Problem: Let $f_n(x)=\displaystyle\frac{1}{1+nx}$ for $x\in(0,\infty)$ and $n\in\mathbb N$.
$\textbf{a)}$ Show that for each $\delta>0$, $\{f_n\}_{n=1}^\infty$ is uniformly convergent on $[\delta,\infty)$, and determine to which function.
This is the easy part. Fix $x\in[\delta,\infty)$. Then $\delta\leq x$ implies that
$$\vert f_n(x)\vert=\left\vert\frac{1}{1+nx}\right\vert\leq\frac{1}{1+n\delta}\to0\quad\text{as }n\to\infty.$$
Since $x\in[\delta,\infty)$ was arbitrary, it follows that $\|f_n-f\|_\infty\to0$ as $n\to\infty$ where $f(x)=0$ for all $x\in[\delta,\infty).$
$\textbf{b)}$ Show that $\{f_n\}_{n=1}^\infty$ is not uniformly convergent on $(0,\infty)$.
Fix $x\in(0,\infty)$ and let $\varepsilon>0$ be given. Choose $N\in\mathbb N$ such that $N>\displaystyle\frac{1}{\varepsilon x}$. Then for all $n\geq N$ we have
$$\vert f_n(x)\vert=\frac{1}{1+nx}\leq\frac{1}{nx}<\varepsilon.$$
Therefore, $f_n\to0$ as $n\to\infty$ pointwise on $(0,\infty).$ So we assume that the convergence is uniform and try to derive a contradiction. Pick $\varepsilon=1/2.$ Then there is some $N(\varepsilon)\in\mathbb N$ such that $\vert f_n(x)\vert<\varepsilon$ for all $n\geq N(\varepsilon)$ and all $x\in(0,\infty)$. Therefore, we may choose $n$ and $x$. Pick $n=N(\varepsilon)$ and $x=1/3N(\varepsilon)$. Then $\vert f_n(x)\vert=\displaystyle\frac{1}{1+1/3}>1/2=\varepsilon$ and we have a contradiction. It follows that $\{f_n\}_{n=1}^\infty$ does not converge uniformly on $(0,\infty).$
Do you agree with my proof above? I am specifically concerned about the last argument in part (b).
Thank you in advance for any help, your time is much appreciated.
Best Answer
A most straightforward argument for part $b)$ is to notice that $$f_n \left( \frac{1}{n}\right) = \frac{1}{2}$$
does not tend to $0$, hence $(f_n)$ cannot converge uniformly.