Convergence of $F_n=x^n\sin(nx)$ on $S=(-1,1)$

Question

let $\{F_n(x)\}$ be a sequence of functions where $F_n=x^n\sin(nx)$ on $S=(-1,1)$. Find $\lim_{n\to \infty}F_n(x)$. Show $F_n$ converges uniformly on closed subset of $S$ but not on $S$.

My attempt:

1. Find $\lim_{n\to \infty}F_n(x)$:

\begin{align} -1&\leq\sin(nx)\leq 1\\ -x^n&\leq x^n\sin(nx)\leq x^n, x\in[0,1),\\ x^n&\leq x^n \sin(nx)\leq -x^n, x\in(-1,0)\\ |x^n \sin(nx)|&\leq |x|^n, x\in(-1,1) \end{align} Since, $|x|<1$ implies $|x|^n\to 0$ as $n \to \infty$, by the squeeze theorem $x^n \sin(nx) \to 0$.

2. Show $F_n$ converges uniformly on closed subset of $S$ but not on $S$:

Let $a,b\in\mathbb{R},-1<a<b<1$ \begin{align} &M:=\sup_{[a,b]}|x^n\sin (nx)|\leq \max(|a|,|b|)^n\to 0~ as~ n\to \infty \end{align} Hence $f_n$ converges uniformly to $f$ on $[a,b]$

On the other hand, If $\{f_n\}$ were to converge uniformly to $f$ on $(-1,1)$, then it should also converges pointwise to some $f$ at $x=1$ and $x=-1$. However, $f_n( 1)= \sin(n)$ implies $\lim_{n\to \infty}f_n$ does not exists. Hence, it does not converge uniformly on $(-1,1)$.

Is my solution correct. Is there another way to show it is not uniformly convergent at $x=\pm1$.

Answer 1

Take $$x\in \left(1- \frac{2\pi}{n},1\right)$$ then $$nx \in \left(n-2\pi,n\right)$$ so $nx$ is contained in a full period of the sine… hence there is an $x_0 \in \left(1- \frac{2\pi}{n},1\right)$ s.t. $|\sin(nx_0)| = 1$

And so: $$\|x^n\sin(nx)\|_\infty \ge |x_0^n\sin(nx_0)| = |x_0|^n \ge \left(1- \frac{2\pi}{n}\right)^n \to e^{-2\pi}$$

So $$\|x^n\sin(nx)\|_\infty \not\to 0$$

Answer 2

In order to show that $(F_n)_n$ does not converge uniformly on the interval $(a,-1)$ with $a\in(-1,1)$ note that $$\limsup_{n\to+\infty }F_n\left(1-\frac{\pi}{2n}\right)= \lim_{n\to+\infty } \left(1-\frac{\pi}{2n}\right)^n\cdot\limsup_{n\to+\infty }\left(-\cos\left(n\right)\right)= e^{-\pi/2}\not=0.$$ For the interval $(-1,a)$ consider the sequence $(F_n\left(-1+\frac{\pi}{2n}\right))_n$.

Moreover to show that $(F_n)_n$ converges uniformly on any closed set $C$ of $(-1,1)$ you have to mention in your proof that there is a closed interval $[a,b]$ such that $C\subseteq [a,b] \subset (-1,1)$.