Convex interaction energy

Question

Does anybody know examples of absolutely continuous probability measures $\mu_0,\mu_1$ on $\mathbb{R}^n$ with finite 2nd moments such that $$ \frac{d^2}{dt^2}\left(\int_{\mathbb{R}^n\times \mathbb{R}^n}|x-y|^2d\mu_t(x) d\mu_t(y)\right)<0, $$ where $\mu_t=(1-t)\mu_0+t\mu_1$? The inside of the above bracket is known as an interaction energy.

Answer 1

See Leo's comment below. I will construct two discrete probability measures on $\mathbb{R}$ for a counterexample.

It seems you can just decompose the measure $\mu_t$ explicitly and expand into three integrals (using symmetry of the integral domain): $$\int_{\mathbb{R}^n \times \mathbb{R}^n} |x -y|^2 [(1-t)^2d\mu_0(x) d\mu_0(y) + t^2 d\mu_1(x) d\mu_1(y) + 2 t(1-t) d\mu_0(x) d\mu_1(y)].$$ Taking second derivative with respect to $t$, you get: $$ 2\int |x-y|^2 (d\mu_0(x) d\mu_0(y) + d\mu_1(x) d\mu_1(y) - 2 d\mu_0(x) d\mu_1(y))\\ = 2 \int |x-y|^2 (d\mu_0(x) - d\mu_1(x))(d\mu_0(y)- d\mu_1(y)).$$

Now let $\nu = \mu_0 - \mu_1$ be the signed measure above. I can take $\nu$ to be the following discrete signed probability on $\{1,-1\} \subset \mathbb{R}$: $\nu(-1) = -1/2$, $\nu(1) = 1/2$. Then the last integral becomes $$\sum_{x \in \{\pm 1\}} \sum_{y \in \{\pm 1\}} |x-y|^2 \nu(x)\nu(y) = 2 * 4 * (-1/4) = -2 < 0.$$

To construct $\nu$, simply take $\mu_0(1) = 3/4$, $\mu_0(-1) = 1/4$ and $\mu_1(1) = 1/4$, $\mu_1(-1) = 3/4$, and verify that $\nu = \mu_0 - \mu_1$.