Let $f:[0,1]\rightarrow\mathbb{R}$ be a function continuous on $[0,1]$ and differentiable on $(0,1)$ with $f(0)=f(1)$ and $f(\alpha)=f(\beta)+1$ for some $\alpha,\beta$ such that $0<\alpha<\beta<1$. Prove that there exists some $\xi\in(0,1)$ such that $\lvert f^{\prime}(\xi)\rvert=2$.
I've come up with some approaches but none give me a complete solution:
If $\beta-\alpha\leq\frac{1}{2}$, then by the mean value theorem, there exists a point with a derivative less than or equal to $-2$, and since $f(0)=f(1)$ by Rolle we also have a point with derivative $0$ so then by Darboux for any number $d$ in $[-2,0]$ there exists a point at which the derivative is exactly equal to $d$. But I don't know how to deal with the case $\beta-\alpha>\frac{1}{2}$. But also any other approach to finding a $c\in (0,1)$ for which $\lvert f^{\prime}(c)\rvert\geq 2$ would be sufficient for Darboux to finish the job...
Let $g(x)=f(x)-2x$ be a function also defined on the segment $[0,1]$ then it is also continuous so by Weierstrass' extremum theorem it must reach its maximum value on its domain. And since $0=g(0)>g(1)=-2$, the maximum isn't achieved at $1$, so if I could also prove that it doesn't achive the max value at $0$, then it would be that a maximum is achieved for some $x$ in $(0,1)$, and from there by Fermat's theorem the derivative of $g$ should vanish, giving the desired claim. But again, I’m stuck, since I don’t know how to prove that the max isn't attained at $0$, or equivalently that there exists $s\in(0,1)$ such that $g(s)>g(0)=0$. I have been able to do the following: since $g(\alpha)-g(\beta)=1+2(\beta-\alpha)$ it must be that
(a) $g(\alpha)\geq\frac{1}{2}+\beta-\alpha$ but then the right-hand side is strictly greater than zero which is sufficient for the claimed result; or
(b) $-g(\beta)\geq\frac{1}{2}+\beta-\alpha$ with which I, once again, have no idea what to do with...
I guess I could have taken $g(x)=f(x)+2x$ to get $f^{\prime}(\xi)=-2$, since a point $\xi$ with slope $2$ will also give a point with slope $-2$, because the function $f$ cannot be monotone because of $f(0)=f(1)$. Right?
I am sorry for such an elementary query, but I've tried this a couple of times and it just won't budge; I’m starting to develop serious self-esteem issues from this. I'm sure there's an elegant and easy solution and I just wish to see it so I can then deal with the frustration of not having come up with it myself. My deepest gratitudes to anyone willing to humour me with this problem.
You were on the right track with your first observation. Let’s write the important bit as a lemma.
Lemma:
For any two points $x,y\in[0,1]$ with $x<y$, $$\left|f(y)-f(x)\right|<2(y-x).$$
Proof:
Suppose otherwise. By the MVT, there’s a point of $f$ with a derivative whose magnitude is greater than $2$. However, there’s also a point with a derivative equal to $0$, by Rolle. Therefore, by Darboux, there’s a point with a derivative of magnitude of exactly $2$. $\square$
For the rest of the proof, we proceed by contradiction, and we consider two cases: when $f(\alpha)\geq f(0)$, and when $f(\alpha)<f(0)$. You’ve already established the case when $\beta-\alpha\leq\frac12$, so we may furthermore suppose $$\beta-\alpha>\frac12.$$ In particular, $\alpha<\frac12$, $\beta>\frac12$.
$\mathbf{f(\alpha)\geq f(0)}$:
By our lemma, $f(\alpha)<f(0)+2\alpha$. Therefore, $$f(\beta)<f(0)+2\alpha-1<f(0).$$ Using $f(0)=f(1)$ and our lemma, this implies that $$2(1-\beta)>\left|f(1)-f(\beta)\right|>1-2\alpha\Rightarrow$$ $$\beta-\alpha<\frac12,$$a contradiction.
$\mathbf{f(\alpha)< f(0)}$:
We have that $f(\beta)=f(\alpha)-1<f(0)-1$. However, again by our lemma, $$f(\beta)>f(1)-2(1-\beta).$$ Therefore, using $f(0)=f(1)$, $$f(0)-1>f(\beta)>f(0)+2\beta-2\Rightarrow$$ $$\beta<\frac12,$$a contradiction.
We conclude what we wished to prove. $\blacksquare$