A function with a point of slope of exactly $2$ 
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:


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*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.
 A: 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. $\blacksquare$
