Clarification needed regarding uniform convergence of a sequence of functions

Question

Given the following sequence of functions: $$f_n(x) = 1 - \cos\Big(\frac{x}{n}\Big)$$

I was requested to show that the sequence converges uniformly on the closed interval $[0, 2\pi]$ to the function $f(x) = 0$.

When attempting the find the limit of $\sup|f_n(x) - f(x)|$. I've noticed that it does not approach zero, as needed in order to show that the sequence converges uniformly.

Here's what I've done: Let $g(x) = |f_n(x) - f(x)| = 1 - \cos(\frac{x}{n})$. We shall now find the supremum of $g(x)$, $g'(x) = \frac{\sin(\frac{x}{n})}{n}$.

And it is easy to see, that for $x = n\cdot\pi\cdot k$, for $k\in\{0, 1 ,2\}$

I've discovered that $x = \pi\cdot n$ is where the function gets its supremum. (This is where I suspect of an error in my calculation, since $\pi\cdot n$ might not be in the domain specified)

And so, I get $\lim_{x\to\infty} \sup|f_n(x) - f(x)|=2\neq 0$

As far as I know, a sequence of functions is uniformly convergent on a closed inverval if and only if $$\lim_{x\to\infty} \sup|f_n(x) - f(x)|=0.$$

I know that the sequence is supposed to converge. Can anyone clarify please?

Thanks in advance!

Answer 1

Start by showing that $0\leq 1-\cos(t)\leq \frac{t^2}{2}$ for all $t\geq 0$, and see what this tells you about the sequence $f_n(x)$.

Answer 2

You can also use the mean value theorem: Let $g(x) = \cos (x/n).$ Then $1-\cos (x/n) = g(0) - g(x) = g'(c_x)(-x).$ That should help.

Answer 3

It's easy to show (using the derivative) that $f_n=|f_n|$ take a local maximum on $n\pi$, and there is no local maximum on $[0,n\pi[$. Therefore, for $n$ big enough, $$\sup_{x\in [0,2\pi]}|f_n|=f_n(2\pi)\underset{n\to \infty }{\longrightarrow }0.$$

Therefore the convergence is uniform.