For a function $f\colon\mathbb{R}^n\to\mathbb{R}$ the following two are equivalent\begin{align*}&\text{(i) there exist $a\in\mathbb{R}^n$, $b\in\mathbb{R}$ so that $f(x)=a^{t}x+b$}\newline&\text{(ii) for all $x,y\in\mathbb{R}^n$ and $\gamma\in(0,1)$ it holds $f(\gamma x +(1-\gamma)y)=\gamma f(x)+(1-\gamma)f(y)$}.\end{align*}
See for example here for a proof
I want to argue that the statement still holds if we consider functions $f:X\to\mathbb{R}$ where $X\subset\mathbb{R}^n$ is a convex set. My intuitive idea: Obviously $(i)\implies (ii)$ still holds in that case. On the other hand, if we assume that (ii) is true, then we could consider the affine space spanned by $X$. Let us call this affine space $Y$. I think it should be true, that the interior of $X$ is open in $Y$. So we could probably define an affine function $\tilde{f}:Y\to\mathbb{R}$ that extends $f$ first. And then extend $\tilde{f}$ to all of $\mathbb{R}^n$ (somehow). I am hoping that someone knows how to do this rigorously.