# Analytic extensions of convex-linear maps

Let $S=\{v_1,\ldots,v_n\}$ be a set of vectors. Let us consider its convex and affine hulls: $${\rm conv}(S)=\left\{\sum_{i=1}^n \lambda_i v_i \left\vert \lambda_i\geq 0, \sum_{i=1}^n \lambda_i=1\right.\right\}$$ and $${\rm aff}(S)=\left\{\sum_{i=1}^n \lambda_i v_i \left\vert \sum_{i=1}^n \lambda_i=1\right.\right\},$$ respectively.

Consider a convex-linear function on $f:{\rm conv}(S)\rightarrow\mathbb{R}$: $$f(w)=(1-\alpha)f(w_1)+\alpha f(w_2), \quad\text{with } 1\geq\alpha\geq0$$ for $w=(1-\alpha)w_1+\alpha w_2$ and $w,w_1,w_2\in{\rm conv}(S)$.

The question is the following: Consider an extension $f^*$ of $f$ from ${\rm conv}(S)$ to ${\rm aff}(S)$: \begin{eqnarray} &f^*:{\rm aff}(S)\rightarrow\mathbb{R},\\ &f^*(w)=f(w) \quad \text{for}\ w\in{\rm conv}(S). \end{eqnarray} If we require $f^*$ to be analytic (infinitely differentiable) on ${\rm aff}(S)$, must $f^*$ be affine?

• Unless the question is missing some requirements, the answer appears to be trivially no; for example $f(x)=|x|$ is a non-affine function which is linear on the interval $[1,2]$. – user357151 Jan 1 '17 at 22:28
• Sorry, I mean to say "infinitely differentiable" – NessunDorma Jan 2 '17 at 16:56
• Okay; then let $f(x)=0$ for $x\ge 0$ and $\exp(1/x)$ for $x<0$. It's $C^\infty$ on the real line. – user357151 Jan 2 '17 at 17:12
• Ok, many thanks. I see there are many examples. I guess a similar one would be $f(x)=x$ for $x\geq0$ and $x\exp(x)$ for $x<0$. – NessunDorma Jan 3 '17 at 15:23