I understand the formal definition of uniform continuity of a function, and how it is different from standard continuity.

My question is: Is there an intuitive way to classify a function on $\mathbb{R}$ as being uniformly continuous, just like there is for regular continuity? I mean, that for a "nice" function $f:\mathbb{R} \to \mathbb{R}$, it is usually easy to tell if it is continuous on an interval by looking at or thinking of the graph of the function on the interval, or on all of $\mathbb{R}$. Can I do the same for uniform continuity? Can I tell that a function is uniformly continuous just by what it looks like? Ideally, this intuition would fit with the Heine-Cantor theorem for compact sets on $\mathbb{R}$.


3 Answers 3


I like to think of the following fact: a $C^1$ function on $\mathbb{R}$ with bounded derivative is uniformly continuous. So in order for a function not to be uniformly continuous, there have to be places where its graph is "arbitrarily steep".

Because of the theorem that any continuous function on a compact set is uniformly continuous, this can only happen if you make the function steeper and steeper as you go off to $\infty$, either unboundedly (like $x^2$) or boundedly (like $\sin(x^2)$).

Edit: As pointed out, the converse is false: a function with unbounded derivative can still be uniformly continuous. $\sin(x^4)/(1+x^2)$ is such an example, I believe. So this may not be great intuition after all.

  • $\begingroup$ Beat me to it by 27 seconds :) $\endgroup$
    – Alex B.
    Nov 23, 2010 at 17:15
  • $\begingroup$ A good guess would be that there is a theorem like this: "Let $X$ be a 'sufficiently nice' subset of $\mathbb{R}$ and let $f \in C(X)$. Then $f$ is uniformly continuous on $X$ if and only if for all $\varepsilon > 0$ there exists a $g \in C^1(X)$ such that $g' \in C_b(X)$ and $\lVert f-g \rVert_\infty < \varepsilon$." $\endgroup$
    – kahen
    Nov 23, 2010 at 18:21
  • $\begingroup$ Isn't $f(x)= \sqrt x$ a uniformly continuos $C^{1}$ function on $(0,1)$ with unbounded derivative ? $\endgroup$
    – Mykie
    Nov 23, 2010 at 18:50
  • 5
    $\begingroup$ It is actually $\textbf{wrong}$ to say function is uniformly continuous iff its derivative is bounded. Only one way conclusion is possible. If the derivative is bounded then the function is uniformly continuous but not the other way around. As Myke pointed out there are other counter examples as well. $\endgroup$
    – user17762
    Nov 23, 2010 at 19:00
  • $\begingroup$ @Sivaram, Myke: Good point. I corrected this. $\endgroup$ Nov 23, 2010 at 21:04

This is slightly tricky because your intuitive picture for "continuous" probably more closely matches "uniformly continuous." That is, if you can actually draw a graph without lifting your pencil then it's uniformly continuous. In order to get something continuous but not uniformly continuous you have to do something that you can't actually draw like going off to infinity on an open interval or oscillating really wildly (as explained in the other two answers).


One possible intuition for uniform continuity is that a function shouldn't oscillate arbitrarily wildly. If the function is differentiable e.g. on an open interval, then the derivative should be bounded in absolute value. That's something that you can see by looking at the graph. For example if you consider the function $sin(1/x)$ on the open interval $(0,1)$, then just by looking at it, you will realise that it's not uniformly continuous, since it's slope becomes arbitrarily steep as you approach $0$ from the right.

The way this intuition ties in with the Heine-Cantor theorem is that on compact sets, bounded functions attain their extrema. If your function is differentiable, then this is in particular true for its derivative, so the slope of the function cannot get arbitrarily steep on a compact interval.

  • $\begingroup$ rtel Thanks. I think you mean that the "Extreme Value Theorem" applies to the derivative if the function is continuously differentiable. $\endgroup$
    – jake
    Nov 23, 2010 at 17:49
  • $\begingroup$ That's exactly what I mean. Thanks. $\endgroup$
    – Alex B.
    Nov 23, 2010 at 17:58

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