Show that if \(u\) is harmonic on \(\RR^2\) and bounded, then \(u\) is constant.
Solution: Using Liouville
Write \(f=u+iv\) for \(v\) a harmonic conjugate of \(u\), then \(f\) is holomorphic on \(\CC\). Now \(e^f = e^{u+iv} = e^u e^{iv}\) and thus \[ \abs{e^f}= \abs{e^u} \abs{e^{iv}} = e^u \] is bounded, so \(e^f\) is a bounded entire function and thus constant by Liouville. Then \(0=(e^f)'=f'e^f\), and \(e^f\) has no zeros, so \(f'\equiv0\) and \(f\) is constant. Hence \(u=\Re f\) is constant.
Solution: Using the mean value property
