Let $f_n(x) = \sin^2(\pi x)(\chi_{[n, n+1]})$.
I need to find its pointwise limit, and show if it converges uniformly. Also I am asked if $f_n(x_n) \to f(x_0) $ $\; \forall x_n\to x_0 $ in R.
I am encountering $\chi$ notation for the first time so I am not sure what is going here. I know that $f_n(x)$ are zero outside $[n,n+1]$
for the pointwise limit:
I am assuming as $n\to \infty $, then $[n, n+1]$ becomes $\phi$ so intuitively $f_n(x) \to 0$ pointwise. But I don't think this is a rigorous proof.
Please help me with this.
Best Answer
For a fixed $x$, choose an $n_{0}$ such that $n_{0}>x$, then for all $n\geq n_{0}$, $x\notin[n,n+1]$, and hence $f_{n}(x)=0$, so $f_{n}\rightarrow 0$ pointwise.
It is not uniform since $\sup_{x\in\mathbb{R}}f_{n}(x)=1$.