Since our lectures were cancelled because of the ongoing situation, I have to essentially self-study for my analysis exam in two months. Understandably this comes with a great deal of trouble, so I would like someone to help me with the following question:
Question: Show that $f_{n} : \mathbb{R}\rightarrow\mathbb{R}$, $f_{n}(x):=\frac{x^{2n}}{1+x^{2n}}$ pointwise converges on $\mathbb{R}$. Examine whether it uniformly converges on $I=[0,2]$, $J=[2,\infty)$.
Pointwise convergence: From what I've understood so far, I have to show $\lim_{n \to \infty}f_n(x) = f(x)$ for every $x$ to prove it. This is my solution:
$lim_{n\rightarrow\infty}\frac{x^{2n}}{1+x^{2n}}=lim_{n\rightarrow\infty}\frac{1}{\frac{1}{x^{2n}}+1}$
$f_n(x)\rightarrow f(x) =\begin{cases}0 &if& |x|<1\\\frac{1}{2} & if&|x| = 1\\1&if&|x|>1\end{cases}$
Uniform convergence: Here, I can show that:
$\big|~f_n(x) - f(x)~\big| = \displaystyle \begin{cases}\displaystyle\frac{x^{2n}}{1+x^{2n}} & \text{if }~ |x|<1\\ 0 &\text{if } |x|=1\\ \frac{x^{2n}}{1+x^{2n}}-1 &\text{if } |x|>1 \end{cases} $
Since $\frac{x^{2n}}{1+x^{2n}}-1$ = $\frac{-1}{1+x^{2n}}$ it would be equal to $0$ for $|x|>1$ as well. This would mean that this function does indeed uniformly converge on $J=[2,\infty)$. On the other hand, $f_{n}(x)$ doesn't converge uniformly on $I=[0,2]$ since it converges to a discontinuous function on this interval.
Is this correct?
