Take a $\mathcal C^2$ potential $V:\mathbb R^d\to \mathbb R$, and assume that it is bounded from below (say $\min V=0$ for simplicity, so that $V\geq 0$).
Consider the autonomous gradient-flow
$$
\dot X\_t=-\nabla V(X\_t)
$$
and let $\Phi(t,X\_0)$ be the corresponding flow.
It is well-known that if $V$ is $\lambda$-convex (i-e the Hessian $D^2V\geq \lambda Id$ in the sense of symmetric matrices) then the flow is exponentially $\lambda$-contractant,
$$
|\Phi(t,X\_0)-\Phi(t,X\_0')|\leq e^{-\lambda t}|X\_0-X\_0'|,
\qquad 
\forall \, X\_0,X\_0'\in \mathbb R^d.
$$
In particular for $\lambda=0$ (convex potential) the flow is just nonexpanding.



> 
> **Question:** is this an equivalence? I-e is it true that if $V$ is smooth and $\Phi(t,.)$ is $1$-Lipschitz for all times then necessarily $V$ must be convex? I am also interested in the corresponding statement for $\lambda>0$, i-e if $\Phi(t,.)$ is $e^{-\lambda t}$-Lipschitz for all $t>0$ is it true that $D^2V\geq \lambda$?
> 
> 
> 


Let me just add a few comments:


* The implication "$D^2V\geq \lambda$ $\Rightarrow$ $\Phi(t,.)$ is $e^{-\lambda t}$-contractant" is classical and easy to prove. As should be clear from my question above, I am only interested in the converse implication.
* I stated the problem in $\mathbb R^d$ and smooth potentials for simplicity, but in reality I'm interested in this kind of problems in infinite dimension, and in fact in the context of abstract gradient-flows in metric spaces. The point is that I want to prove that some potential $V$ is geodesically convex, assuming only that the generated flow is contractant. But I want to check that this is not a complete equivalence. For example sometimes it is easy to see that the flow map is well-behaved by pure "1st-order calculus" PDE arguments (I mean, taking just one derivative in time along solutions), but characterizing the convexity requires 2nd order calculus and is therefore more delicate to justify rigorously in infinite dimensions so the mutual implications between both concepts may not be totally clear
* This is related to my previous posts:
	+ [Heat flow, decay of the Fisher information, and $\lambda$-displacement convexity](https://mathoverflow.net/questions/358929/)
	+ [Universal decay rate of the Fisher information along the heat flow](https://mathoverflow.net/questions/311257/)
	+ [improved regularization for $\lambda$-convex gradient flows](https://mathoverflow.net/questions/358828/)



Gradient flows: convex potential vs. contractive flow?
Doesn't this follow from dependence on initial data? 


Consider the flow mapping $\Phi(t,X)$ which solves 


$$ \frac{d}{dt}\Phi(t,X) = - \nabla V(\Phi(t,X)) $$


so taking the derivative in $X$ we have


$$ \frac{d}{dt} \partial\_X \Phi(t,X) = - \nabla^2 V(\Phi(t,X)) \cdot \partial\_X \Phi(t,X) \\= - \nabla^2 V(X) \cdot \partial\_X \Phi(t,X) + O(t) \cdot \partial\_X \Phi(t,X)$$


So if $-\nabla^2 V(X\_0)$ has negative eigenvalue $-\lambda\_0$ with eigenvector $v\_0$, taking the partial in the $v\_0$ direction gives 


$$ \partial\_{v\_0} \Phi(t,X\_0) = e^{\lambda\_0 t} v\_0 + O(t^2) $$


For $t>0$ sufficiently small you guarantee that 
$$ |\partial\_{v\_0} \Phi(t,X\_0) | \geq (1 + \frac{\lambda\_0}{2}t) |v\_0| $$
showing that the solution map cannot be 1 Lipschitz. 


