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:

2 Answers 2


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.

  • $\begingroup$ Yes indeed! I should have figured that out sooner. Thank you @Willie Wong. Actually I would be interested in a "non-differential" proof, since I want to go to infinite dimension and metric spaces. $\endgroup$ Apr 30, 2020 at 18:05
  • $\begingroup$ I think the same argument can be made through differences if you are careful with your epsilons and deltas. You have that the difference between solutions $X$ and $Y$ satisfy $$ (X-Y)' = -\nabla V(X) + \nabla V(Y) $$ and if you start with $X(0) - Y(0)$ sufficiently small and only look at sufficiently small times you can approximate using Taylor series and everything goes through. The only part you may have to modify your assumption is when you work in infinite dimensions, and continuity does not guarantee local uniform continuity. This means that the term I denoted by $O(t)$ in the... $\endgroup$ Apr 30, 2020 at 18:12
  • $\begingroup$ ... answer may require quite a bit more work/slightly strengthened assumptions beyond $C^2$. $\endgroup$ Apr 30, 2020 at 18:15
  • $\begingroup$ yes yes, I get the idea, but the problem is that in abstract metric gradient flows even the $\nabla V(X)$ object does not really exist, only its magnitude somehow (metric slope and upper gradients). So I cannot even try Taylor-expanding anything. Life is hard. $\endgroup$ Apr 30, 2020 at 18:15
  • 1
    $\begingroup$ Ah, I understand. Sorry, cannot help more. I do still think that this is a local and not a global problem, and the key is to understanding the variation of your flow with respect to changes in initial data. $\endgroup$ Apr 30, 2020 at 18:20

It should be noticed that already on $R^d$ equipped with a non-Euclidean norm $\|.\|$ the answer to your question is no. Ohta-Sturm [1] proved the following: let $\lambda\in R$ and consider the classes of $\lambda$-convex functions on $R^d$ on one side and the class of functions whose gradient flows are $\lambda$-contractive (they call these functions skew-convex). Then these classes coincide if and only if the norm $\|.\|$ (which is relevant for the notion of gradient flow) comes from a scalar product.

This shows that "convexity" and "contractivity" are really related only on Riemannian-like world, whereas on Finsler-like ones they are different concepts.

In particular, the answer in general metric spaces is no.

However, you may have better luck in spaces which are Hilbert-behaved on small scales (whatever this means). Judging by the other questions that you linked it seems that you are particularly interested in the Wasserstein space built over a Riemannian manifold. In this case, the paper by Otto-Westdickenberg [2] might be the reference you are looking for. They basically make the same argument Willie Wong gave above in $R^d$ but in $(P_2(M),W_2)$.

Inspired by [2], Daneri-Savare' [3] proved that if you interpret gradient flows in the $EVI_\lambda$ sense (a condition that implies $\lambda$-contractivity but that in practical situation one is often able to obtain if s/he is able to get contractivity - but beware of the results in [1]), then if a functional admits gradient flows in such sense, it must be $\lambda$-convex. This holds in arbitrary metric spaces.

[1] https://arxiv.org/abs/1009.2312

[2] http://www.instmath.rwth-aachen.de/~mwest/files/OttoWest.pdf

[3] https://arxiv.org/abs/0801.2455


oh, small world indeed :) Let me add some comments:

  • one of the greatest intuitions in optimal transport is Otto’s interpretation of $(P_2(M),W_2)$ (here $M$ is a Riemannian manifold) as a sort of "infinite dimensional Riemannian manifold" (this answers your question). Quotation marks are needed because this is not really a Riemannian manifold in any reasonable sense (not even a Hilbert manifold), but still it resembles it a lot. This intuition drove the research on optimal transport (and in particular on gradient flows over the Wasserstein space) in the last 20 years, thus my advice is that if you are not familiar with this, try to build some intuition for it.

  • Yes, EVI is stronger than contractivity (as said above, it implies both contractivity and convexity). Now, the question "how the EVI formalism means Riemannian on small scale" is of course tricky as nobody knows what "Riemannian on small scale" means. Still, let me mention some results pointing in this direction:

[4] shows that if we have many EVIs in a Banach space then the space is Hilbert

[5] shows that gradient flows of convex functionals on CAT(k) spaces satisfy the EVI. Thus it goes the other way around by showing how to derive the EVI from some geometric property of the space, but it is interesting to see what is the key property used. They call it "commutativity of the distance" and it is a nice exercise to check that a Banach space is "commutative" in this sense if and only if it is Hilbert.

[4] https://www.sciencedirect.com/science/article/pii/S0022247X11006391

[5] https://www.researchgate.net/publication/267983274_Gradient_flows_and_a_Trotter--Kato_formula_of_semi-convex_functions_on_CAT1-spaces

  • $\begingroup$ Thanks Nicola, when I asked the question I was not aware of your reference [2]. I'm working on this issue with your coauthor L. Tamanini (small world!) One thing still puzzles, me, though: you said that "convexity" and "contractivity" are really related only on Riemannian-like world, or at least in situations which are Riemannian-like on small scales. But then how should I interpret the results of [2], which give precisely this type of equivalence? I really don't see how the EVI formalism means Riemannian on small scale. Is it just that EVI is much much stronger than $\lambda$-contractivity? $\endgroup$ Jul 14, 2020 at 19:47

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