Pointwise convergence of $f_n$

Question

Let $(f_n)_{n \ge 1}$ be a sequence of functions defined by $$ f_n(x) = \begin{cases} 1, &\text{if } -1 \le x < -\frac{1}{n}, \\ \frac{n}{2} \left(\frac{1}{n}-x\right), &\text{if } -\frac{1}{n} \le x < \frac{1}{n}, \\ 0, &\text{if } \frac{1}{n} \le x \le 1. \end{cases} $$ Show that $(f_n)$ is pointwise convergent to a function $f$. What is $f$?

My attempt: I want to show that $f_n(x) \to f(x)$, where $f$ is defined by: $$ f(x) = \begin{cases} 1, &\text{if } x < 0, \\ 0, &\text{if } x > 0, \\ \frac{1}{2}, &\text{if } x = 0. \end{cases} $$ It is clear that when $x < -\frac{1}{n}$, $f_n(x) \to 1$ on $\left[-1,-\frac{1}{n} \right)$; and when $x > \frac{1}{n}$, $f_n(x) \to 0$ on $\left( \frac{1}{n}, 0 \right]$.

Let $\epsilon > 0$. When $-\frac{1}{n} < x < 0$, we want to find integer $N(\epsilon,x) > 0$ such that whenever $n \ge N$, we have $|f_n(x)-1| < \epsilon$. $$ |f_n(x)-1| = \left| \frac{1}{2} - \frac{nx}{2} - 1 \right| \le \frac{1 - nx}{2}. $$ I'm getting stuck here. I can't connect this expression with $N$ to have the whole thing less than $\epsilon$. What am I missing? What should I do?

Thank you very much.

Answer 1

If $-1 \leq x<0$ then we have $-1\leq x<-\frac 1n$ as long as $n \geq -\frac 1 x$. Take $N=[-\frac 1 x]+1$. Then $f_n(x)=1$ whenever $n >N$ so $|f_n(x)-1|=0<\epsilon$.

Answer 2

When you talk about pointwise convergence, the $n$ in "$x<-\frac{1}{n}$" varies. So just notice for any $x<0$, there always exists a sufficiently large $N$ satisfy $x<-\frac{1}{n}$ whenever $n>N$. Then $f_n(x)$ equals 1 ever after. In other words $x$ do not belong to $-\frac{1}{n}<x<\frac{1}{n}$ for every $n$ so you cannot use the expression of $f_n$ in the middle forever.