Tie, Fall 2009
Show that \(f(z) = z^2\) is uniformly continuous in any open disk \(|z| < R\), where \(R>0\) is fixed, but it is not uniformly continuous on \(\mathbb C\).
Solution
A direct computation: fix \(\eps>0\) and suppose \(\abs{z-w} < R\). Then \[ \abs{z^2-w^2} &= \abs{z-w}\abs{z+w} \\ &\leq \delta \qty{\abs z + \abs w} \\ &\leq \delta \cdot 2R ,\] so choose \(\delta < { \eps \over 2R}\) to get uniform continuity on \(\DD_{R/2}(0)\).
To see \(f\) can’t be uniformly continuous on \(\CC\), take \(\eps \da c\) any constant and suppose the appropriate \(\delta\) exists. We’ll look for a bad pair of \(z, w\), so take \(w = z + {1\over 2}\delta\). This would imply \[ \abs{z^2 - w^2} &= \abs{z^2 - (z+\delta)^2} \\ &= \abs{-2z\delta - \delta^2} \\ &= \abs{2z\delta + \delta^2} \\ &= \delta \abs{2z + \delta} \\ &\convergesto{\abs{z}\to\infty}\infty ,\] using the \(\delta = \delta(\eps)\) can’t depend on \(z\) or \(w\), and is thus constant in this expression. This contradicts that \(\abs{z^2-w^2} < \eps = c < \infty\).