The definition of uniform continuity of a real-valued function states:

A function $f\colon A\mapsto\mathbb{R}$ is uniformly continuous on $A$ iff for every $\varepsilon \gt 0$ there exists a $\delta \gt 0$ such that for every $x$ and $y$ in $A$, whenever $y \in \left(x-\delta,x+\delta\right)$, it is the case that $f\left(y\right) \in \left(f\left(x\right)-\varepsilon,f\left(x\right)+\varepsilon\right)$.

Basically how my book distinguishes this from point-wise continuity is that there exists a single $\delta$ that works for every point in the domain, so once we find that $\delta$, we know it works everywhere. On the other hand, point-wise continuity says that given a $c\in A$, there exists a $\delta$ such that the function is continuous at $c$, but all these $\delta$s may be different, perhaps depending on $c$, and we might not be able to find just one $\delta$ that works for all $c$s everywhere.

I interpret this definition a completely different way, and I want to see if my conjecture is correct. I think that functions which have bounded derivatives are uniformly continuous. That is, if the "steepness" and "shallowness" of a function is limited to a certain minimum and maximum, then the there's a sufficiently small enough $\delta$ that we can use, particularly at the steepest part of the function (say at $x_0$), such that the output stays within the $\varepsilon$-neighborhood of $f\left(x_0\right)$. Is that correct? How can I prove/disprove it?

The converse states that the derivative of a uniformly continuous function is bounded. Is that also true?

  • 3
    $\begingroup$ Your first conjecture is true (to prove it think of the mean value theorem). For the second think of $f(x)=\sqrt{x}$. $\endgroup$
    – Jose27
    Jan 31, 2013 at 6:21
  • $\begingroup$ Note that continuous does not imply differentiable. $\endgroup$
    – Calvin Lin
    Jan 31, 2013 at 6:21
  • $\begingroup$ @CalvinLin uniformly continuous does not imply differentiable? or just point-wise continuous does not imply differentiable? $\endgroup$
    – chharvey
    Jan 31, 2013 at 6:42
  • $\begingroup$ @TestSubject528491: Neither does: Look up Weierstrass' nowhere differentiable function. With this in mind, my previous comment assumed that you were working with an a-priori differentiable function. $\endgroup$
    – Jose27
    Jan 31, 2013 at 6:44
  • $\begingroup$ @Jose27, so what makes $\sqrt{x}$ different from $x^2$? Why is the former uniform while the latter is not? Is it because the derivative of $\sqrt{x}$ is decreasing while the derivative of $x^2$ is increasing? $\endgroup$
    – chharvey
    Feb 1, 2013 at 14:48

1 Answer 1


As Jose27 noted, uniformly continuous functions need not be differentiable even at a single point.

It is true that if $f$ is defined on an interval in $\mathbb R$ and is everywhere differentiable with bounded derivative, then $f$ is uniformly continuous. In fact, it follows from the Mean Value Theorem that such an $f$ is Lipschitz, which is much stronger.

However, if $f$ is uniformly continuous and everywhere differentiable, then $f$ need not have bounded derivative. Jose27 mentions $\sqrt x$, which would work as an example on the interval $(0,\infty)$. The function $$f\left(x\right)=\begin{cases}x^2\sin\left(\frac{1}{x^2}\right) & :x\neq 0\\ 0 &:x=0\end{cases}$$ is uniformly continuous on any bounded interval such as $(-1,1)$, but has unbounded derivative near $0$. There are also examples where $f'$ is bounded on bounded intervals, but unbounded on $\mathbb R$, while $f$ is uniformly continuous. You can show that any continuous function $f$ on $\mathbb R$ such that $\lim\limits_{|x|\to \infty}f(x)=0$ is uniformly continuous, and using this fact you can see that Nate Eldredge's example here of $\sin(x^4)/(1+x^2)$ provides such an example. Another source of examples is the question

Why if $f'$ is unbounded, then $f$ isn't uniformly continuous?

  • $\begingroup$ great answer; just one question, did you mean $\sin\left(\frac{x^4}{1+x^2}\right)$ or did you mean $\frac{\sin\left(x^4\right)}{1+x^2}$? $\endgroup$
    – chharvey
    Feb 1, 2013 at 15:09
  • $\begingroup$ $\dfrac{\sin(x^4)}{1+x^2}$. $\endgroup$ Feb 1, 2013 at 15:34

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