I'm having trouble on a problem about sequences of functions.
Considering the sequence of functions $f_n(x)\colon\mathbb{R}\to\mathbb{R}$ such that $f_n(x)=1$ if $x$ belongs to $[-n,n]$ and $f_n(x)=0$ if $x$ doesn't belong to $[-n,n]$.
I seem to think that it will converge pointwise to $f(x)$, that is defined just like $f_n(x)$, meaning $f(x)=1$ if $x$ belongs to $[-n,n]$ and $f(x)=0$ if $x$ doesn't belong to $[-n,n]$.
My doubt is in figuring out $M_n$. I know that it is supposed to be the supremum of $|f_n(x)-f(x)|$, however i do not know which interval to choose from, and because of that i don't know what to substitute $f(x)$ for.
Any help would be appreciated, thanks :)
You have the wrong limit function (it should not depend on $n$).
If we define the function $f$ as the limit of the sequence of functions $\{f_n\}$, then the function $f\colon\mathbb{R}\to\mathbb{R}$ is defined by $f(x)=1$ for every $x\in \mathbb{R}$.
At a specific point $x\in \mathbb{R}$, the limit of the sequence of a function is defined at the point $x$ as the limit of the sequence of numbers $\{f_n(x)\}$.
In this situation, the function $f_n(x)=1$ if $|x|\leq n$, and it equals $0$ otherwise. For a real number $x$, there exists a natural number $N$ such that $|x|<N$ and for every $n\geq N$, $|x|<N\leq n$ so there exists an $N\in \mathbb{N}$ such that $f_n(x)=1$ for every $n\geq N$ at every point $x\in \mathbb{R}$. The limit function $f(x)=1$ for every point $x$ in the real line.
To study the uniform convergence of the sequence, we want to see if the sequence $\{M_n\}$ converges to $0$ where $$M_n=\sup_{x\in \mathbb{R}} |f_n(x)-f(x)|$$
However, $M_n=1$ for every $n$ because there exists there is always a real number $x$ greater than $n$ and $M_n\geq |f_n(x)-f(x)|=|0-1|=1$. The sequence does not converge to $0$ so the sequence of function is not uniformly converging.