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Actually, I've already asked a similar question but I specified my question here.

Consider a max-affine function $f:K \rightarrow \mathbb{R}$ such that $f(x) = \max_{i \in I}f_i(x)$ where $K \subset \mathbb{R}^n$ is a compact set and $f_i(x) = a_i^T x + b_i, \forall i \in I$ are affine functions.

  1. (Assumption) Suppose that $a_i$ and $b_i$ are bounded by $M > 0$ for all $i \in I$.

  2. (Statement) The max-affine function $f$ is continuous on $K$.

(Proof) Since the max-affine function $f$ is convex (see this) and (Assumption) is assumed, $f$ is a proper convex function on $K$. Furthermore, $dom(f) = \mathbb{R}^n$ in this case. By Theorem 10.4 in R.T. Rockafella, "Convex Analysis", a proper convex function in $ri(dom(f))$ is continuous, it concludes the statement.

I wonder if 1) the proof is correct, 2) the assumption is necessary to make sure that $f$ is proper, and 3) this holds even if $I$ is uncountable.

Jinrae Kim
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1 Answers1

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  1. The proof is correct, although for the sake of argument you should clarify what $ri(dom(f))$ is in this case.

  2. Yes, you need this assumption. It is automatically satisfied when $I$ is finite. When $I$ is not finite and you do not make this assumption, you could take $a_i=i$ resulting in $\sup_{i \in \mathbb{N}} f_i(x) = \infty$. Bounded should be in absolute sense, so $|a_i|\leq M$ and $|b_i|\leq M$.

  3. When $I$ is uncountable you may need $\sup$ instead of $\max$, but the proof is still valid.

LinAlg
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