I have a question about the convergence of $(f_n)_{n \in \mathbb{N}}$ with
$$
f_n(x) := \begin{cases} (x-n+1)(n+1-x): n-1 < x < n+1\\ 0: x \leq n-1 \lor x \geq n+1 \end{cases}.
$$
Does this sequence of functions converge and if so what is its limit function? Does it converge point wise or also uniformly?
What I have done so far:
I plotted some elements of the sequence to see how it behaves (I know this is also kind of obvious already from the definition but I did it anyways)
and I realized that the sequence consists just of these "hills" that look like $f(x)=-x^2+1$ and shoot off to positive infinity. What I don't get about this is how this could every possibly converge to anything since it clearly does not "stop" sliding off to the right. But of course the functions values always stay the same and I am sure herein lies the "trick" somewhere.
I thought of trying to prove that
$$
f_n \to f = \begin{cases} -x^2+1: -1 < x < 1\\ 0: x \leq -1 \lor x \geq 1 \end{cases} \ (n \to \infty)
$$
but I got stuck when taking $x \in [1, \infty)$ because even if $f=0$ at that point I cannot conclude the same for any $f_n(x)$. The same problem occurs if I assume $-1 < x < 1$. What about the values of $f_n(x)$ on that interval?
Thanks for any help in advance!
It is quite clear that $\{f_n\}$ converges pointwise to $0$. This is because, for any given $x\in \mathbb{R}$, choosing $N_x\in \mathbb{N}$ such that $x<N_x-1$, then for all $n\geq N_x$, we would have $x < n-1$ and hence by definition $f_n(x)=0$, for all $n\geq N_x$, implying that $\{f_n(x)\}$ converges to $0$.
However, the convergence is not uniform. This is because, for each $n$, we have $n\in (n-1, n+1)$, forcing by definition that $$f_n(n) =1.$$
Hence, for any $\epsilon <1$, it would not hold good that there exists an $N_{\epsilon}$ such that for all $n\geq N_{\epsilon}$, we would have $$|f_n(x)| < \epsilon,$$ for all $x$.