# Convex combinations of function values

Suppose $X \subset \mathbb{R}^n$ is a convex set and $f: X \rightarrow \mathbb{R}$ is a function which is bounded from below. Let $x_0 \in X$ be a point.

Define $$a = \inf \{ \alpha f(x_1) + (1 - \alpha) f(x_2) | x_1, x_2 \in X,\,\, \alpha \in [0,1], \text{ and } \alpha x_1 + (1 - \alpha) x_2 = x_0\}$$ and $$b = \inf \{ \sum_{i = 1}^k \alpha_i f(x_i)| k \geq 1, \,\, x_1, \ldots, x_k \in X,\,\, \alpha_i \in [0,1], \,\, \sum_i \alpha_i = 1, \text{ and } \sum_i \alpha_i x_i = x_0\}.$$ Of course, $a \geq b$. My question is whether $a = b$? If $n = 1$ (so $X \subset \mathbb{R}$, I can show that $a = b$. Since I extensively used properties of convex combinations in $\mathbb{R}$, I am not sure it is also true in higher dimensional spaces. I can neither construct a counter-example. Can someone help me out?

Thank you!

I have deleted my previous 1/2 baked answer to this question.

This question is covered by section 17 of Rockafellar's Convex Analysis, in particular Corollary 17.1.5 on p.157. What the OP calls $b$ is what Rockafellar calls $(\operatorname{conv}f)(x_0);$ the OP's formula for $b$ appears at the bottom of R's p.36. The correct formula for $a$ in the case $n>1$ is the conclusion of Cor. 17.1.5: $$(\operatorname{conv}f)(x) = \inf\left\{\sum_{i=1}^{n+1}\lambda_i f(x_i)\,\big| \sum_{i=1}^{n+1} \lambda_i x_i = x\right\}.\tag{*}$$

Here $\operatorname{conv}f$ is the pointwise supremum of all convex functions pointwise less than $f$.

The following counterexamples, based on Ashkan's example, show that the convex combinations in (*) need $n+1$ terms. Let $X\subset\mathbb R^n$ be a regular $n$-simplex, with vertex set $V=\{v_1,\dots,v_{n+1}\}$ and centroid $m=\sum_{i=1}^{n+1} v_i/ (n+1)$. Note that $m$ is at positive distance from each of the facets of $X$. Now define $f(x)=1$ for all $x\in X\backslash V$ and $f(v)=0$ for all $v\in V.$ Clearly $g = \operatorname{conv}f$ is the constant zero function. Now we try to approximate $g(m)$ with $n$-term sums of the form $S=\sum_{i=1}^n \lambda_i f(x_i),$ with $m=\sum_{i=1}^n \lambda_i x_i.$ If there is a $\lambda$ for which $S<\epsilon$, all but $\epsilon$ of the mass of $\lambda$ is concentrated on $V$. But $\lambda$ has only $n$ terms, so all but $\epsilon$ of the mass is concentrated on $n$ of the points of $V$. Hence $m$ is within distance $O(\epsilon)$ of one of the facets of $X$. Since $m$ is at positive distance from the facets, $S$ cannot be made arbitrarily close to $0.$

I found a counter example where $f$ defined on a compact convex set.

Let $f: \Bbb B \to \Bbb R$ and with

$$f(x)= \left\{\begin{matrix} -1 & x\in C \\ 1 & x \notin C \end{matrix}\right.$$

where $X= \Bbb B$ is the closed Unit ball in $\Bbb R^2$ and $C$ is the open left semicircle i.e. $$C= \{ (x,y) \in \Bbb R^2 ~ | ~ x^2 + y^2 = 1, ~~ x <0 \}$$

Then for $x_0 = 0$ observe that $a=0$ and $b < 0$

P.S: Note that

$$b \leq \frac{1}{3}f(\frac{-\sqrt2}{2}, \frac{+\sqrt2}{2} ) + \frac{1}{3} f (\frac{-\sqrt2}{2}, \frac{-\sqrt2}{2}, ) + \frac{1}{3} f(1, 0) = - \frac{1}{3},$$

Maybe you need assume continuity, however still I feel claim is not true!