Consider $$ f_n : \begin{cases} \mathbb{R} & \longrightarrow \mathbb{R} \\ x & \longmapsto \displaystyle \frac{1}{1+(x-n)^2} \end{cases}$$ and verify piecewise convergence on $\mathbb{R}$, also show that $f_n$ is not uniformly convergent on $\mathbb{R}$ but it is uniformly convergent on any given Interval $I$ of the form $I=[a.b] \subset \mathbb{R}$
Since this is a homework question, I will provide my full solution and would appreciate corrections where they are necessary
My Attempt:
For point wise convergence let $x_0 \in \mathbb{R}$, such that: $$ \lim_{n \to \infty} \frac{1}{1+(x_0-n)^2}=\lim_{n \to \infty} \frac{\frac{1}{n^2}}{1-\frac{2x_0}{n}+\frac{x_0^2}{n^2}+\frac{1}{n^2}}=\frac{0}{1}=0 \\ \implies \text{ point wise convergence for }x_0 \in \mathbb{R} \\ f \equiv 0 \ \text{limit function}$$
For uniform convergence on $\mathbb{R}$ $$\lim_{n \to \infty} \sup_{x \in \mathbb{R}} \left|\frac{1}{1+(x-n)^2}-0 \right|=1$$ Because (please verify my reasoning) $$\sup_{x \in \mathbb{R}}\left|\frac{1}{1+(x-n)^2}\right|$$ takes on its supremum for $x=n \in \mathbb{R}$ such that $\displaystyle \lim_{n \to \infty} 1 =1 $
For the second part:
Let $I:= [a,b] \subset \mathbb{R}$ with $a < b, \forall a,b \in I$ such that $f_n(x)$ is monotone decreasing in $[a,b]$ if $a,b$ are positive and monotone increasing if $a,b$ are negative. For monotone decreasing we have: $$ \lim_{n \to \infty} \sup_{x \in [a,b]} \left|\frac{1}{1+(x-n)^2} \right|=\lim_{n \to \infty} \frac{1}{1+(a-n)^2}=0$$ and same argument for monotone increasing $$ \lim_{n \to \infty} \sup_{x \in [a,b]} \left|\frac{1}{1+(x-n)^2} \right|=\lim_{n \to \infty} \frac{1}{1+(b-n)^2}=0$$
Especially in the second part of my homework I am not sure if the steps I apply are correct or require more mathematical rigor to be valid. I'd appreciate some supervision over my steps and verification.
Update: Do I need to mention that $n \notin [a,b]$ for the above to be true? Because otherwise it seems to me like we would obtain again, that $f_n(x)$ is not uniformly convergent on $[a,b]$
Update 2: How does that look?
Let $M:= \max \lbrace |a|, |b| \rbrace$ and $J: [-M,M]$ such that $I \subset J$ if we can prove uniform convergence for $J$, we have also proven it for $I$. Let $n \geq 2M$ and $\forall x \in J \implies |x| \leq M$
My idea now was to find an estimate for the $(x-n)^2$ term in the denominator with the above assumptions $$|x-n|=|n-x| \geq |n|-|x| \geq n-M \geq n - \frac{n}{2}=\frac{n}{2}$$ such that $$\forall n \geq 2M, \sup_{x \in I} \left| \frac{1}{1+(x-n)^2}\right| \leq \sup_{x \in J} \left| \frac{1}{1+(x-n)^2} \right| \leq \sup_{x \in J} \left| \frac{1}{1+\frac{n}{2}}\right|=\frac{1}{1+\frac{n}{2}} $$ such that: $$\lim_{n \to \infty} \frac{1}{1+\frac{n}{2}}=0$$ which would imply that $f_n(x)$ is uniformly convergent on $J$ and therefore also uniformly convergent on $I$