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.

  • $\begingroup$ 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. $\endgroup$
    – jacques
    Commented Mar 14, 2019 at 14:19
  • 1
    $\begingroup$ 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 $x_k\in K$. $\endgroup$
    – MaoWao
    Commented Mar 14, 2019 at 14:25
  • $\begingroup$ 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. $\endgroup$
    – copper.hat
    Commented Mar 14, 2019 at 16:07

1 Answer 1


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)<n\ $. 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'\ $.


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