Determine if $f_n(x)=x^n\sin{(nx)}$ is uniform convergent on the following intervals.

Question

The intervals are:

$I_1=\{ x\in\mathbb{R}:a\leq x\leq b\}$ with $-1<a<b<1$.

$I_2=\{ x\in\mathbb{R}:-1< x< 1\}$.

I've shown $f_n$ is pointwise convergent to $0$ on $I_1$ and $I_2$. But what is the difference for showing uniform convergence on these two intervals? I have a few other questions that have very similar intervals and I'm not sure what I should be doing different.

Basically, what is different in showing uniform convergence on $I_1$ and $I_2$?

Answer 1

The sequence of function $(f_n)$ is unifom convergent to $f$ in the interval $I$ if: $$\sup_{x\in I}|f_n(x)-f(x)|=0.$$

1. We have $$ \forall x\in I_1,|f_n(x)-0|\leq \max(|a|^n,|b|^n)\to0,$$ so the sequence $(f_n)$ is uniform convergent to the zero function on $I_1$.

2. The sequence $(f_n)$ is pointwise convergent to the zero function, so if we have the uniform convergence, it must be to the zero function also.

In another way, we have $|f_n(1-\frac{1}{n})|=(1-\frac{1}{n})^n|\sin(n-1)|,$ and this sequence has not a limit, so $(f_n)$ is not uniform convergent to the zero function on $I_2$.

Answer 2

To show uniform convergence to a function $f$ on an interval $I,$ you must show that $$\lim_{n\to\infty}\left[\sup_{x\in I}|f_n(x)-f(x)|\right]=0.$$ In this particular case, you must show that $$\lim_{n\to\infty}\left[\sup_{x\in I}|f_n(x)|\right]=0.$$

Since $|f_n(x)|\leq |x^n|$ for all $x$, then to see uniform convergence on $I_1$, it suffices to show that $$\lim_{n\to\infty}\left[\sup_{x\in I_1}|x^n|\right]=0,$$ which is fairly straightforward.

To see that there isn't uniform convergence on $I_2$, consider the sequence $x_n=1-\frac1n$, and note that $|f_n(x_n)|$ fails to converge at all, which would not be the case if we had uniform convergence to $0$. (Why?)