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

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

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

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

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?

"for $$0"
Whatever follows from here works only for $$0 but not for $$x=0$$. So it doesn't converge to $$0$$ "for all x simultaneously".
• Just to add onto this a bit: even for $x > 0$, the closer to 0 we get with $x$, the larger the value of $n$ needed to make $1/(nx) < \varepsilon$ for any fixed $\varepsilon > 0$. For example, if $\epsilon = 1/2$ then we only need $n = 3$ when $x > 2/3$, since $1/(3x) < 1/2$. If $x = 1/100$, we need $n = 201$. Commented Nov 14, 2019 at 17:28
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 $$\sup_{x\in[0,1]} f_n(x) \ge f_n(1/n) = \frac{n(1/n)}{1+n^2(1/n^2)} = 1/2.$$