Uniform convergence of $f_n = \frac{nx}{1+nx^2}$.

Question

let $f_n(x)$ = $\frac{nx}{1+nx^2}$ for $x\ge 0$,

(a) show that $\{f_n\}$ converges uniformly on $[M,\infty]$ for any $M>0$

(b) show that $\{f_n\}$ does not converge uniformly on $[0,\infty)$.

I tried solving this problem by using Weierstrass $M_n$ test for sequence of functions:

$$f_n(x) \to f(x)=\frac{1}{x}.$$

Now, I am stuck in finding

$$M_n=\sup\{\left\lvert f_n(x)-f(x)\right\rvert\},$$ where the supremum is taken over $[M,\infty)$ for (a) or $[0,\infty)$ for (b).

Also, is it true in general that if $M_n$ does not tend to $0$, then $\{f_n\}$ does not uniformly converge to $f$?

Answer 1

Hint 1: For $|x|\ge M$, $$ \begin{align} \left|\frac{nx}{1+nx^2}-\frac1x\right| &=\frac1{x(1+nx^2)}\\ &\le\frac1{nM^3} \end{align} $$ Hint 2: The Uniform Limit Theorem.

Answer 2

Instead of using the uniform limit theorem, we show directly that $f_n$ does not converge uniformly on $(0,M)$ for all $M>0$. Clearly the pointwise limit is

$$ g(x) = \begin{cases} \frac{1}{x} & x>0, \\ 0 & x =0 \end{cases}.$$

Then when $x>0$,

$$ |f_n(x) - g(x) | =\frac{1}{x(1+nx^2)}.$$

And this implies

$$ \left|f_n\left(\frac{1}{\sqrt{n}}\right) - g\left(\frac{1}{\sqrt{n}}\right)\right| = \frac{\sqrt n}{2} \ge \frac{1}{2}$$ for all $n$. Then for any $M>0$,

$$\sup_{x\in (0,M)}|f_n(x) - g(x)| \ge \frac 12 $$ for all $n\ge 1/M^2$. Thus $\{f_n\}$ does not converge uniformly to $g$ on $(0,M)$ for any $M>0$.