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A face of convex set $C$ is a convex subset $F$ of $C$ such that for $x,y\in C$ and some $\lambda\in \langle 0,1\rangle, \lambda x+(1-\lambda)y\in F$ implies $x,y\in F$.

I was wondering if every face was closed set? If not, can someone please give me an example of non-closed face. What about faces in infinite dimensional spaces like function spaces? Could those be non-closed as well? Any example or a hint is welcome.

Dee
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  • This may come too late, but convex faces of a closed convex set are closed. In the counter-example below, the only convex faces are singletons in the boundary of the set. – William M. Jul 06 '22 at 20:59

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If $C$ is the open unit ball, then $C$ is convex. Moreover, $F=C\subseteq C$ is a face of $C$. However, it is not closed.

More generally, every convex set is a face of itself, but need not be closed.

If we restrict ourself to proper faces, then it is still not true. For example consider $C=\{(x,y)\mid x\geq 0, y>0\}$, then $\{0\}\times ]0,\infty[$ is a face of $C$ but is not closed.

Surb
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To complete the accepted answer: In a finite dimensional vector space, any face of a closed convex set is closed.

This is proved in Corollary 18.1.1 of Convex Analysis, R.T. Rockafellar, 1996.

Guillaume
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  • I checked Rockafellar and what he proves is that in a normed space the convex faces of a convex set are closed if the convex set is closed. There is no finite dimension assumption. – William M. Jul 06 '22 at 20:57
  • Rockafellar's entire book is set in Euclidean spaces, so I quote strictly what is proved there.

    Maybe what you mean instead is that Rockafellar's proof could be easily translated to normed spaces? I don't know if this is true, and too lazy to check. But my intuition is that there is need to be careful given that relative interiors can be empty in such general settings.

     – Guillaume Jul 18 '22 at 11:52
  • Fair enough. However, it is quite obvious when he uses finite dimensionality: hyperplanes are closed, algebraic interior = topological interior, linear functions are continuous, or when he uses that $\mathbf{R}$ is canonically embedded into the underlying vector space (e.g. when forms convex sets taking suprema or infima). Face is a vector-space definition and the his proof uses none of the aforementioned finite-dimensional assumptions. I suspect Rockafellar wanted to avoid Banach-Hahn theorem at all costs so he "only" works with $\mathbf{R}^d.$ – William M. Jul 18 '22 at 14:54