If a function $f$ of real numbers is globally continuous on its domain, then ... If a function $f$ of real numbers is globally continuous on its domain, then is it also true that for every $\delta > 0$ and every $c$ in the domain of $f$, there is an $\epsilon > 0$ such that $|f(x) - f(c)| < \epsilon$ for all $x$ in the domain of $f$ such that $|x-c| < \delta$?
I raise this question because of the typical pattern of an $\epsilon - \delta$ proof. For example, in proving $x^{2}$ is continuous on $\mathbb{R}$ the choice $\delta = \epsilon/3|c|$ for $c \neq 0$ and $|x| < 2|c|$ and the choice $\delta = \sqrt{\epsilon}$ for $c=0$ suffice. But since these two representations can also be written as $\epsilon = 3|c|\delta$ and $\epsilon = \delta^{2}$, we conclude that given any $\delta > 0$ there is an $\epsilon > 0$ for which the required condition holds.
 A: If $c$ and $\delta$ are such that $[c-\delta,c+\delta]$ is in the domain of $f$ then you have $f(x)-f(c)$ is continuous on the compact set $[c-\delta,c+\delta]$ and so admits a bound $\epsilon>0$ (that depends on $c$ and $\delta$). In particular, $f(x)-f(c)$ is bounded on $(c-\delta,c+\delta)$ by the same $\epsilon$.
On the other hand, if $[c-\delta,c+\delta]$ isn't in the domain of $f$, then things can get problematic. Take $f:\mathbb{R}_{++}\to\mathbb{R}_{++}$ defined as $f(x)=\frac{1}{x}$. Pick any $c>0$ in the domain of $f$ and let $\delta=c$. Then, for $x\in(c-\delta,c+\delta)=(0,2c)\subset\mathbb{R}_{++}$, it's clear that $|f(x)-f(c)|$ can get arbitrarily large.
The second paragraph above implies the proposed property in this question doesn't work in general.
A: A function satisfies your property if each value of $\delta$ is small enough for $\varepsilon$-fine control at every point, provided $\varepsilon$ is large enough. Uniform continuity implies this, in that you can pick a modulus of continuity $\delta(\varepsilon)$ which works at every point, modify it (by decreasing it) to be an increasing function, and then invert it.
Your property is weaker than uniform continuity, as your example of $f(x)=x^2$ on the whole line shows. However, it is stronger than continuity, because if $\delta=\delta(x,\varepsilon)$ is the modulus of continuity of $f$ and $\lim_{x \to c} \delta(x,\varepsilon)=0$ for some $c,\varepsilon$ (where $c$ is in the closure of the domain), then your property will not hold. In particular, a continuous function which is not continuous on the closure of its domain will not have your property, in that given each $\delta$, you can pick a point which is within $\delta$ of the boundary.
I think the universal phenomenon that you are seeing as follows. If $\delta(x,\varepsilon)$ is the modulus of continuity of $f$, then you can modify $\delta$ (by decreasing it) to be an increasing function of $\varepsilon$, and then invert it to get $\varepsilon(x,\delta)$. This function tells you how small $|f(x)-f(c)|$ will be if $|x-c|<\delta$. Most of the time, we're going the other way: we find a $\varepsilon(x,\delta)$, modify it (by increasing it) to be an increasing function of $\delta$, and then invert it.
