Bi-Lipschitz constant of arc-length parametrisation of convex curve Assume that $f:[0,2\pi]\to [0,2\pi]$ is a an increasing diffeomorphism, and let $\underline f = \min f'$ and $\overline f = \max f'$ and define $$g(s) = \int_0^s e^{if(t)} dt.$$ Assume that $g(0)=g(2\pi)$. It seems that $$\underline f \le \frac{|g(x)-g(y)|}{|e^{ix}-e^{iy}|}\le \overline f,$$ but I do not have the proof jet. 
 A: Take $f(x)=k\log(x+1)$ where $k=2\pi/\log(2\pi+1)$, hence $f$ is a diffeomorphism. The maximum of the derivate is atained when $x=0$ and it is $k\approx 3.1644$. On the other hand, taking $x=\pi/3$, $y=\pi/3-1$ we obtain
$$ \frac{|g(x)-g(y)|}{|e^{\mathrm{i}x}-e^{\mathrm{i}y}|}> \frac{2}{0.2}=10.$$
Edit: As Chistian noted in a comment, with the function $f$ defined above the function $g$ don't satisfy $g(0)=g(2\pi)$.
Now, we can choose $f(x)=-x+2\pi$. For $g(2\pi)$ we get
$$g(2\pi)=\int_0^{2\pi}e^{-\mathrm{i}x+2\pi\mathrm{i}}=\int_0^{2\pi}e^{-\mathrm{i}x}=0.$$
The derivate is $-1$, and
$$ \frac{|g(x)-g(y)|}{|e^{\mathrm{i}x}-e^{\mathrm{i}y}|}=1.$$
This is not a surprise if we think the above quotient as 
$$ \left|\frac{g(x)-g(y)}{x-y}\frac{x-y}{e^{\mathrm{i}x}-e^{\mathrm{i}y}}\right|.$$
If the estimate is true, then $\underline{f}\leq|g^\prime(x)|\leq \overline{f}$ for every $x\in(0,2\pi)$. But $|g^\prime(x)|=|f^\prime(x)|$. I will think about the case where $\underline{f}=\min |f^\prime|$ and $\overline{f}=\max |f^\prime|$.
A: This is not really a complete solution, it contains possibly challenging exercises to the reader, but I have some hope that one could make a full proof out of this.
I'll also only address the lower bound.
The function $g(t), 0\le t\le 2\pi$, defines a closed curve in the plane. It is parametrized with respect to arc length, and its curvature is given by $f'(t)$. By assumption, $a\le f'\le b$ (so the curve is also convex).
For the lower bound, we are interested in how small $|g(x)-g(0)|$ can become under these restrictions: curvature $\ge a$, the piece in question has length $x$, and the total length equals $2\pi$.
It seems intuitively plausible that the minimum is achieved for an American football type curve that we obtain by gluing together two circular arcs of radius $R=1/a$ and length $\pi$ each. (I don't know how to prove it formally though.) This curve is not smooth at the gluing points, but of course we can approximate by smooth curves, so this is not an issue.
The minimal distance $|g(x)-g(0)|$ is achieved by considering distances parallel to the small axis of symmetry of our football. Notice also that if we brought these two points $g(0),g(x)$ closer together still by moving (parts of) our circular arcs around, then it would not be possible to close the curve with another arc of length $2\pi -x$ and at the same time keep the curvature $\ge a$ throughout. (We of course can't prove the claim without making use of this non-local feature of the problem.)
If all this is swallowed, then the rest follows by direct computation. By elementary geometry,
$$
G(x):=|g(x)-g(0)| = 2R \left( \cos\frac{\pi -x}{2R} - \cos \frac{\pi}{2R} \right) .
$$
We want to show that this is $\ge$
$$
H(x):=a|e^{ix}-1| = \frac{2}{R}\, \sin (x/2) .
$$
For $x=\pi$, this boils down to showing that
$$
2R^2 \sin^2 \frac{\pi}{4R} \ge 1 ,
$$
or, equivalently, $\sin\delta\ge (\sqrt{8}/\pi)\delta$ for $0<\delta\le \pi/4$, which (just about) works. The upper bound on $\delta$ came from the fact that $R=1/a\ge 1$.
I have shown that $H(\pi)\ge G(\pi)$, and to obtain that $H(x)\ge G(x)$ also for general $0<x<\pi$, I can compare the derivatives: it's not hard to check that $H'\le G'$ in this range for arbitrary $R\ge 1$.
