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 $$\forall \varepsilon>0 \ \exists\delta=\delta(\varepsilon) : 0<d_X(x_0,x)<\delta \Rightarrow d_Y(f(x_0),f(x))<\varepsilon$$ with $(X,d_X), (Y,d_Y)$ proper metric spaces.

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

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 $\delta=\delta(\varepsilon, x_0)=\frac{1}{2}\min\bigg\{1,\frac{\varepsilon}{1+2|x_0|}\bigg\}$.)


1 Answer 1


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 $\epsilon$.


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