Restriction of a convex function to {0, 1}^n

Suppose I have a real-valued convex function $$f$$ on the unit hypercube $$[0,1]^n$$, and let $$\bar{f}$$ be its restriction to the integer points $$\{0,1\}^n$$. Does $$\bar{f}$$ satisfy any properties, or can any function on $$\{0,1\}^n$$ be obtained as a restriction of a convex function?

• Any real valued function $g$ defined on $\{0,1\}^n$ can be extended to a convex function over $[0,1]^n$ (it is called the convex closure). The implication for your question is that indeed $\bar{f}$ will not have any specific properties. Apr 22 at 20:27
• Converted comment into an answer and added link to Dughmi's survey so that the question can essentially be closed. Apr 23 at 22:29

Any real valued function $$g$$ defined on $$\{0,1\}^n$$ can be extended to a convex function over $$[0,1]^n$$ (it is called the convex closure). See Dughmi's nice survey. The implication for your question is that indeed $$\bar{f}$$ will not have any specific properties.
• @EmilJeřábek I think you're right. I thought I recalled some paper with the condition on an $f:\{0,1\}^n\to\mathbb{R}$ having a convex extension, but now I think it was probably any such convex extension. Apr 24 at 11:15
• One should be able to extend the convex function from $[0,1]^n$ to all of the space by making it become very steep outside the cube - this is typically done in convex optimization though there may be some technical details to be a bit careful about. Apr 24 at 20:41
• Yeah, in this case we can directly extend it to a convex function on all of $\mathbb{R}^n$ because the convex closure will be "polyhedral" -- the pointwise maximum of a finite set of affine functions, i.e. $\bar{f}(x) = \max_{j \in J} h_j(x)$. Here affine = linear + constant. Each of those affine functions $h_j$ extends to all of $\mathbb{R}^n$, so the definition of $\bar{f}$ does too.