# On the relation between simple and uniform continuity

In the Real and Functional Analysis course, we were introduced to the concept of uniform continuity (UC) without much elaboration. I am currently exploring the nuances of uniform continuity compared to the conventional notion of continuity. Here are the definitions we have been provided with:

Def. (Continuity) $$f$$ is said to be continuous at $$x_0$$ if either:

• $$x_0$$ is an isolated point,
• or, $$x_0$$ is an accumulation point and $$\lim_{x\rightarrow x_0}f(x)=f(x_0)$$. Namely $$\forall \varepsilon>0 \ \exists\delta=\delta(\varepsilon) : 0 with $$(X,d_X), (Y,d_Y)$$ proper metric spaces.

Def. (UC) $$f$$ is said to be uniformly continuous in $$D\subset X$$ if: $$\forall \varepsilon>0 \ \exists\delta=\delta(\varepsilon) : \forall x,x'\in D \ \text{with} \ 0.

My understanding is that in the continuity definition $$\delta = \delta(\varepsilon, x_0)$$, while in UC $$\delta = \delta(\varepsilon)$$. In this view, my guess is that if from the continuity definition I get a $$\delta$$ depending on $$x_0$$, then the function cannot be uniformly continuous on $$D(\ni x_0)$$ unbounded. Is this correct?

(E.g. considering $$f(x)=x^2$$, which is not UC on $$\mathbb{R}^+$$, by considering the continuity definition we have $$\delta=\delta(\varepsilon, x_0)=\frac{1}{2}\min\bigg\{1,\frac{\varepsilon}{1+2|x_0|}\bigg\}$$.)

This reasoning contains a correct idea, but must be clarified. After all, one could certainly take a uniformly continuous function like the identity on the interval $$(0,1)$$ and could still inadvisedly choose $$\delta$$ to depend on $$x$$, e.g., $$\delta(\epsilon, x)=\min(\epsilon, x)$$.
To fail uniform continuity, it has to be the case that the definition of continuity always gives you a function depending on $$x$$, no matter how hard you try to eliminate that dependence. So looking at a particular $$\delta(x_0,\epsilon)$$ that you cooked up to satisfy regular continuity won't tell you that the function can't be uniformly continuous, since you may have simply picked a suboptimal $$\delta$$.
What you can say is, if $$\hat{\delta}(x_0,\epsilon)$$ is the (possibly infinite) pointwise supremum of all the functions $$\delta(x_0,\epsilon)$$ that fit the definition of continuity, then $$f$$ is uniformly continuous if and only if the function $$x_0\mapsto\hat{\delta}(x_0,\epsilon)$$ is bounded away from $$0$$ for every fixed $$\epsilon$$.