Problem 1805 (difficulty: 7/10)

The function \(\displaystyle u\) is harmonic inside the unit disk and continuous along the set \(\displaystyle \overline{B}(0,1)\setminus\{1\}\), and \(\displaystyle u=0\) at the points of the unit circle. Does it follow that \(\displaystyle u\equiv0\)?

Solution:

The answer is NO. Consider the function

\(\displaystyle u(z) = \re\frac{1+z}{1-z}. \)

This is defined everywhere except for \(\displaystyle z=1\), and, being the real part of the holomorphic function, it is harmonic.

The fractional linear transform \(\displaystyle \frac{1+z}{1-z}\) maps the unit disk to the right half plane, so \(\displaystyle u=1\) along the unit circle (with the exception \(\displaystyle z=1\)).

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