Using Darboux's Theorem to prove a derivative doesn't exist

Question

We have the following function:

\begin{equation*} g(x) = \begin{cases} x,&x<1\\ x-1, &x\geq 1. \end{cases} \end{equation*}

Prove that this function cannot be a derivative using Darboux's Theorem.

My confusion is this: Darboux's Theorem basically says that the derivative has to obey the intermediate value theorem, so if we have an input interval $ s < t $ in which $ f(s) < f(t) $, then $f$ has to take every value between $f(s)$ and $f(t)$ as well on this interval. I don't see how this function doesn't fulfill that? If you choose an interval in which the two pieces overlap, like $(0,2)$, then $f$ takes on some values twice, but it never "skips" a value.

What am I missing here? I'm pulling my hair out over this one, and the only other thread I've found on this question doesn't have the specific answer I'm looking for regarding the intermediate value theorem.

EDIT: To remove confusion, the definition of Darboux's I was given for this problem was this:

If $f $ is differentiable at every point of an open interval $I$, $s$ and $t$ are any numbers in $I$ such that $s < t$, and $f ' (s) < f' (t)$, then for every real number y such that $f' (s) < y < f' (t)$, there is a $c ∈ (s, t)$ such that $f ' (c) = y$.

So my problem is that according to the above defintion, if $f'(s) > f'(t)$, I can't even apply Darboux's at all.

My issue with the linked thread is that I don't understand the answer given based on "images", I'd like an answer to point out specifically where $g(x)$ betrays the Intermediate Value Theorem as that is the specific point I'm confused on.

G. Sassatelli - My confusion with your comment is if I choose an interval that's too small, then I can't even apply Darboux's because $f(t)$ isn't greater than $f(s)$
@DHMO I saw the old thread, but I made this one because your answer on it doesn't clear my confusion. I'm wondering specifically about how this function doesn't fulfill the IVT. The explanation using images - I don't know what that means, and even if I did I'd still want an answer that responds directly to the IVT. — Bookie, Apr 17 '17 at 09:12
@Bookie actually, now I see the point. I was actually under a false impression myself. — , Apr 17 '17 at 09:13
@Bookie If you are basing this thread on the old thread, then you should at least link to the old thread and elaborate on what you are not clear about... — DHMO, Apr 17 '17 at 09:18
@Bookie (To your new thread) why do you write $f(s) < f(t)$? When is this a part of the intermediate value theorem? — DHMO, Apr 17 '17 at 09:19
@G. Sassatelli Would you mind explaining fully? My issue is this: let's say we take the interval $(.5, 1.25)$. Pretty close. But then $f$ returns $(.5,.25) .5 > .25$ and so Darboux's can't even be applied, let alone used to disprove. — Bookie, Apr 17 '17 at 09:20
@Bookie Basically your problem is that you are under the false impression that IVP is "for each pair of points $s<t$ such that $f(s)<f(t)$ and for all $\chi\in(f(s),f(t))$ there is $\xi\in(s,t)$ such that $f(\xi)=\chi$". If this were the case, then all weakly increasing functions would have the IVP. The correct form would be "for each pair of points $s<t$ such that $f(s)<f(t)$ and for all $\chi\in(f(s),f(t))$ there is $\xi\in(s,t)$ such that $f(\xi)=\chi$ and for each pair of points $s<t$ such that $f(s)>f(t)$ and for all $\chi\in(f(t),f(s))$ there is $\xi\in(s,t)$ such that $f(\xi)=\chi$". — , Apr 17 '17 at 09:31
@ G. Sassatelli Comparing the definitions of Darboux's and the IVT given in class, I can see that. Just to make sure I'm on the same with you, what you're saying is that the definition I gave of Darboux's in the OP is a bit misleading, because $f(s)$ does not strictly have to be less than $f(t)$ because $s < t$? — Bookie, Apr 17 '17 at 09:36
@Bookie I think that's your problem here. In fact, here the counterexamples satisfy $s<t$ and $f(s)>f(t)$. — , Apr 17 '17 at 09:39
@ G. Sassatelli Alright thanks, I appreciate it. Sorry I can't accept you as an answerer due to the comment section thing, if you were to drop a quick official answer then I would certainly accept it (if that matters to you). Otherwise, I appreciate the help! — Bookie, Apr 17 '17 at 09:41

Paramanand Singh · Accepted Answer · 2017-04-17T11:41:15.577

The function $g$ you have given has a jump discontinuity at $x = 1$ because the left hand limit is $1$ and right hand limit is $0$. Now there is a result which says that derivatives can't have jump discontinuity and hence $g$ can't be the derivative of any function $f$. The linked answer shows two proofs one of which is based on Darboux theorem.

If you wish to solve the problem directly using Darboux theorem then let's assume that there is a function $f$ such that $f' = g$ so that $f'(1) = 0$. Now consider the function $g$ in $[c, 1]$ where $0 < c < 1$. Clearly $g$ has intermediate value property by Darboux theorem. Now $g(1) = 0$ and $g(c) = c$ and take $d$ such that $0 < d < c$. There is no number $x \in [c, 1]$ such that $f(x) = d$. So we get a contradiction.

More generally you can see that if $f'(a) = A$ and $\lim_{x \to a^{-}}f'(x) = B$ then $A = B$ because if $A \neq B$ we can choose midpoint $M = (A + B)/2$ such that all values of $f'$ in $(a - h, a)$ (for some $h > 0$) are between $B$ and $M$ and hence no values between $A$ and $M$ are taken by $f'$ in $(a - h, a]$ which contradicts Darboux theorem. Similarly if $\lim_{x \to a^{+}}f'(x) = C$ then $A = C$.

Using Darboux's Theorem to prove a derivative doesn't exist

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