A sequence \(\{f_n:\mathbb R\to\mathbb R\}_{n=1}^{\infty}\) converges uniformly to \(f:\mathbb R\to\mathbb R\) if, for every \(\varepsilon>0\), there is a positive integer \(N\) such that \[ n>N\quad\Longrightarrow\quad |f_n(x)-f(x)|<\varepsilon \] for all \(x\in\mathbb R\).
What must one check to show that a sequence does not converge uniformly?
Give an example such that \(f_n(x)\) converges to \(f(x)\) for every \(x\in\mathbb R\), but the sequence does not converge uniformly.
Solution
The negation of uniform convergence is \[ \exists\varepsilon_0>0\ \forall N\in\mathbb N\ \exists n>N\ \exists x\in\mathbb R: |f_n(x)-f(x)|\geq\varepsilon_0. \]
For example, let \(f_n(x)=x/n\) and \(f(x)=0\). For each fixed \(x\), one has \(x/n\to0\). The convergence is not uniform. Take \(\varepsilon_0=1\). For every \(N\), choose \(n=N+1\) and \(x=n\). Then \[ |f_n(x)-f(x)|=1. \]