Let $f_n(x)=\frac{1}{1+n^2x^2}$ for $n \in \mathbb{N} , x \in \mathbb{R}$
Is $f_n$ uniformly convergent on $ \ [\frac{1}{2},1\ ] $?
1) If $x=\frac{1}{n}$, $f_n(x)=\frac{1}{2}$ but $x=\frac{1}{n}$ does not belong to the set $ \ [\frac{1}{2},1\ ] $
2) I know that $f_n$ converges pointwise to $f(x)=0$ on $ \ [\frac{1}{2},1\ ] $
3) $f_n$ is defined on a closed and bounded interval so supremum of $\vert f_n(x)-f(x)\vert $ is attained for each x.
To find sup, let $f'_n(x)=0$. This gives $\frac{-2n^2x}{(1+n^2x^2)^2}=0$ which gives $x=0.$ How to conclude for supremum?
How do we prove it converges uniformly on $ \ [\frac{1}{2},1\ ] $?
Hint:
$$|f_n(x) -0| \leqslant \frac{1}{1 + n^2(1/2)^2}$$