Question:
If $f$ is an affine map from $\mathbb{R}^n$ to $\mathbb{R}$ then must $f$ be supported on some affine subspace of positive dimension if and only if $f\neq 0$? So in particular $f$ is never compactly supported unless it is $0$?
In the multivariate case, I define the "support" of a measurable function $f:\mathbb{R}^n\rightarrow \mathbb{R}^m$ as $$ \operatorname{supp}(f) \triangleq \overline{\left\{ x \in \mathbb{R}^n:\, \|f(x)\|_{\mathbb{R}^m}>0 \right\}}. $$
Here's what I've got (so far...):
Let $f(x) = Ax +b$ for some $n\times m$ matrix $A$ and some $B \in \mathbb{R}^n$. If $A\neq 0$, then the support of $f$, denoted $\operatorname{supp}(f)$, is equal to $$ \operatorname{supp}(f) = \mathbb{R}^n-[\ker(A)-b], $$ where $X-b:=\{x\in \mathbb{R}^n:\, x+b \in X\}$ and $\ker(A)$ is the kernel of the linear map $A$. In particular, $\operatorname{supp}(f)$ is a closed (affine) subspace of $\mathbb{R}^n$.
By rank-nullity (or the splitting Lemma really...) and the fact that $A \neq 0$ (so $Im(A)\neq 0$) we have $$ \dim(\ker(A)) = n - \dim(Im(A)) < n. $$ So $\operatorname{supp}(f)$ is a closed affine subspace of $\mathbb{R}^n$ of (topological) dimension $n - \dim(Im(A))>0$ (and of dimesional $0$ iff $A=0$ and $b=0$).
So $f$ is compactly-supported if and only if $f=0$, in which case $\operatorname{supp}(f)=\emptyset$.