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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 $f'(x) = \lambda$.

Proof:

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?

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  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".
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  • $\begingroup$ 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

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