Is distance function defined on a convex set is always convex? I am looking for an answer to the following question: 
Is the distance function defined on a convex set always convex?
Obviously the convex set in question is metric. In particular I am interested in the case when the convex set is the set of probability measures.
In formal terms: Let $\Delta(X)$ be the space of probability measures over $X$ (and assume for the sake of simplicity that $X$ is just a compact convex subset of $\mathbb{R}$). Distance is a function: $d:\Delta(X)\times \Delta(X)\rightarrow [0,1]$ that verifies the standard conditions, including the triangle inequality (and it metrizes a topology on $\Delta(X)$). 
My question is then if the following is true for any $d$ on $\Delta(X)$:
$$d(\alpha \mu_1+(1-\alpha)\mu_2,\mu_3)\leq \alpha d (\mu_1,\mu_3)+(1-\alpha)d(\mu_2,\mu_3)$$
Thanks for any hints or references.
 A: $\renewcommand{\Re}{\mathbb{R}}$In general, a metric $d:\Omega\times\Omega\to\Re_+$ may not be convex even when $\Omega$ is convex. Most of the metric used, however, are convex; so you have to examine each case separately.
Take for example a probability space $(\Omega, \mathcal{F}, \mathrm{P})$ and the space of random variables defined thereon, that is measurable functions $X:\Omega\to\Re$ where $\Re$ is endowed with the Borel sigma algebra. Consider the space
$$
\mathcal{L}_{0.5}(\Omega, \mathcal{F}, \mathrm{P}) = \left\{X:\Omega\to\Re, \text{measurable}, \int_{\Omega} \sqrt{|X(\omega)|}\mathrm{P}(\mathrm{d}\omega)<\infty \right\},
$$
and consider the following norm-like function on $\mathcal{L}_{0.5}(\Omega, \mathcal{F}, \mathrm{P})$:
$$
\|X\|_{0.5} = \int_{\Omega} \sqrt{|X(\omega)|}\mathrm{P}(\mathrm{d}\omega).
$$
Now take the distance function $d(X,Y)$ on $\mathcal{L}_{0.5}(\Omega, \mathcal{F}, \mathrm{P})$ induced by this norm, that is
$$
d(X,Y)=\|X-Y\|_{0.5}.
$$
This metric is not convex. Having done this for random variables we can extend this to measures. Given a random variable $X_1\in\mathcal{L}_{0.5}(\Omega, \mathcal{F}, \mathrm{P})$ we may define a measure
$$
\mu_1(A) = \int_A X_1\mathrm{dP}
$$
Construct, analogously, another measure
$$
\mu_2(A) = \int_A X_2\mathrm{dP}
$$
for some $X_2\in\mathcal{L}_{0.5}(\Omega, \mathcal{F}, \mathrm{P})$ and define their distance $D(\mu_1, \mu_2)$ to be
$$
D(\mu_1, \mu_2) = d(X_1, X_2).
$$
Update 1. There is a simple way to construct a weak metric on $\Re\times \Re$, $d(x,y)$ which is convex in the first argument but not in the second. This is
$$
d(x,y) = \begin{cases}
2(y-x),&x<y\\
2(x-y),&0\leq y\leq x\\
2x-y,&y<0\leq x\\
x-y,&y\leq x < 0
\end{cases}
$$
This function satisfies (i) $d(x,y)\geq 0$, (ii) $d(x,x)=0$, (iii) $d(x,y)\leq d(x,z) + d(x,y)$, (iv) $d$ is convex in the first argument. However, $d$ does not satisfy $d(x,y)=0\Rightarrow x=y$ and is not symmetric.
And this is how it looks like:
 
Update 2. (The Hilbert metric). In Example 1 in this article the authors prove that the Hilbert metric is not convex.
