Does $(f_n)=(n\sin(\frac{x}{n})-x)$ converge uniformly on $[-a,a]$ for $a\geq0$?

Question

I'm trying to solve the next problem: Let $\left(f_{n}\right)_{n\in\mathbb{N}}$ be a sequence of functions such that $f_{n}\colon\mathbb{R}\to\mathbb{R}$ is given by $f_{n}\left(x\right)=n\sin\left(\frac{x}{n}\right)-x$, for all $n\in\mathbb{N}$.

$1.$ Prove that $\left(f_{n}\right)_{n\in\mathbb{N}}$ is pointwise convergent.

$2.$ Prove that $\left(f_{n}\right)_{n\in\mathbb{N}}$ doesn't converge uniformly.

$3.$ Does $\left(f_{n}\right)_{n\in\mathbb{N}}$ converge uniformly on $\left[-a,a\right]$ for $a\geq0$?

For $1.$ I proved that $\left(f_{n}\right)_{n\in\mathbb{N}}$ coneverges pointwise to 0. Given $x\in\mathbb{R}$, using L'Hopital's rule we have that: $$ \lim_{n\to\infty}\frac{\sin(\frac{x}{n})}{\frac{1}{n}}=\lim_{n\to\infty}\frac{\cos(\frac{x}{n})\cdot (-\frac{x}{n^{2}})}{-\frac{1}{n^{2}}}=x\cdot\lim_{n\to\infty}\cos(\frac{x}{n})=x\cdot\cos(0)=x. $$

And therefore $\lim_{n\to\infty}f_{n}\left(x\right)=\lim_{n\to\infty}n\sin\left(\frac{x}{n}\right)-\lim_{n\to\infty}x=x-x=0.$ Hence $\left(f_{n}\right)_{n\in\mathbb{N}}$ converges pointwise to 0.

In $2.$ I took $\varepsilon=1$. Then given $n\in\mathbb{N}$, we can take $x=n\pi\in\mathbb{R}$ and we have that $$ \mid f_{n}\left(x\right)-0\mid=\mid n\sin\left(\frac{n\pi}{n}\right)-n\pi\mid=n\pi>1=\varepsilon. $$

And with this we can conclude that $\left(f_{n}\right)_{n\in\mathbb{N}}$ doesn't converge uniformly in $\mathbb{R}$.

I don't know for $3.$ if the answer is yes or no. I'm thinking that may be the answer could be yes, since we are taking a compact set and $f_{n}$ is continuous for all $n\in\mathbb{N}$, but I couldn't prove this. Could you help me or give me some idea for this problem?

Thanks.

Answer 1

It is known that $\lim\limits_{y \to 0}\frac{\sin y}{y} =1$.

Take $\epsilon >0$. You can pick up $\delta >0$ such that $\left\vert \frac{\sin y}{y} -1 \right\vert \le \epsilon$ for $\vert y \vert \le \delta$. For $n >a/\delta$ and $\vert x \vert \le a$, you have $\left\vert \frac{x}{n} \right\vert \le \delta$ and therefore

$$\left\vert x \frac{\sin(\frac{x}{n})}{\frac{x}{n}}-x\right\vert =\vert f_n(x)\vert \le a \epsilon$$

proving that the function converges uniformly on each compact $[-a,a]$.

Answer 2

Note that, assuming that $x\in[-a,a]$,$$\left\lvert n\sin\left(\frac xn\right)-x\right\rvert=\left\lvert x\frac{\sin\left(\frac xn\right)}{\frac xn}-x\right\rvert=\lvert x\rvert\cdot\left\lvert\frac{\sin\left(\frac xn\right)}{\frac xn}-1\right\rvert\leqslant a\left\lvert\frac{\sin\left(\frac xn\right)}{\frac xn}-1\right\rvert.$$Now, use the fact that $\lim_{t\to0}\frac{\sin t}t=1$ and that$$(\forall x\in[-a,a]):\left\lvert\frac xn\right\vert\leqslant\frac an.$$