# Definition of convexity via probability measure

I found this alternative definition of convexity in a set of lecture notes on convex optimization:

A subset $$K \subset \mathbb R^n$$ is convex if and only if for every probability measure $$\mathbb P$$ supported on $$K$$, $$\mathbb E [x] \in K$$.

Here, $$x$$ is the coordinate function (i.e. the identity).

I was unable to find any more information on this definition. I would like to know if it is equivalent (also: when is it equivalent? What about infinite-dimensional Banach spaces?) and how equivalence is proven.

It's clear that this definition implies convexity defined in the usual way.

For the converse, I tried a bit myself and came up with the following proof relying on the martingale convergence theorem, but it only shows that a compact set is convex if and only it fulfills the above condition. Also, I'm assuming that $$\mathbb P$$ is a Borel measure.

Let $$\mathcal F_n := \sigma \left( \lbrace K \cap ( m_1 2^{-n},(m_1+1) 2^{-k}] \times \dots \times (m_n 2^{-n}, (m_n+1) 2^{-n}] \vert m_1,\dots ,m_n\in\mathbb Z \rbrace \right)$$. Then $$\mathcal F_1 \subset \mathcal F_2 \subset \dots \subset \mathcal F_\infty \subset \mathcal B(K)$$ is a filtration, w.r.t. which $$X_n := \mathbb E[x \vert \mathcal F_n]$$ is a martingale, which, due to boundedness of $$K$$, is uniformly bounded in the $$L^2 (\mathbb P)$$ norm. Hence, it converges almost surely and in $$L^2$$ to $$X_n \to X_\infty = \mathbb E[x \vert \mathcal F_\infty ] = x$$ Therefore, $$\mathbb E[X_n]=\mathbb E[\mathbb E[x\vert \mathcal F_n]]\in K$$ (due to convexity of $$K$$) converges to $$\mathbb E[x]$$. Since $$K$$ is closed, it implies that $$\mathbb E[x] \in K$$.

I'm grateful for all references.

• Thank you for your comment! I should have been clearer in the post: It's clear to me why this definition of convexity implies convexity according to the usual definition. It's the converse that is not clear to me. I will edit. Commented Mar 14, 2019 at 14:19
• If $K$ is not compact, then $\mathbb{E}[x]$ may not even by well-defined. If it is, you can write $\mathbb{E}[x]$ as limit of convex combinations of $x_k\in K$. Commented Mar 14, 2019 at 14:25
• One direction is trivial. If $K$ is closed the other direction is straightforward to prove. Otherwise, the only proof I could come up with is an awkward induction on dimension. Commented Mar 14, 2019 at 16:07

If $$\ K\subsetneq \mathbb{R}^n\$$ is convex (according to the standard definition), $$\ \mathbb{P}\$$ a probability measure supported on $$\ K\$$, $$\ \mathbb{e}=\mathbb{E}_\mathbb{P}\hspace{-0.2em}\left(x\right)\$$, $$\ A=\mathrm{aff}\left(\mathrm{supp}\left(\mathbb{P}\right)\right)$$, and $$\ K'=A\cap K\$$, then $$\ K'\subseteq K\$$ is convex, and $$\ \mathbb{P}\$$ is supported on $$\ K'\$$.
If $$\ K' = A\$$, then necessarily $$\ d=\mathrm{dim}\left(A\right). Let $$\ M\$$ be an $$\ \left(n-d\right)\times n\$$ matrix, and $$\ a\in\mathbb{R}^{n-d}\$$ such that $$\ A=\left\{x\in\mathbb{R}^n\,\vert\,Mx=a\,\right\}\$$. Then $$\ a=\mathbb{E}_\mathbb{P}\hspace{-0.2em}\left(Mx\right)\ = M\mathbb{e}$$, so $$\ \mathbb{e}\in A=K'\$$
Otherwise, let $$\ \left\{\,x\in A\,\vert\, \langle\lambda,x\rangle\le\beta\,\right\}\$$ be any closed tangent half-space to $$\ K'\$$ in $$\ A\$$, where $$\ \lambda\not\in A^\perp\$$. Since $$\ \left\{\,x\in A\,\vert\, \langle\lambda,x\rangle=\beta\,\right\}\$$ is a proper affine subspace of $$\ A\$$, then $$\ \mathrm{supp}\left(\mathbb{P}\right)\$$ cannot lie entirely inside it, so $$\ \mathbb{P}\left(\left\{\,x\in K'\,\vert\, \langle\lambda,x\rangle<\beta\,\right\}\right)>0\$$, and $$\ \mathbb{E}_\mathbb{P}\hspace{-0.2em}\left(\langle\lambda, x\rangle\right)= \langle\lambda,\mathbb{e}\rangle<\beta\$$. Thus $$\ \mathbb{e}\$$ lies in every open tangent half-space to $$\ K'\$$ in $$\ A\$$, and therefore in its relative interior, $$\ \mathrm{ri}\left(K'\right)\subseteq K'\$$.