Here is the question:
For $x\in(0,1)$ and $n\in \mathbb{N}$,define
$f_n(x)=\begin{cases}0&0<x<\frac{1}{n+1},\\\sin^2(\frac{\pi}{x})&\frac{1}{n+1}\le x\le \frac{1}{n},\\0&\frac{1}{n}<x<1.\end{cases}$
Prove that there is a continuous function $f$ such that $f_n\to f$ pointwise, but $f_n$ does not converge uniformly to $f$.
The Proof says that $f_n(x)\to 0$ for all $x$ , and I could not figure out why?
let $x=\frac{2}{2n+1}$ ,here $x\in[\frac{1}{n+1},\frac{1}{n}]$ and $f_n(x)=1$ , for every $n$ this $x$ exists. So why $f_n(x)\to 0$ ? Is the Proof wrong?
I could think that when $n\to \infty$ $\mu([\frac{1}{n+1},\frac{1}{n}])\to 0$ here $\mu$ denote the measure, but if the function kick out the zero measure part, it converge to $0$ for almost everywhere($a.e.$) but can converge to $0$ $ a.e.$ can be consider as converge to $0$ ? I think there exist some differences.
If anyone can help, please leave your answer :)
The proof is pretty simple and is based on a basic property of $\mathbb{N}$ in $\mathbb{R}$, called the Archemedian property. If we consider any point $x \in \left( 0, 1 \right)$, then $\exists n_0 \in \mathbb{N}$ such that $n_0 x > 1$ or $\dfrac{1}{n_0} < x$.
Now, for $n \geq n_0$, we have $\dfrac{1}{n} \leq \dfrac{1}{n_0} < x < 1$. Therefore, after certain stage, the point $x$ lies in the interval $\left( \dfrac{1}{n}, 1 \right)$ so that after the same stage, we have $f_n \left( x \right) = 0$.
Therefore, given any $x \in \left( 0, 1 \right)$, the sequence of real numbers $\left\lbrace f_{n} \left( x \right) \right\rbrace_{n \in \mathbb{N}}$ is eventually constant and hence convergent. Also, eventually, it takes the constant value $0$ so that $f_n \left( x \right) \rightarrow 0$ as $n \rightarrow \infty$.