Functional distance in a manifold In Riemannian Geometry of Peter Petersen, he gave a definition of functional distance in a manifold $M$ as following:
$$
d_F(p,q)=\sup\{|f(p)-f(q)|\ |f:M\to \mathbb{R} \text{ has } |\nabla f|\leq 1 \text{ on } M\}.
$$
He said that the distance is always smaller than the arclength distance. But I can't understand it.
 A: It is for the same reason as this would hold in $\mathbb{R}^n$. I also assume that your manifold is Riemannian due to the book you are referring to. The answer to your question follows from the estimate
$$
|f(p)-f(q)| \leq |\nabla f| |p-q| \leq 1 \cdot |p-q|
$$
for all $p,q \in M$.
It only remains to check why this estimate is actually true. For that, consider a minizing geodesic $\gamma:[0,t] \to M$ such that $\gamma(0)=p$ and $\gamma(t)=q$.
By Cauchy-Schwarz we have the estimate
$$
|g(\nabla f(\gamma(t)),\gamma'(t)) | \leq |\nabla f (\gamma(t))| \cdot |\gamma '(t)| \leq |\nabla f | |\gamma '(t)| 
$$
and we can start calculating:
$$
|f(p)-f(q)|=|\int_0^t \partial_t f(\gamma(t)) dt|= |\int_0^t g(\nabla f(\gamma(t)),\gamma'(t))| dt \leq   \\
\int_0^t |\nabla f| |\gamma'(t)| dt \leq |\nabla f| \int_0^t|\gamma'(t)|dt=|\nabla f| |x-y|.
$$
In the last equality, I have used that $\gamma$ is a geodesic. Furthermore, note that $|\nabla f|$ does not depend on the curve (or more explicitly on $t$), so we can just drag it out of the integral.
