Uniform convergence of a sequence of functions 4

Question

Prove that the sequence $\left((nx)/(1+4n^2x^2)\right)_{n\in\mathbb N}$ is not uniformly convergent on $(-a,a)$, where $a > 0$

My attempt:

$\lim_{n\to\infty}(nx)/(1+4n^2x^2) = 0 = f(x)$

Now, compute$$M_n = \sup_{-a<x<a}|f_n(x) - f(x)| = \sup_{-a<x<a}\left|(nx)/(1+4n^2x^2) - 0\right| = \sup_{-a<x<a}(nx)/(1+4n^2x^2)$$

Finally, compute $\lim_{n\to\infty} M_n$

The theorem says that if $\lim_{n\to\infty}$ $M_n$ = $0$, then $\bigl(f_n(x)\bigr)_{n\in\mathbb N}$ is uniformly convergent. And the question is to prove that it is not. Any help please? What is the supremum of $f_n(x)$, i.e what is $M_n$?

Answer 1

You can compute $\sup f_n$ by the standard method of computing $f_n'$ and determining where it is equal to $0$. It turns out that $f_n'(x)=0\iff x=\pm\frac1{2n}$ and that $f_n\left(\pm\frac1{2n}\right)=\pm\frac14$. So, $\sup\lvert f_n\rvert=\frac14$, and therefore your sequence does not converge uniformly to the null function.

Answer 2

We have $f_n(\frac{1}{n})=\frac{1}{5}$ for all $n$. Hence $M_n \ge \frac{1}{5}$ for all $n$.