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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?

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

2 Answers 2

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$$\lim_{n\to\infty}\left\lVert f_n - f \right\rVert = \lim_{n\to\infty}\left\lVert (\sin\frac{x}{n}, \cos\frac{x}{n}) - (0, 1) \right\rVert$$

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 $(0,1)$.

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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.

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