Convergence Locally in Uniform Convergence

Question

As far as I understand when we say uniformly, it means we are indicating an area where the function converges, isn't is always local when something converges uniformly? Please explain what does locally convergence mean. This question is related to my previous questions my previous questions (click here)

Answer 1

As defined here, local uniform convergence of a sequence of functions $(f_n)$ in a metric space $E$ means that for every $x \in E$ there exists an open ball $B(x,r)$ such that the sequence converges uniformly on $B(x,r) \cap E$. Uniform convergence on $E$ implies local convergence, but the converse may not be true.

For example, the well known series $e^x = \sum_{k=0}^\infty \frac{x^k}{k!}$ converges uniformly locally in $(0,\infty)$. For any $x \in (0,\infty)$ and $r > 0$, we have $\frac{|y|^k}{k!} < \frac{(x+r)^k}{k!}$ for all $y \in (x-r,x+r) \cap (0,\infty)$ and uniform convergence follows by the Weierstrass M-test. However, convergence is not uniform for all $x \in (0,\infty)$ since

$$\sup_{x \in (0,\infty)}\left|\sum_{k=n+1}^\infty \frac{x^k}{k!}\right| \geqslant\sup_{x \in (0,\infty)}\sum_{k=n+1}^{2n} \frac{x^k}{k!}\geqslant \sup_{x \in (0,\infty)}\frac{nx^n}{(2n)!} = +\infty $$