Let \(f_n(x) = \frac{nx}{n+x}\) for \(x \in [0, \infty)\) and \(n \in \mathbb{N}\).
Find a function \(f\) such that \(\{f_n\}\) converges to \(f\) pointwise on \([0, \infty)\).
Is the convergence uniform on \([0, \infty)\)? Justify your answer.
Solution
For fixed \(x\ge0\), \[ f_n(x)=\frac{nx}{n+x}=\frac{x}{1+x/n}\longrightarrow x. \] Thus the pointwise limit is \[ f(x)=x. \]
The convergence is not uniform on \([0,\infty)\). Indeed, \[ |f_n(x)-f(x)| =\left|\frac{nx}{n+x}-x\right| =\frac{x^2}{n+x}. \] Taking \(x=n\) gives \[ |f_n(n)-f(n)|=\frac{n}{2}, \] which does not tend to \(0\). Therefore \[ \sup_{x\ge0}|f_n(x)-f(x)|\not\to0, \] so convergence is not uniform.