Derivatives, the Intermediate Value Property, and Jump Discontinuities

Question

I am trying to understand the connection between derivatives, the intermediate value property (IVP) and jump discontinuities.

In class we proved that if $f$ is a differentiable function on the interval $[a, b]$ such that $f'(a)f'(b) < 0$, then there exists a point $c \in (a, b)$ such that $f'(c) = 0$. we used this to prove darboux's theorem - under the same hypotheses as before, if i pick $\alpha$ between $f'(a)$ and $f'(b)$ then there exists $c \in (a, b)$ such that $f'(c) = \alpha$. this is the intermediate value property for derivatives.

Based on this result, it should follow that derivatives cannot have jump discontinuties. however, the IVP is not enough to claim a function does not have jump discontinuities - see: \begin{cases} 2 & 0 \leq x\leq 2 \\ -x+4 & 2 < x < 3 \\ 2 & x \geq 3 \end{cases} which satisfies IVP in $[0, 4]$ but has a jump discontinuity at $x = 3$. (Note: as was pointed out in the comments this might've been a bit ambiguous - here I specifically refer to the function $f(x)$ and not to its derivative)

my question is:

• why does the IVP imply no jump discontinuities specifically for derivatives (and not in general, as in: why can't i make such a counterexample for derivatives)?

The only answer i could come up with is that in the general case, a function may satisfy the IVP in a "large enough" interval and not in a "smaller" interval - in the example i've made, the IVP is satisfied in $[0, 4]$ but not in $[2.5, 4]$ for instance. but if i'm working with a function which is differentiable on $[a, b]$, then it should also be differentiable on $[c, d] \subseteq [a, b]$, thus it should satisfy the IVP on any subinterval. i'm not sure this explanation satisfies me, though.

if my answer to the above question is correct, then the follow-up question is: are there any mistakes in this proof that $f$ satisfies the IVP in any subinterval $\implies$ $f$ has no jump discontinuities?

Suppose that $f$ has a jump discontinuity at $x_0$. Let $\lim_{x \rightarrow x_0^-} f(x) = L^-$ and $\lim_{x \rightarrow x_0^+} f(x) = L^+$ and suppose $L^- < L^+$. Then, for any $\epsilon_1$, $\epsilon_2$ there exists a left (right) neighborhood of $x_0$ where $|f(x) - L^-| < \epsilon_1$ ($|f(x) - L^+| < \epsilon_2$). Let $\alpha \in (L^-, L^+)\setminus{f(x_0)}$, and set $\epsilon_1 = \alpha - L^-$, $\epsilon_2 = L^+ - \alpha$. Then for all $x \in (x_0 - \delta_1, x_0)$ and $x \in (x_0, x_0 + \delta_2)$ we have $2L^- - \alpha < f(x) < \alpha$ and $\alpha < f(x) < 2L^+ + \alpha$. Setting $\delta = \min(\delta_1, \delta_2)$ yields $f(x) \neq \alpha$ for all $x \in (x_0 - \delta, x_0 + \delta)$, $x \neq x_0$. Since $f(x_0) \neq \alpha$, we have a contradiction.

thanks for your time!! i'm sure i have either been long winded or ambiguous in some of this question so i will apologise in advance. i am also aware the proof itself is very badly written, i'd appreciate any tips on that...

Answer 1

If $f$ is everywhere differentiable on the interval $(a,b)$, then any discontinuities of $f'$ must be of the oscillatory type. That is, if the right limit $f'(x+):=\lim_{t\to x, t>x}f'(t)$ and the left limit $f'(x-):=\lim_{t\to x, t<x}f'(t)$ both exist for a certain $x\in (a,b)$, then $f'(x+)=f'(x-)=f'(x)$. For example, if $f'(x+)<f'(x-)$, choose $\epsilon=[f'(x-)-f'(x+)]/3>0$. Then there is a $\delta>0$ so small that if $x-\delta<t<x$ then $|f'(t)-f'(x-)|<\epsilon$; and also if $x<t<x+\delta$ then $|f'(t)-f'(x+)|<\epsilon$. Thus, no real number in the interval $(x-\delta,x+\delta)$ (with the possible exception of $x$) takes values in $(f(x+)+\epsilon,f(x-)-\epsilon)$, in violation of the Intermediate Value Property of $f'$. The other ways in which $f'(x+), f'(x-), f'(x)$ could differ are handled similarly.