Compute $\|f_n\|_E\quad$ for $\quad f_n(x)=\frac{nx}{1+n^2x^2}\quad$ where $\quad E=[0,1],\quad$ and $\quad E=[1,\infty].\quad$ In which case does $f_n$ converge uniformly to $0$?
Below are my calculations for the interval $[0,1]$-
For $x=0, \quad f_n(0)=0.\quad$For $x>0,\quad$ $f(x)=\lim_{n\to\infty}f_n(x)=0$ $\quad\implies\quad f_n(x)\longrightarrow 0 \ $ pointwise for each $x>0.$
Now $$\|f_n\|_{[0,1]}= \sup\{|f_n(x)|;0 \leq x \leq1\}.$$
We have $$f_n(x)-f(x)=f_n(x)=\frac{nx}{1+n^2x^2}.$$
As $$\quad \frac{d}{dx}\bigg[\frac{nx}{1+n^2x^2}\bigg]=\frac{n-n^3x^2}{(1+n^2x^2)^2}=0\quad \text{when}\quad x=\frac{1}{n}$$
Thus$$\|f_n-f\|_{[0,1]}=\bigg|f_n\bigg(\frac{1}{n}\bigg)-f\bigg(\frac{1}{n}\bigg)\bigg|=\frac{1}{2}\quad\implies\quad\text{not uniformly convergent}. $$
Is this correct? And assuming it is correct I do not see how the result would be different for the interval $[1,\infty],$ but the solutions manual states that the interval $[1,\infty]$ is uniformly convergent on this interval.What am I missing?
As regards the interval $[0,1]$ you are correct. Looking back to your calculations $$f'_n(x)=\frac{n(1-(nx)^2)}{(1+n^2x^2)^2}<0\quad \mbox{for $x>1/n$},$$ that is $f_n$ is decreasing (and positive) in the interval $[1,+\infty)$. Hence $$\|f_n\|_{[1,+\infty]}= f_n(1)=\frac{n}{1+n^2}\to 0$$ and the sequence $(f_n)_n$ is uniformly convergent to $f=0$ in $[1,+\infty)$. Note that the maximum point $1/n$ belongs to $[0,1]$ and not to $[1,+\infty)$ for $n>1$.