# Complex Variables Problem

1. Mar 3, 2008

### nicksauce

1. The problem statement, all variables and given/known data
Let f be analytic in the disk |z| <= 1. Prove that for any 0 < r < 1,

$$|f(0)|^2 <= \frac{1}{\pi r^2} \int \int_{x^2 + y^2 <= r^2} |f(z)|^2 dxdy$$

2. Relevant equations
The hint is apply the Gauss mean value theorem on $$f^2(z)$$

3. The attempt at a solution
Having difficulty starting this one. Any hints?

All I've got is

$$f^2(0) = \frac{1}{2\pi} \int(f^2(z))d\theta$$

By applying the Gauss mean value theorem. Then I'm stuck.

Last edited: Mar 3, 2008
2. Mar 3, 2008

### Dick

You mean to have an f^2(z) inside of your first integral as well, correct? Otherwise, it's obviously false.

3. Mar 3, 2008

### nicksauce

Sorry, fixed it now.

4. Mar 3, 2008

### Dick

Sorry to be dense. I'm running a little slow this time of night. But the quantity on the right hand side of the inequality is complex. The number on the left is real. It doesn't make much sense to say a real is less than a complex. Actually, my confusion should be giving you some hints.

5. Mar 3, 2008

### nicksauce

lol I'm the one being dense. Fixed it for good this time.

6. Mar 3, 2008

### Dick

My point is that there HAS to be an absolute value on both sides. Another hint is that if you leave the absolute values out, both sides are actually equal. Put the absolute values back in and show the inequality.

7. Mar 3, 2008

### Dick

The absolute value of the integral of a complex function over a domain is less than or equal to the integral of the absolute value of the complex function over the domain. Wink, wink. Nudge, nudge.

8. Mar 4, 2008

### nicksauce

Ok thanks I'll have another go at it.