How to preform a uniform convergence test on a sequence of functions with intervals that depend on $n$?

Question

Let $$ f_{n}(x) = \begin{cases} \dfrac{1}{x} & \dfrac{1}{n}\leq x \leq 2 \\[4pt] 0 & 0 < x < \dfrac{1}{n} \end{cases} $$

$f_{n}(x)$ converges point-wise to $f(x)=\frac{1}{x}$, $x\in(0,2]$. How to test whether this convergence is uniform?

Answer 1

For uniform convergence, given $\epsilon > 0$, we need to find $N$ such that $|f(x)-f_n(x)|<\epsilon$ for all $x$ and $n\geq N$. Every $f_n(x)$ differs from $f(x)=1/x$ on the interval $(0,1/n)$ where $f_n(x)=0$ and so the difference is $1/x$. This difference is not bounded at all for any $n$ (becoming infinite as $x\to 0$), so it can't be bounded by some $\epsilon$.

Answer 2

Uniform convergence of $(f_n)_n$ to $f$ on a domain $D$ is equivalent, by definition, to $\lim_{n\to \infty} \sup_{x\in D}|f_n(x)-f(x)|=0.$ In the Q, with $D=(0,2]$ we have $\sup_{x\in D}|f_n(x)-f(x)|=\sup_{0<x<1/n}1/x=\infty$ for every $n$.