For each \(n\ge1\), let \(f_n:\RR\to\RR\) be differentiable with \[ |f_n'(x)|\le1 \] for all \(n,x\). Suppose \(f_n(x)\to g(x)\) for every \(x\in\RR\). Prove that \(g\) is continuous.
Solution
1. Every \(f_n\) is \(1\)-Lipschitz:\[ |f_n(x)-f_n(y)|\le|x-y| \]for all \(x,y\in\RR\).
Fix \(n\) and \(x\ne y\). By the mean value theorem, there is a point \(c\) between \(x\) and \(y\) such that \[ f_n(x)-f_n(y)=f_n'(c)(x-y). \] Hence \[ |f_n(x)-f_n(y)| \le |x-y| \] because \(|f_n'(c)|\le1\). The case \(x=y\) is immediate.
2. The pointwise limit \(g\) is also \(1\)-Lipschitz:\[ |g(x)-g(y)|\le|x-y| \]for all \(x,y\in\RR\).
Fix \(x,y\in\RR\). By pointwise convergence, \[ f_n(x)-f_n(y)\longrightarrow g(x)-g(y). \] Taking limits in the inequality from step 1 and using continuity of the absolute-value function gives \[ |g(x)-g(y)|\le|x-y|. \]
3. The function \(g\) is continuous on \(\RR\).
Fix \(x\in\RR\) and \(\varepsilon>0\). If \[ |y-x|<\varepsilon, \] then step 2 gives \[ |g(y)-g(x)|\le|y-x|<\varepsilon. \] Thus \(g\) is continuous at every \(x\).
4. Q.E.D.
Step 3 is the required conclusion.