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.

The following is exercise 11.8 in Rudin's Real and Complex Analysis:

Suppose $\Omega$ is a region, $f_n \in H(\Omega)$ for $n = 1, 2, 3, \ldots$. $u_n$ is the real part of $f_n$. $\{u_n\}$ converges uniformly on compact subsets of $\Omega$, and $\{f_n(z)\}$ converges for at least one $z \in \Omega$. Prove that then $\{f_n\}$ converges uniformly on compact subsets of $\Omega$.

My thoughts: By Harnack's theorem, the limit $u$ of $\{u_n\}$ is harmonic. Thus it's the real part of a holomorphic function $f$ on $\Omega$ defined up to an imaginary constant. Using the limit of $\{f_n(z)\}$, we can find this constant. Thus $f$ is well-defined.

What's left is to show that the imaginary part of $f_n$ converges to that of $f$. I suspect I need to use the Cauchy-Riemann equations for this, but I cannot apply the familiar uniform convergence theorems with partial derivatives. What should I do?

Note: A region is a connected open subset of the complex plain.

share|improve this question
    
Are you familiar with the interior gradient estimate for harmonic functions? Using it, the proof goes like this: $u_n-u_m$ is small on compact subsets, therefore $\nabla (u_n-u_m)$ is small too, which by the Cauchy-Riemann equations translates into smallness of $f_n'-f_m'$. In other words, $f_n'$ converge uniformly on compact subsets, and from there it's easy. –  user53153 Feb 7 '13 at 3:21
    
@5PM Thanks. I wasn't familiar with this estimate but now I am. I wouldn't be surprised if Rudin expected the reader to come up with it on his/her own. It makes sense now. I can probably mimic the proof of this theorem to finish your proof. Is there an easier way? Would you like to post this as an answer for me to accept it? –  PeterM Feb 7 '13 at 13:02
    
I'm also interested in seeing a proof of this result, but possibly avoiding the technique for harmonic functions suggested by 5pm in the earlier comment. –  user44532 Feb 20 '13 at 19:04

2 Answers 2

up vote 1 down vote accepted

The following inequality helps: if $D$ is a disk centered at $z_0$, $f$ is holomorphic in $D$ and continuous on $\overline D$, then $$\sup_{D'} |f|\le \operatorname{Im}f(z_0) + 3\max_{\partial D}|\operatorname{Re}f| \tag1$$ where $D'$ is the disk concentric with $D$ and half the radius.

Proof of (1) is immediate from the Schwarz integral formula: $$f(z)= i \operatorname{Im}f(z_0) + \frac{1}{2\pi i} \int_{\partial D}\frac{\zeta+z}{\zeta-z} \,\operatorname{Re}f(\zeta)\,\frac{d\zeta}{\zeta} \tag2$$ where $$\left|\frac{\zeta+z}{\zeta-z}\right|\le 3,\qquad \zeta\in\partial D,\quad z\in D' \tag2$$

Returning to the problem at hand, an application of (1) to the differences $f_n-f_m$ yields uniform convergence of $f_n$ on any disk $D$ centered at $z_0$ and compactly contained in $\Omega$. Now (1) can be applied to disks whose centers lie in the set in which uniform convergence is already known. Repeating in this fashion, we can cover any compact subset of $\Omega$ in finitely many steps.

share|improve this answer

I'm not entirely sure about this answer but I'll give it a try and people can point out any mistakes.

Let $z_{0}=x_{0}+iy_{0}$ be the point in $\Omega$ where the sequence $(f_{n})$ converges. We write $f_{n}(z)=2u_{n}(\frac{z+\bar{z_{0}}}{2},\frac{z-\bar{z_{0}}}{2i})-u_{n}(x_{0},y_{0})+ic_{n}$ where $c_{n}\in\mathbb{R}$. Letting $z=z_{0}$, we see that $c_{n}=\mathrm{Im}f_{n}(z_{0})$. Then uniform convergence of $(u_{n})$ on compact subsets and convergence of $(f_{n}(z_{0}))$ imply uniform convergence of $(f_{n})$ on compact subsets.

share|improve this answer
    
try to explain why you can write $f_n$ that way. –  peter Jun 11 at 12:31

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.