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$f(z)$ is analytic in the unit circle, and $u=Re(f), v=Im(f)$.

Please prove that if $u(0)=v(0)$, then $\int_0^{2\pi}(u(re^{i\theta}))^2d\theta=\int_0^{2\pi}(v(re^{i\theta}))^2d\theta$ for every $0<r<1$.

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2 Answers 2

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Answer to the original question:

If $f(z) = 1$, then $\int_0^{2\pi}(u(re^{i\theta}))^2d\theta=2 \pi$, $\int_0^{2\pi}(v(re^{i\theta}))^2d\theta = 0$, so it cannot be true.

Answer to the updated question:

Let $g(z) = f(z)-f(0)$. The power series for $g$ shows that $\eta(z) = \frac{g(z)}{z}$ and $\phi(z) = \frac{g^2(z)}{z}$ have a removable singularity at $z=0$, and hence may be taken to be analytic on $B(0,1)$.

Letting $\gamma(\theta) = r e^{i\theta}$, with $r\in (0,1)$ we have $\int_\gamma \eta = 0$ and $\int_\gamma \phi = 0$.

Since $\int_\gamma \frac{h(z)}{z} d z = 0$ is equivalent to $\int_0^{2 \pi} h(re^{i \theta}) d \theta = 0$, we have $\int_0^{2 \pi} g(re^{i \theta}) d \theta = 0$ and $\int_0^{2 \pi} g^2(re^{i \theta}) d \theta = 0$.

Expanding these gives $\int_0^{2 \pi} f(re^{i \theta}) d \theta = 2 \pi f(0)$, and $\int_0^{2 \pi} f^2(re^{i \theta}) d \theta = \int_0^{2 \pi} (g^2(re^{i \theta}) +f(0)g(re^{i \theta}) + f^2(0)) d \theta =2\pi f^2(0)$. Since $u(0) = v(0)$, we see that $2\pi f^2(0) = -4 \pi i u^2(0)$, which is purely imaginary, and hence $\text{Re} (\int_0^{2 \pi} f^2(re^{i \theta}) d \theta) = 0$.

Since $\text{Re}(f^2(r e^{i\theta})) = u^2(r e^{i\theta})-v^2(r e^{i\theta})$, we conclude that $\int_0^{2 \pi} u^2(r e^{i\theta}) d \theta = \int_0^{2 \pi} v^2(r e^{i\theta}) d \theta$.

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Sorry, I corrected the question. –  Ran Kashtan Dec 30 '12 at 18:27
    
That condition was added later, @julien –  DonAntonio Dec 30 '12 at 18:29
    
I think questions & answers should have version numbers, and comments include the relevant version :-). –  copper.hat Dec 30 '12 at 18:32
    
Yes, this is one of the serious flaws in this site... –  DonAntonio Dec 30 '12 at 18:42
    
So... Are there any answers :)? –  Ran Kashtan Dec 30 '12 at 19:12

Since $f^2$ is analytic, its real part $Re\:f^2=u^2-v^2$ is harmonic.

In particular, $Re\: f$ has the Mean Value Property at $0$. This yields, for every $0<r<1$: $Re\:f(0)=\frac{1}{2\pi} \int_0^{2\pi} Re\:f(re^{i\theta})d\theta$.

Now since $Re\:f(0)=u^2(0)-v^2(0)=0$, this implies $\int_0^{2\pi} (u^2(re^{i\theta})-v^2(re^{i\theta}))d\theta=0$. And the result follows.

PS: For a statement and a proof of the Mean Value Property for harmonic functions, see Theorem 4 here, for instance: http://www.maths.qmul.ac.uk/~boris/potential_th_notes.pdf

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Can you please expand more? –  Ran Kashtan Dec 30 '12 at 20:08
    
@Eliran Done. Do you need more details? –  1015 Dec 30 '12 at 20:18
    
Yes please, we didn't learn about harmonic function in this stage :) –  Ran Kashtan Dec 31 '12 at 22:30

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