# uniform convergence on (0, 1) but not convergent

Let $X = (0, 1)$.

${f_n}(x) = \frac {1}{x}$ and ${g_n}(x) = \frac {x}{1+nx^2}$. We are to prove that $\{f_n\}$ and $\{g_n\}$ converge uniformly on X but $\{f_ng_n\}$. I know how to prove ${g_n} \to g$ uniformly but how does ${f_n}(x) = \frac {1}{x}$ converge uniformly on $X=(0, 1)$?

For second part of the proof, I was under the impression that if ${f_n} \to f$ and ${g_n} \to g$ uniformly, then $f_n g_n \to fg$ uniformly.

If ${f_n}(x) = \frac {1}{x}$ then its pointwise limit is also ${f}(x) = \frac {1}{x}$ which means that $|f_n(x) - f(x)| = 0 <\epsilon$ for every x and an arbitrary $\epsilon$ greater than 0. Thus it is uniformly convergent.(You can choose N to be 1,then $|f_n(x) - f(x)| = 0 <\epsilon$ for every $n>N$.)
$f_n(x)=\frac1x$ is independent of $n$, hence it converges to $\frac1x$ uniformly.
• uniform convergence means that for any $\epsilon > 0, \exists N >0, n \geq N \implies |f_n(x) - f(x)| < \epsilon, \forall x \in X$. Now if $f_n(x)=f(x)$, i.e. $f_n(x)$ is independent of $n$, we have $|f_n(x)-f(x)|=0$. Hence we can choose $N=1$ regardless of $\epsilon$. – Siong Thye Goh Dec 6 '17 at 8:09
Let $f(x)=\dfrac{1}{x}$, so $|f_{n}(x)-f(x)|=0$ for all $x\in(0,1)$ and $n$, so $\|f_{n}-f\|_{\infty}=0\rightarrow 0$.