I am so confused with the first part because $\frac{f(z)}{z}$ has a z in the denominator but in your second expression there is no z in the denominator. I know that you are using Cauchy's integral formula but then how do you get rid of the z in the denominator? @copper.hat – Tim Jul 08 '23 at 13:57
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    @John Just explicitly evaluate the integral on the curve $\theta \mapsto e^{ i \theta } $. – copper.hat Jul 08 '23 at 15:11
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    @John I added an addendum to elaborate. – copper.hat Jul 08 '23 at 16:48
  • Now I see, thought you were using Cauchy's and the assumption in the first line, my fault! Thanks! Sorry, the last question, using the identity theorem you are saying it is real everywhere? I know the fact that it is real on the unit circle will suffice, but just to make sure. @copper.hat – Tim Jul 08 '23 at 18:58
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    @John If $\operatorname{re} z = |z|$ then $z$ must be real. – copper.hat Jul 08 '23 at 19:30