MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Assume that $f$ is harmonic in the unit disk $|z|<1$, with boundary function of bounded variation, such that $$\lim_{r\to 1}f(re^{it})= 0$$ for $t\in[0,\pi]\setminus \mathbf{Q}$, where $\mathbf{Q}$ are rational numbers. Can we then state the following $$\lim_{r\to 1}f(re^{it})= 0,\ \ \ t\in[0,\pi]. $$

share|cite|improve this question
    
What do you exactly mean by "boundary function"? – Alexandre Eremenko Mar 9 '13 at 3:18
    
$f=P[g]$, where $g(t):[0,2\pi]\to \gamma\subset \mathbf{C}$ is of bounded variation. – djoke Mar 9 '13 at 10:17
    
You can use the fact that Fourier series of a function of bounded variation converges everywhere to arithmetic mean of left and right limits. – Mateusz Wasilewski Mar 9 '13 at 12:41

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.