For \(n\geq1\), show that \(f(x)=x^n\) is uniformly continuous on any interval \([-M,M]\).
Solution
If \(M=0\), the claim is immediate. Assume \(M>0\). For \(x,y\in[-M,M]\), \[ \abs{x^n-y^n} &=\abs{x-y}\abs{\sum_{k=0}^{n-1}x^{n-1-k}y^k} \\ &\leq nM^{n-1}\abs{x-y}. \] Thus \(f\) is Lipschitz on \([-M,M]\), hence uniformly continuous. Explicitly, given \(\varepsilon>0\), take \(\delta=\varepsilon/(nM^{n-1})\).