Given sequence is $${f_n(x)}=\frac{log(1+n^2x^2)}{n^2}$$, where $x\in[0,1]$.
To test whether the sequence is or is not uniformly convergent my method is-
$$f(x)=\lim_{n\to\infty} f_n(x), \mid form \frac{\infty}{\infty}$$
$$=\lim_{n\to\infty}\frac{1}{n^2}\frac{2nx^2}{1+n^2x^2}=0$$
Now, $$\mid{f_n(x)}-f(x)\mid=\frac{log(1+n^2x^2)}{n^2}<\frac{(1+n^2x^2)}{n^2}<\epsilon $$
Which gives, $$n^2>\frac{1}{\epsilon-x^2}\Rightarrow n>(\epsilon-x^2)^\frac{-1}{2}$$
As $x$ varies from $0$ to $1$ we get almost a finite value $\epsilon$ where there is no $x$ terms are attached
The ${f_n(x)}$ is continuous over $[0,1]$ and is
$$1+n^2x^2<e^{n^2} \Rightarrow log(1+n^2x^2)<n^2 \Rightarrow \frac{log(1+n^2x^2)}{n^2}<1$$, whenever $x\in [0,1]$ bounded.
Is it enough or correct step to prove the sequence uniformly convergent?
Another approach, if $f(x)=0$ then $$f'(x)=0 \Rightarrow x=0$$ which shows that $f''(x)=2$, i.e. $f'(x)$ has no maxima. I am confused what to do there at next step?
Any help is appreciated.
You can use more simple techniques.
For all $n \in \mathbb N$ and $x \in [0,1]$, tou have $$\vert f_n(x) \vert = \left\vert \frac{\ln(1+x^2n^2)}{n^2} \right\vert \le \frac{\ln(1+n^2)}{n^2},$$
as the function $x \mapsto \ln(1+n^2x^2)$ is increasing on $[0,1]$. The sequence $(\frac{\ln(1+n^2)}{n^2})$ converges to zero. Therefore the sequence $(f_n)$ converges uniformly to the always vanishing function.
Moreover the series $\sum \frac{\ln(1+n^2)}{n^2}$ converges. Therefore, $\sum f_n(n)$ converges uniformly according to Weierstrass M-test.