Let $X,Y$ be Riemannian manifolds. Let $f:X \to Y$ be an everywhere differentiable map, such that the operator norm $\|df_p\|_{op}$ is globally bounded (from above) by $L$.
Is it true that $f$ is Lipchitz? (what about $L$-Lipschitzity?)
I am interested in the general question in the case of low regularity, i.e I do not assume $f$ is $C^1$ (then it's quite easy). In particular this means that for a smooth path $\alpha:I \to X$, we do not know a-priori that $f\circ \alpha$ is Lipschitz, so we cannot necessarily write its length as the integral of its speed.
Note that in the case where $X,Y$ are Euclidean spaces, the answer is positive and is known as the "mean value inequality". However, the standard proof does not seem to pass over to the general case.
Updated "conjecture":
I guess $f$ does have to be Lipschitz, and even $L$-Lipschitz.
First, since locally the Riemannian metric is "close" to Euclidean metric, I guess one could claim local-Lipschitzity certainly holds (admittedly, some details are probably required to justify this rigorously).
Then, for any smooth path $\alpha$ in $X$, $f \circ \alpha$ will also be locally-Lipchitz and (compactness?) even Lipchitz. So we can express its length as the integral of its speed, so $f$ will be $L$-Lipschitz by the usual argument.
Yes, this is indeed true. Proving that $f$ is $L$-Lipschitz is a local problem. Consider a path $p: [a,b]\to X$ satisfying $|p'(t)|=1$ and the composition $q=f\circ p$.
Set $y_0:= q(a)$ and consider also the distance function $d_{y_0}: Y\to {\mathbb R}$, given by $d(y_0, y)$. Since the problem is local, it suffices to consider the case when the image of $q$ is contained in a convex neighborhood $U$ of $y_0$; in this neighborhood the distance function $d_{y_0}$ is differentiable away from $y_0$; it is also 1-Lipschitz (by the triangle inequality). Hence, the norm of its gradient is $\le 1$ in $U$ (away from $y_0$ of course).
Now, consider the composition $h(t):= d_{y_0}\circ q(t)$. This function is differentiable on the interval $(a,b)$ and continuous on $[a,b]$; it also satisfies $|h'(t)|\le L$ on $(a,b)$ (the Chain rule). Hence, by the Mean Value Theorem, $$ d(q(a), q(b))= h(b)\le L|b-a|. $$ It follows that the map $f$ is $L$-Lipschitz.