$\sum_{ n=0}^{\infty} a_n(x)$ 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.
UPDATE:
Seems the proposition does follow from this basic definition:
Only that I'm not sure why the $\sup_{x\in\Bbb R}$ comes up. Can anyone explain?
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 $1$.