exercise on pointwise convergence of an (easy) function.

Question

Exercise 6.2.5. Taken from understanding analysis of Stephen Abbott

For each n $\in N$, define $f_n on \ R$ by

$$f_n(x) = \begin{cases} 1, & \mbox{if} \ |x| \ge 1/n \\ n|x|, & \mbox{if} \ |x| < 1/n \end{cases}$$

Find the pointwise limit of $(f_n)$ on $R$.

I would like to know how to properly think about this, because thus far to demonstrate pointwise convergence I have always taken the limit as n goes to infinity of a function. The Definition of pointwise convergence made it sensible to use this approach.

If I naively try to apply that to the conditional statements on x, I obtain $$f(x) =\begin{cases} 1, & \mbox{if} \ |x| \ge 0 \\ n|x|, & \mbox{if} \ |x| < 0 \end{cases}$$

But this seems wrong because no absolute value of x $\in R$ can be less than 0. Moreover I am unsure how to treat $n|x|$ because $n$ goes to infinity but $|x|$ is always smaller than n. So what is the right way to think about this problem? and how does it agree with the def. of pointwise convergence?

Thank you in advance

Answer 1

You are almost correct.

$|x|$ can never be less than $0$, by definition.

Now, for $x=0$, what is $f_n(0)?$

Also, if $|x|>0$, then there always exists $N$ such that for all $n>N$, $|x|\geq 1/n$ (This follows from the fact that the limit of $1/n$ is 0).

So, $f_n(x)=1$ for all $n>N$, which means $\cdots$?

Answer 2

Note that when passing to the limit in expressions like abs(x) < 1/n the inequality typically weakens; in this case, you get abs(x) ≤ 0 not the impossible abs(x) < 0.