Question about baby rudin theorem 5.12 corollary

Question

The corollary says if $ f $ is differentiable on $ [a,b] $ then $ f' $ cannot have any simple discontinuities on $[a,b] $.

I just don't how to prove it.

I think it should be proved on both two cases of simple discontinuities(first type and second type of simple discontinuities).

Thanks in advance.

What is the definition of a $\textbf{simple discontinuity}$? — Faraad Armwood, Jun 30 '17 at 11:22
Does theorem 5.12 say that derivatives have the intermediate value property? — Daniel Fischer, Jun 30 '17 at 11:36
Definition 4.26 in baby rudin. Let $f$ be defined on $(a, b)$. If $f$ is discontinuous at a point $x$, and if $f(x+)$ and $f(x-)$ exist, then $f$ is said to have a discontinuity of the first kind, or a simple discontinuity, at $x$.@FaraadArmwood — Johnny Ji, Jun 30 '17 at 12:28
And the simple discontinuity includes Removable discontinuity and Jump discontinuity.@FaraadArmwood — Johnny Ji, Jun 30 '17 at 12:30
Yes, the theorem 5.12 in baby rudin is that derivatives have the intermediate value property @DanielFischer — Johnny Ji, Jun 30 '17 at 12:33
So try to prove that a function with a removable or jump discontinuity doesn't have the intermediate value property. — Daniel Fischer, Jun 30 '17 at 12:43

score 1 · Answer 1 · answered Jun 30 '17 at 18:07

Let $g \colon (a,b) \to \mathbb{R}$ a function. If $g$ has a simple discontinuity at $c \in (a,b)$, then $g$ doesn't have the intermediate value property.

Let's look at the case of a jump discontinuity. Replacing $g$ with $-g$ if necessary, we can assume that

$$L := g(c^-) < R := g(c^+).$$

Let $\varepsilon = \frac{R-L}{3}$. By definition of the one-sided limits, there are $\delta^-, \delta^+ > 0$ such that

\begin{align} c - \delta^- < x < c &\implies \lvert g(x) - L\rvert < \varepsilon\qquad\text{and} \\ c < x < c + \delta^+ &\implies \lvert g(x) - R\rvert < \varepsilon. \end{align}

With $\delta = \min \: \{\delta^-,\delta^+\}$, we thus have

$$g(x) \in (L-\varepsilon, L+ \varepsilon) \cup (R-\varepsilon, R+\varepsilon)$$

for $0 < \lvert x-c\rvert < \delta$. By the choice of $\varepsilon$, we have $L + \varepsilon < R - \varepsilon$, hence there is

$$v \in [L+\varepsilon,R-\varepsilon] \setminus \{g(c)\},$$

and there is no $x \in [c-\delta/2, c+\delta/2]$ with $g(x) = v$, but of course

$$g(c-\delta/2) < L+\varepsilon \leqslant v \leqslant R-\varepsilon < g(c + \delta/2).$$

Thus $g$ doesn't have the intermediate value property.

The argument for a removable discontinuity - that means $g(c^-) = g(c^+) \neq g(c)$ - is proved quite similarly.

Since derivatives have the intermediate value property (theorem 5.12), it follows that derivatives cannot have simple discontinuities.

this answer settles this question. – Bijesh K.S Jan 06 '18 at 17:24 — Bijesh K.S, Jan 06 '18 at 17:24

Question about baby rudin theorem 5.12 corollary

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