It is well known that the internal energy (see, e.g., Definition 3.32 in and Proposition 3.33 in [1](https://www.math.umd.edu/~yanir/OT/AmbrosioGigliDec2011.pdf)) is geodesically convex with the 2-Wasserstein distance. I was wondering under what condition, the internal energy functional is $\lambda$-convex where $\lambda>0$ (see the top of Page 55 in [1](https://www.math.umd.edu/~yanir/OT/AmbrosioGigliDec2011.pdf) for the definition of a $\lambda$-convex functional)?



Strong convexity of internal energy with respect to Wasserstein metric
This is a very delicate topics and requires a curvature condition on the underlying Polish space $X$ (upon which the probability space $\mathcal P(X)$ is based). This is known nowadays as the "Sturm-Lott-Villani" synthetic theory of curvature, see part III in Villani's (big) book.
Long story short: requiring that $H(\mu)=\int\_X\mu\log\mu\, \mathrm d vol$ be $\lambda$-convex is Sturm-Lott-Villani's generalized definition for the Ricci lower bound $R(X)\geq \lambda$.


For a more practical answer to your question: in the flat space $\mathbb R^d$ (or any reasonable subdomain thereof) the internal energy will in general not be $\lambda$-convex for ANY $\lambda>0$. In the sphere (with uniformly positive Ricci curvature), you can get a positive modulus of convexity.


