# Let $h_n:[0,\infty]\to\mathbb{R}$ be given by $h_n(x)=\frac{nx}{x+n}$, and let $h:[0,\infty)\to \mathbb{R}$ be given by $h(x)=x$.

Problem: Let $$h_n:[0,\infty]\to\mathbb{R}$$ be given by $$h_n(x)=\frac{nx}{x+n}$$, and let $$h:[0,\infty)\to \mathbb{R}$$ be given by $$h(x)=x$$. (a) Prove that for each $$c>0$$, $$(h_n)$$ convergences uniformly to $$h$$ on the interval $$[0,c]$$.
(b) Prove that $$(h_n)$$ does not converge uniformly to $$h$$ on $$[0,\infty)$$

My solution/proof: (a) Let $$m_n=\sup|h_n(x)-h(x)|$$.
Consider \begin{align}|h_n(x)-h(x)|&=\left|\frac{hx}{x+n}-x\right|\\ &=\left|\frac{nx-x(x+n)}{x+n}\right|\\ &=\left|\frac{nx-x^2-nx}{x+n}\right|\\ &=\left|\frac{-x^2}{x+n}\right|\\ &=\frac{|x|^2}{x+n}.\end{align} Now, $$x\in[0,c]\to\frac{|x|^2}{|x+n|}\leq\frac{c^2}{n}$$.
Therefore, $$m_n=\frac{c^2}{n}$$. As $$n\to \infty$$, $$m_n\to0$$. Now $$m_n=\sup{|h_n(x)-h(x)|:x\in[0,c]}$$and $$m_n\to 0$$ as $$n\to \infty$$. From supremum test, $$h_n(x)$$ converges to $$h(x)$$ uniformly on $$[0,c]$$.

(b) $$h_n(x)$$ does not converge uniformly to $$h$$ on $$[0,\infty)$$. Let $$x_n=n\in[0,\infty)$$. Now, \begin{align}|f_n(x_n)-h(x_n)|&=\left|\frac{nx_n}{x_n+n}-x_n\right|\\ &=\left|\frac{n*n}{n+n}-n\right|\\ &=\left|\frac{n^2}{2n}-n\right|\\ &=\left|\frac{n}{2}-n\right|\\ &=\left|\frac{n}{2}\right|\\&\geq\frac{1}{2}\to|h_n(x_n)-h(x)|\geq \frac{1}{2}\end{align} from Cauchy's Criterion, $$h_n$$ does not converge uniformly to $$h(x)$$ on $$(0,\infty)$$.

Please tell me if my proof is correct. Give an alternate proof if it's not correct. Thank you!

• Presumably you meant $[0,\infty)$ rather than $[0,\infty].$ Commented Apr 28, 2021 at 23:36
• Technically, you’ve only shown that $m_n\leq \frac{c^2}n,$ not equality. It’s good enough, though. Commented Apr 28, 2021 at 23:44
• In your last inequality, it has to be $h(x_n)$ instead of $h(x)$. Commented Apr 28, 2021 at 23:52

Other proof. For $$x\ge 0$$ and $$n\ge 1$$, put $$g_n(x)=|h_n(x)-h(x)|=\frac{x^2}{x+n}.$$
$$\lim_{x\to+\infty}g_n(x)=+\infty$$ thus $$\sup_{x\ge 0}g_n(x)=+\infty$$ or $$g_n$$ is unbounded . the convergence is not uniform at $$\Bbb R^+$$.