# Disprove uniform convergence of sequence of functions $f_n: \mathbb{R} \to \mathbb{R}^2$, $f_n(x) = (\sin\frac{x}{n}, \cos\frac{x}{n})$

I have already found the pointwise limit of $$f_n(x) = (0, 1)$$.

I have a theorem that states "Let $$D\subset\mathbb{R}^q$$, and $$D$$ compact. Let $$f, f_n:D\to\mathbb{R}^p$$ and $$f_n$$ continuous for all $$n\in\mathbb{N}$$. Then $$f_n$$ converges uniformly to $$f$$ if and only if $$\lim_{n\to\infty}\left\lVert f_n - f \right\rVert_D = 0$$."

So, following an example from my professor, I let $$f = (0, 1)$$ and found that $$\lim_{n\to\infty}\left\lVert f_n - f \right\rVert = \lim_{n\to\infty}\left\lVert (\sin\frac{x}{n}, \cos\frac{x}{n}) - (0, 1) \right\rVert = \lim_{n\to\infty}\left\lVert (\sin\frac{x}{n}, \cos\frac{x}{n} - 1) \right\rVert = \lim_{n\to\infty} \sqrt{\sin^2\frac{x}{n} + (\cos\frac{x}{n} - 1)^2} = \sqrt{0 + (1-1)^2} = 0$$

Shouldn't this prove that in fact the function DOES converge uniformly? I'm supposed to prove that it does not. What am I missing here?

• $\mathbb{R}$ is not compact Mar 12, 2019 at 4:25

$$\lim_{n\to\infty}\left\lVert f_n - f \right\rVert = \lim_{n\to\infty}\left\lVert (\sin\frac{x}{n}, \cos\frac{x}{n}) - (0, 1) \right\rVert$$
This is incorrect. The right hand-side is the pointwise limit, evaluated at some particular $$x \in \mathbb{R}$$, which we already know is $$0$$. But the usual definition for $$||f_n - f||$$ is $$\sup_{x \in \mathbb{R}} |f_n(x) - f(x)|,$$ the supremum over all possible $$x$$.
Your mission is to find an $$\epsilon$$ so that, given an arbitrarily large $$n$$, there is a point $$x \in \mathbb{R}$$ so that $$|f_n(x) - f(x)| > \epsilon$$. (Here $$f$$ is the constant function $$f(x) = (0,1)$$ for every $$x$$.) This shows that the supremum is greater than $$\epsilon$$. This should be easy, since the image of each $$f_n$$ is the unit circle. If convergence were uniform, the images would be sets that are shrinking closer and closer to $$(0,1)$$.
What you have proved is pointwise convergence, not uniform convergence. Suppose the convegence is uniform. Then there must be an integer $$m$$ such that $$\|(\sin (\frac x n),\cos (\frac x n))-(0,1)\| <\frac 1 2$$ for all $$x \in \mathbb R$$ for all $$n \geq m$$. Take $$x=\frac {m\pi} 2$$ and $$n=m$$ to get a contradiction.