$W^1(μ,\frac{δ_{x_0}+δ_{x_1}}{2})=\frac{1}{2}(∫d(x_0,y)μ(dy)+∫d(x_1,y)μ(dy))$?

Question

I was looking at this post to try to understand a bit better the Wasserstein distance. In the thread that $$W^1(μ,δ_{x_0})=∫d(x_0,y)μ(dy).$$ But I was wondering if this remains true for $$W^1(μ,\frac{δ_{x_0}+δ_{x_1}}{2})=\frac{1}{2}(∫d(x_0,y)μ(dy)+∫d(x_1,y)μ(dy)).$$ I know we have the $\le$-inequality because of the triangle inequality but I am not sure if we have the equality.

Answer 1

No, usually not. For example, if $\mu=\frac 1 2(\delta_{x_0}+\delta_{x_1})$, then the left side is of course $0$, while the right side is $\frac 1 2 d(x_0,x_1)$.