Potentially elementary question on affine functions on Banach spaces

Question

In Measures Which Agree On Balls by Hoffmann-Jørgensen, it is claimed that the function defined on $T(x)$, the set of normals to the unit sphere at $x$, given by

$ \varphi(x^*) = \left\{ \begin{array}{ll} \sum_{i=1}^n a_i \lambda_i & \text{if } x^* = \sum_{i=1}^n \lambda_i x_i^* \in G \\ +\infty & \text{if } x^* \in T(x) \backslash G \end{array} \right. $

is affine if $G$ is a face of $T(x)$, that is, $G$ is convex and for all $x\in G$, $y \in T(x)\setminus G$, we have the line segment $(x,y] \cap G=\varnothing$. I do not understand this claim, and I don't know if I even understand the definition of affine. Does affine just mean that $\phi(x^*)=L(x^*)+b$, where $L$ is linear and $b \in \mathbb{R}$? How can such a function attain a value of infinity? Not sure if this is more appropriate here or on MSE.

Answer 1

A function $f$ from a linear (or, more generally, affine) space $A$ is affine (see e.g. this post) if for all $a$ and $b$ in $A$ and all real $t$ one has $f((1-t)a+tb)=(1-t)f(a)+tf(b)$. If $f$ is affine and $f(a)=\infty$ for some $a\in A$, then (taking, say, $t=2$, we have) $$f(a)=f((1-2)a+2a)=(1-2)f(a)+2f(a)=-\infty+\infty,$$ which cannot have a meaning.

So, you are right: an affine function cannot possibly take the value $\infty$.

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Also, without the condition (missing in the paper) that the $x^*_i$ be pairwise distinct, the function $\varphi$ will not be well defined. (Take e.g. the case when $n=2$, $x^*_1=x^*_2$, but $a_1\ne a_2$.)