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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.

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Asaf Shachar
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    I don't see how, even in the Euclidean case, you are concluding that $f$ is Lipschitz. It works on sets of bounded diameter, but then the Lipschitz constant tends toward $\infty$ as the diameter gets large. I would expect that your result is true if $X$ is compact, and that the Lipschitz constant depends on the diameter of $X$. – treble Dec 13 '16 at 02:50
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    But yes, if you can do it locally then a partition of unity argument together with the Lebesgue number lemma will give you a global result. – treble Dec 13 '16 at 02:52
  • @treble Yes, you are probably right. I was implicitly thinking on convex domains. Since Riemannian manifolds are length spaces this should pose no problem in the general case. – Asaf Shachar Dec 13 '16 at 09:25

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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.

Moishe Kohan
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  • Thanks! Two points: (1) Do you know a good reference on the differentiability of the distance function $d_p$? (I have searched quite hard, and found several things, but no that proof $d_p$ is differentiable at the points $q$ where there is a unique minimizing geodesic from $p$ to $q$. It is easy to see that if $d_p$ is differentiable at $q$, then there is at most one minimizing geodesic connecting them but the other direction seems less trivial for me). (2) I am still wondering if the way I suggested (in the question) works, – Asaf Shachar Dec 20 '16 at 16:12
  • namely it is supposed to be easy to show $f$ is Lipschitz locally (since everything is "close" to be Euclidean) and hence we can use the expression for the length as an integral, from where $L$-Lipschitzity follows immediately. (I somehow feel this is more elementary since we do not need to use any known facts about differentiability of $d_p$). – Asaf Shachar Dec 20 '16 at 16:12
  • Why is the problem local? The points are are trying to compare $f$ at could be far apart, no? – Ryan Unger Jan 17 '17 at 00:05
  • @0celo7 think about the effect of f on the length of a curve connecting your points. – Moishe Kohan Jan 19 '17 at 04:03