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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