Let \(f_n: \mathbb{R} \to \mathbb{R}\) be the function
\[ f_n(x) = \frac{1}{n^3[x - (1/n)]^2 + 1}. \]
Let \(f: \mathbb{R} \to \mathbb{R}\) be the zero function.
Show that \(f_n(x) \to f(x)\) for each \(x \in \mathbb{R}\).
Show that \(f_n\) does not converge uniformly to \(f\). (This shows that the converse of Theorem 21.6 does not hold; the limit function \(f\) may be continuous even though the convergence is not uniform.)
Solution
Fix \(x\in\mathbb R\). If \(x=0\), then \[ f_n(0)=\frac1{n+1}\longrightarrow0. \] If \(x\ne0\), then \(x-1/n\to x\), so for all sufficiently large \(n\), \[ |x-1/n|\ge |x|/2. \] Hence \[ 0\le f_n(x)\le\frac1{n^3x^2/4+1}\longrightarrow0. \] Thus \(f_n(x)\to0\) for every fixed \(x\).
At the moving point \(x=1/n\), \[ f_n(1/n)=1. \] Therefore \[ \sup_{x\in\mathbb R}|f_n(x)-0|=1 \] for every \(n\). In particular the supremum error does not tend to zero, so the convergence is not uniform.