I don't think I have too much difficulty in understanding and showing that simple function converge pointwise or uniformly, the few "standard" examples, such as $f_n(x) = x^n$, $f_n(x)=\frac{\sin(x)}{n}$ or $f_n(x)=\frac{x}{n^2}$, require almost no more than a simple application of the definition, by finding a suitable convergence rank (paying attention of course to the definition domain).
For example, for $f_n(x)=\frac{\sin(x)}{n}$, we show without too much difficulty with the squeeze theorem that $f_n(x)$ converges pointwise to the null function, and to check if there is uniform convergence, since $f(x)$ is $0$, we show that $|\frac{\sin(x)}{n} - 0| \leq \frac{1}{n}$, then we could apply $\sup_{x\in\mathbb{R}}|f_n(x)-f(x)|\leq \frac{1}{n}$, so $f_n(x)$ does converge uniformly to $f(x)=0$ on the reals.
Our teacher, however, gives us many exercises in which the sequences of functions are defined by parts, here is the example I would like to focus on here:
$$ f_n : \mathbb{R} \to \mathbb{R} : x \mapsto \begin{cases} \frac{x}{n^2} + \frac{1-n}{n} & n^2-n \leq x \leq n^2 \\ -\frac{x}{n^2} + \frac{1+n}{n} & n^2 < x \leq n^2+n \\ 0 & \text{otherwise} \end{cases} $$
and I don't have much idea how to deal with this problem, although the methods should probably be close to the normally defined function sequences. So I imagine that there will be cases to deal with to show (or not) pointwise and uniform convergence.
First, intuitively, it seems clear that this sequence converges pointwise to the null function. A small plot of this sequence shows indeed that the functions collapse as $n$ becomes larger. We can also notice that $f_n$ will always be between $0$ and $1$ inclusive.
What confuses me is that the definition conditions use both $n$ and $x$, so I don't know what to fix because the pieces of the function depend on $n$ but also on $x$, so I don't know which cases it would be interesting to me to treat.
My question is then how to treat pointwise and uniform convergence for this [kind of] sequence, I have indeed some difficulties to know what to actually treat in this problem.
PS: If you think I'm giving you my homework, you're probably right, but I have a whole list of exercises and getting on track with one of them will surely be very useful to keep practicing.
Yes, $(f_n)_{n\in\Bbb N}$ converges pointwise to the null function. That's so because, for each $x\in\Bbb R$, if $N\in\Bbb N$ is such that $N^2-N\geqslant x$, then$$n\geqslant N\implies n^2-n\geqslant N^2-N\implies f_n(x)=0.$$Actually, the convergenc is uniform, since $(\forall x\in\Bbb R):f_n(x)\leqslant\frac1n\left(=f\left(n^2\right)\right)$. And so$$\lim_{n\to\infty}\sup|f_n-0|=\lim_{n\to\infty}\frac1n=0.$$