I need to show that if $f:\mathbb{R} \rightarrow \mathbb{R}$ is differentiable on $[a,b]$ and $'f$ is monotone on that interval, then $f'$ must be continuous.

Is this proof correct?

I will use the following theorems from Baby Rudin (restated quickly):

4.29: If $f$ is monotonically increasing on $(a,b)$, then f(x+) and f(x-) (the right and left handed limits, respectively) exist at every point of $(a,b)$, and $f(x-) \leq f(x) \leq f(x+)$.

5.12 Suppose f is a real differentiable function on $[a,b]$, and $f'(a) \leq \lambda \leq f'(b)$. Then $\exists~ x \in (a,b)$ for which $f'(x) = \lambda$.


Assume without loss of generality that $f'$ is monotone increasing in particular. Because $f'$ is monotone, we know that $f(x+)$ and $f(x-)$ exist everywhere on the interval.

Suppose then that $f'$ has a simple discontinuity at $x_0$. Let $f(x-) = C,~f(x+) = D$, and note that $C \leq f(x_0) \leq D$.

Because $f'$ is discontinuous by assumption, there exists $\epsilon >0$ such that for all $\delta>0, ~ |x-x_0|< \delta \Rightarrow |f'(x)-f'(x_0)|>\epsilon$.

Choose $\epsilon < \min(|f'(x_0)-C|, |f'(x_0) -D|)$. Then for any $\lambda$ s.t. $|f'(x_0)-\lambda|<\epsilon, ~f'(x) \neq \lambda$ for any $x\in(a,b)$.

But by theorem 5.12, the existence of $f'$ implies that such an $x$ must exist. This is a contradiction. Thus $f'$ must be continuous on $[a,b]$.

Strictly speaking, I think there is some issue surrounding the possible existence of discontinuities of the second kind; I think the fact that the derivative is defined on the compact set disallows that, right?

  1. Start your proof with "Assume $f'$ is discontinuous" and "without loss of generality assume that $f$ is monotone increasing".
  2. Change "st" to "such that" or at least "s.t.".
  3. Your sentence that starts with "Choose" could be filled out a little. But mainly you need $f'(x_0)$ where you have written $f(x_0)$ and in the previous line that begins with "Because".
  • $\begingroup$ fixed! All else good? $\endgroup$ – BenL Apr 4 '17 at 15:49
  • $\begingroup$ Yes, I have to teach a class right now, but I will get back to you later when I can about filling out that line involving $\lambda$. $\endgroup$ – Paul Sundheim Apr 4 '17 at 15:54
  • $\begingroup$ The second line of the proof, $x-$ needs to be $x_0 -$ and similarly with $x+$. By the way this is often written $x_0 ^-$ and $x_0 ^+$. $\endgroup$ – Paul Sundheim Apr 4 '17 at 17:08
  • $\begingroup$ And now that I have more time to think about it, the proof looks fine to me. $\endgroup$ – Paul Sundheim Apr 4 '17 at 17:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.