Uniform convergence of sequence of function cos(x/n)

Question

$f_n(x)=\cos(\frac{x}{n})$

I know that its pointwise convergence is 1. Now I need uniform convergence and my range is $[0,2\pi]$ :

(1)$\limsup_{n\to \infty} \lvert \cos(\frac{x}{n})-1\rvert$

(2)$\limsup_{n \to \infty} \left( 1-\cos(\frac{x}{n})\right)$

$g_n(x)=1-\cos(\frac{x}{n})$

$(g_n)'=\frac{1}{n}\sin(\frac{x}{n})$

$g_n(x)' = 0 \Rightarrow x_{max}=\pi n$

So : $\lim\limits_{n \to \infty} g_n(x_{max}) \not= 0$ Its not correct and i dont' know where i am wrong. Thanks for helping me.

Answer 1

If $g(x)=1-\cos(x/n)$, then $g'(x)=\frac{\sin(x/n)}{n}=0$ at $x=0$ only for $n>2$. And $g(0)=0$ is the minimum, not the maximum.

Since $g(x)$ is increasing in $x$ for $n>1$, its supremum on $[0,2\pi]$ is $g(2\pi)=1-\cos(2\pi/n)$ and this approaches $0$ independent of $x$. So, the convergence is uniform.

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Alternatively, note that

$$\begin{align} \left| 1-\cos(x/n)\right|&=2\sin^2(x/2n)\\\\ &\le \frac{x^2}{2n^2}\\\\ &<\frac{2\pi^2}{n^2}\\\\ &<\epsilon \end{align}$$

whenever $n>\frac{\sqrt{2}\pi}{\sqrt{\epsilon}}$.

Answer 2

You can use the Mean Value Theorem. For any $x \in [0,2\pi]$ we know that $$|1-\cos\left(\frac{x}{n}\right)|=|\cos(0)-\cos\left(\frac{x}{n}\right)|=\frac{|x|}{n}|\sin(\xi)|$$ for some $\xi \in (0,2\pi)$. Since $|\sin\left(\xi\right)| \leq 1$ and $|x|\leq 2\pi$, we have

$$|1-\cos\left(\frac{x}{n}\right)| \leq \frac{2\pi}{n}$$ independently of $x \in [0,2\pi]$.