Proving uniform convergence of f_n(x) = x/(1+nx²) on [0,1]

[Lee33]

Homework Statement

Is the sequence of function $f_1, f_2,f_3,\ldots$ on $[0,1]$ uniformly convergent if $f_n(x) = \frac{x}{1+nx^2}$?

2. The attempt at a solution

I got the following but I think I did it wrong.

For $f_n(x) = \frac{x}{1+nx^2}$, I got if $f_n \to0$ then we must find $\epsilon>0$ an $N$ such that for $n>N$ implies $|f_n-0|<\epsilon.$ So $f_n(x) = \frac{x}{1+nx^2}$; $\lim_{n\to\infty}f_n(x) =0$. Then for $\epsilon>0$ we have $|f_n(x)-f(x)| = |\frac{x}{1+nx^2}|\le |\frac{1}{1+n}|<|\frac{1}{n}|<\epsilon$ thus $N = \frac{1}{\epsilon}$. But I think this is wrong since $|1+nx^2|<|1+n|$? How can I show it is uniformly convergent?

 

[jbunniii]

$f_n \rightarrow 0$ uniformly if and only if $\sup f_n \rightarrow 0$. So I would suggest that you start by finding the maximum value of $f_n$.

 

[Lee33]

jbunniii -

We haven't been taught that way yet. We still haven't defined "derivative" or proved $f_n \to 0$ uniformly iff $\sup f_n \to 0$ so I can't use it to prove my problem. What I was thinking is how can I divide into two regions?

 

[jbunniii]

Consider the denominator: $1 + nx^2$. This is not quite a perfect square, but it would be if we added the missing term (actually, subtracting is more useful here): $1 - 2\sqrt{n} x + nx^2 = (1 - \sqrt{n}x)^2$. Now this is a square, so it is nonnegative. What can you conclude?

 

[Lee33]

Thanks, jbunniii for the hint!