Mean value theorem for the circle. Consider the circle $S^1$ as $\mathbb R/\mathbb Z$, and endow it with the metric $d(x,y)=\min\{|x-y|,1-|x-y|\}$. Do we have some analogue of the MVT on $(S^1,d)$? We can think of a function $f:S^1\to S^1$ as a periodic function on the line, and so can consider $f$ to be differentiable if it is differentiable as a function $\mathbb R\to \mathbb R$ (note this means that $f'$ is not a selfmap of the circle anymore). So if we take such a differentiable $f$, the usual MVT then gives, for any $x,y\in S^1$ with $x<y$ on $[0,1)$, that there exists a $c\in(x,y)$ such that
$$|f'(c)||x-y|=|f(x)-f(y)|.$$
I would like to be able to generalize this to there existing a $c\in (x,y)$ such that $|f'(c)|d(x,y)=d(f(x),f(y))$, but we cannot do this straight away, because $d(x,y)$ may equal $|x-y|$, but $d(f(x),f(y))$ equal $1-|f(x)-f(y)|$. As the expanding circle map $E_m(x)=mx\mod 1$ ($m>1$) shows, this is impossible for all $x,y\in S^1$, but intuitively I feel that it should be possible to have such an equality for $d(x,y)$ small enough. In the case of the circle map choosing $x,y$ such that $d(x,y)\leq 1/(2m)$ suffices to get the desired equality. 
For simplicity let us consider the case where $f'$ is strictly positive, and greater than $1$, so we mimic the case of the expanding circle map which does have the desired equality. Maybe a proof for the general case can be achieved via appropriate shifts of problematic points, or by broadening the concept of interval to "wrap-around" intervals. 
So far I have been unable to come up with a rigorous argument, and unfortunately have deadlines for assignments so cannot spend much more time on this at the moment. So does anyone know whether there is the MVT I'm thinking of on $(S^1,d)$, or at least a close analogue of it?
 A: Here is a general statement. Suppose $X_1$ and $X_2$ are metric spaces, and $f:X_1\to X_2$ is a continuous map. Suppose further that 


*

*each $X_j$ contains a subset $\Gamma_j$ that is isometric to an interval, meaning that there exists a distance-preserving map $\gamma_j:[a_j,b_j]\to \Gamma_j$.

*$f(\Gamma_1)\subset \Gamma_2$

*The composition $g:=\gamma_2^{-1}\circ f\circ \gamma_1$, which is a map from $[a_1,b_1]$ into $[a_2, b_2]$, is differentiable on $(a_1, b_1)$.


Define $f' :=  g'\circ \gamma_1^{-1}$ on $\Gamma_1$; this is consistent with your definition of $f'$ up to sign. 
Claim: For any points $x,y\in \Gamma_1$ there exists $z$ between $x,y$ on the curve $\Gamma_1$ such that 
$$d_{X_2}(f(x), f(y)) = |f'(z)| d_{X_1}(x, y) \tag1$$
In the special case of $f$ being $m$ times winding map, any arc of length at most $1/(2m)$ will  work as $\Gamma_1$, since its image is isometric to an interval. Longer arcs do not work. 
Unfortunately, there is nothing new or really interesting in the proof of (1): it amounts to applying the usual MVT to the function $g$.
Fortunately, this illustrates a useful fact of Riemannian geometry: when we  do calculus on manifolds, restricting attention to (globally minimizing) geodesics makes the computation look Euclidean. 
