$\sum_{ n=0}^{\infty} a_n(x)$ converges uniformly if and only if $$\lim_{N\to \infty}\sup_{x\in\Bbb R} \left|\sum_{ n=N}^{\infty} a_n(x)\right|=0$$

This proposition is not in the book I'm using - I'm only aware of the cauchy criterion and Weiserstrass m-test for uniform convergence. Can someone tell me whether this is a proof or definition, and how it is derived. In particular I'm not sure what $\sup_{x\in\Bbb R}$ means. Also an example of the proposition used in a short proof would be helpful.


Seems the proposition does follow from this basic definition:

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Only that I'm not sure why the $\sup_{x\in\Bbb R}$ comes up. Can anyone explain?

  • $\begingroup$ That is trivial. just give us what definition your book has. By the way answer below is correct $\endgroup$ – Guy Fsone Nov 23 '17 at 7:18
  • $\begingroup$ $\sup_{x \in \R}$ means the supremum over the reals. The supremum is the same as the least upper bound of a set of real numbers. In this case, for each N, you’re looking at the set of all absolute values of the tail ends of the series that start at N, each element depending on x, and you want to know what the least upper bound is of this set, i.e., the supremum of the set. $\endgroup$ – gorzardfu Nov 23 '17 at 7:23
  • $\begingroup$ @gorzardfu Hi, I updated my post. I still have trouble with the $\sup_{x\in\Bbb R}$. I understand what you said in terms of what it means. But I can't see why its there - the definition I'm given has no mention of supremums. $\endgroup$ – helios321 Nov 23 '17 at 8:03

If you define $u_n(x) = \sum_{k=1}^n a_k(x)$ and the limit as $u(x) = \sum_{k=1}^\infty a_k(x) = \lim _{n\to\infty} u_n(x)$, then $u_n$ converges uniformly to $u$ in $\Bbb R$ if it converges in $\|\cdot\|_\infty$ norm on $\Bbb R$, i.e. $$ 0 = \lim_{n\to\infty} \|u_n - u\|_\infty = \lim_{n\to\infty} \sup_{x\in \Bbb R} \left|\sum_{k=1}^n a_k(x) - \sum_{k=1}^\infty a_{k}(x)\right| = \lim_{n\to\infty} \sup_{x\in \Bbb R} \left|\sum_{k=n+1}^\infty a_k(x)\right|, $$ and notice that the subscript $k=n+1$ can be displaced by $1$.


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