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 $$\lim_{r\to 1}f(re^{it})= 0,\ \ \ t\in[0,\pi]. $$
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$\begingroup$ What do you exactly mean by "boundary function"? $\endgroup$– Alexandre EremenkoMar 9, 2013 at 3:18
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$\begingroup$ $f=P[g]$, where $g(t):[0,2\pi]\to \gamma\subset \mathbf{C}$ is of bounded variation. $\endgroup$– djokeMar 9, 2013 at 10:17
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$\begingroup$ You can use the fact that Fourier series of a function of bounded variation converges everywhere to arithmetic mean of left and right limits. $\endgroup$– Mateusz WasilewskiMar 9, 2013 at 12:41
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