Uniform convergence of $f_n(x) = \frac{nx}{1+ n^2 x^2}$ - Where is the flaw?

Question

I want to find out whether $||f_n||_\infty \to 0$ or not on $[0,1]$ where $f_n(x) = \frac{nx}{1+ n^2 x^2}$

$$||f_n||_\infty= \sup_{x\in[0,1]}\frac{nx}{1+ n^2 x^2}$$

for $x=0 \rightarrow \frac{nx}{1+ n^2 x^2}=0 $

As the supremum is $\geq0$, we therefore discard the point $0$ and consider only $(0,1]$

$\rightarrow \sup_{x\in[0,1]}\frac{nx}{1+ n^2 x^2} = \sup_{x\in(0,1]}\frac{nx}{1+ n^2 x^2}$

for $0<x\leq1$

$$\frac{nx}{1+ n^2 x^2}\leq \frac{nx}{n^2 x^2} = \frac{1}{nx}$$

$$\rightarrow \sup_{x\in(0,1]}\frac{nx}{1+ n^2 x^2}\leq \frac1{nx}\to0 \text{ as } n \to \infty$$

so $||f_n||_\infty \to 0$

But this is not true as $f_n$ do not converge uniformly to $0$ on $[0,1]$. So where is the flaw?

Answer 1

The flaw is right here:

"for $0<x\leq1$"

Whatever follows from here works only for $0<x\leq1$ but not for $x=0$. So it doesn't converge to $0$ "for all x simultaneously".

That's the requirement for uniform convergence.

Answer 2

For uniform convergence to $0$, you would need $$0 = \lim_{n\to \infty}\|f_n - 0\|_\infty = \lim_{n\to \infty} \sup_{x \in [0,1]} f_n(x).$$ But we see that $$\sup_{x\in[0,1]} f_n(x) \ge f_n(1/n) = \frac{n(1/n)}{1+n^2(1/n^2)} = 1/2.$$