We have to show that $f_n(x)=\frac{nx}{1+n^{2}x^{2}}$ is uniform convergent on $[a,\infty),a \gt 0$ but not on $[0,\infty)$
I am trying to prove uniform covegence on $[a, \infty]$ by using the result a sequence $(f_n)$ of bounded functions on $A \subset \Bbb R$ converges uniformly on $A$ iff $ \Vert f_n-f \Vert _A \to 0$.
Here I got $f(x)=0$ and for proving sequence $(f_n)$ is sequence of bounded functions i tried to prove that $f_n(x)$ is a decreasing function and have maxima at $x=a$. For this I differentiated $f_n(x)$ and got $f_n^{'}(x)=n(1-n^{2}x^{2})/(1+n^{2}x^{2})^{2}$ but don't know how to move further.
Can anyone tell is this the right direction if yes then how to proceed further?
Let $n \in \mathbb{N}$ be large enough so that $\frac1n < a$. Then for any $x \in [a, +\infty)$ we have $x > \frac1n$ or $n^2x^2 > 1$.
Therefore
$$\|f_n\|_{[a, +\infty)} = \sup_{x \in [a, +\infty)} \frac{nx}{1+n^2x^2} \le \sup_{x \in [a, +\infty)} \frac{nx}{2n^2x^2} = \sup_{x \in [a, +\infty)} \frac{1}{2nx} \le \frac1{2na} \xrightarrow{n\to\infty} 0$$
so $f_n \to 0$ uniformly on $[a, +\infty)$.
On the other hand, for $[0, +\infty)$ we have
$$\|f_n\|_{[0, +\infty)} = \sup_{x \in [0, +\infty)} f_n(x) \ge f_n\left(\frac1n\right) =\frac12 \not\to 0$$
so $f_n \not\to 0$ uniformly on $[0, +\infty)$.