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Crossposted at Mathematics SE


Let us define the following class of functions:

$$ \mathcal{F} := \left\{ f : \mathbb{R}^d \to \mathbb{R} \,\middle|\, f(x) = \sum_{i=1}^n \varphi_i(\xi_i^\top x),\ \varphi_i : \mathbb{R} \to \mathbb{R} \text{ convex},\ \xi_i \in \mathbb{R}^d,\ n \in \mathbb{N} \right\}.$$

That is, $\mathcal{F}$ consists of finite sums of $n$ convex univariate functions composed with linear forms (convex ridge functions). I am interested in the closed convex hull of $\mathcal{F}$ where the closure is taken in the topology of uniform convergence on compact subsets of $\mathbb{R}^d$. I think that

$$\overline{\mathrm{conv}}(\mathcal{F}) = \left\{ f : \mathbb{R}^d \to \mathbb{R} \,\middle|\, f \text{ is convex} \right\}$$

I tried with the max-affine representation of convex function but then I am stuck with the max. Any idea?

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  • $\begingroup$ Do you have a response to the answer below? $\endgroup$ Commented Apr 14 at 17:18

1 Answer 1

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$\newcommand\de{\delta}\newcommand\F{\mathcal F}\newcommand\R{\mathbb R}\newcommand\ol{\overline}\newcommand\co{\operatorname{conv}}$It is not true that $\ol\co\F$ is the set of all convex functions $f\colon\R^d\to\R$.

Indeed, here without loss of generality $d=2$. Let $$\nu=\frac16\,(3\de_{(x+y+z)/3}+\de_x+\de_y+\de_z),\quad \mu=\frac13\,(\de_{(x+y)/2}+\de_{(y+z)/2}+\de_{(x+z)/2}) $$ for some $x,y,z$ in $\R^2$, where $\de_w$ is the Dirac measure supported on the set $\{w\}$. Then, as shown in this previous answer, $$\int f\,d\mu\le\int f\,d\nu \tag{1}\label{1}$$ for all $f\in\F$ and hence for all $f\in\ol\co\F$.

However, as was also noted in that previous answer, \eqref{1} fails to hold for the convex function $f\colon\R^2\to\R$ given by the formula $f(w)=\max(0,\xi_1,\xi_2)$ for all $w=(\xi_1,\xi_2)\in\R^2$ and $$x=(0,-1),\quad y=(-1,0),\quad z=(2,2).$$
So, the latter convex function $f$ is not in $\ol\co\F$. $\quad\Box$

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  • $\begingroup$ Would it be possible to find an example where $\mu$ is a multivariate Gaussian centered in $(0,0)$? $\endgroup$ Commented Apr 10 at 8:44
  • $\begingroup$ @Aristodog : At this point, I don't know, but I will have this in mind. $\endgroup$ Commented Apr 10 at 12:16

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