I'm supposed to determine whether the function sequence $f_n(x)=b(x-n)$ converges pointwise or uniformly on the following for $N\in \mathbb{N}$ and $x\in \mathbb{R}$:
$ b(x)= \begin{cases} 1-x^2 & \text{if}\, x\in[-1,1]\\ 0 & \text{if}\, x\lt -1, \text{or } x\gt 1 \end{cases} $
I'm not sure where to start. Am I trying to find if there's uniform convergence of $f_n(x)$ to $b(x)$ in those 2 piecewise cases? If $f_n(x)=1-(x-n)^2$, then that's a "shift" from $1-x^2$ so how would it converge at all to $b(x)$ then?
I was also given a couple propositions to help avoid having to prove these things from definition:
Prop 14.6: Let $f_n$ be pointwise convergent to $f$ on $D$. Then $f_n$ converges uniformly on $D$ to $f\iff \exists a_n\to 0, a_n\ge 0$ such that $|f_n(x)-f(x)|\le a_n$ $\forall x\in D$ for all $n\in \mathbb{N}$ large enough.
Prop 14.7: Assume $f_n$ be converges pointwise to $f$ on $D$. If there exists a sequence $x_n\in D$ such that $|f_n(x_n)-f(x_n)|\ge C_n$ and $C_n\to C\gt 0$, then $f_n$ does not converge uniformly.
Edit: The intermediate questions are only helpful when find out if I'm correct about them. I think I have been ignoring the $0$ part of $b(x)$ by accident. $\lim_{n\to \infty}f_n(x)=0$ since $x-n\lt -1$ as $n\to \infty$