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I have a function $f:X \to \mathbb{R}$ strictly convex, where $X \subset \mathbb{R}^{n}$ is a convex compact space. Also, given $A \in \mathbb{R}^{n \times m}$ and $b \in \mathbb{R}^{n}$, consider an affine mapping $u \mapsto Au + b$ for $u \in \mathbb{R}^{m}$. Then, can we say $u \mapsto f(Au +b)$ is strictly convex?

Jinrae Kim
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  • Have you tried to prove this yourself? – Deane Jun 23 '23 at 01:43
  • @Deane Yes, and I've studied for a bit after posting this question. I'll leave what I've found here:
    1. I found that at least it is not "always" true since $Au_{1}$ and $Au_{2}$ might be the same for different $u_{1}$ and $u_{2}$.
    2. My concern is actually about log-sum-exp with affine mapping. According to this paper (https://arxiv.org/abs/1803.07225), the log-sum-exp with affine mapping for stochastic sampling settings for integration estimation is strictly convex under (mild) assumptions. Here, $n$ corresponds to the number of sampling.
     – Jinrae Kim Jun 24 '23 at 04:20
  • (Cont'd) Therefore, if $n$ is large and $A$ is "regular", we can expect that the function is likely strictly convex (not guaranteed for all cases tho). – Jinrae Kim Jun 24 '23 at 04:22
  • A function $f: X \rightarrow \mathbb{R}$ is convex if for any $x_0, x_1 \in X$ and $0 \le t \le 1$, $$ f(x_0 + t(x_1-x_0) \le f(x_0) + t(f(x_1)-f(x_0)). $$ It is straightforward, using this, to show that the function $$g(u) = f(Au+b)$$ is convex. – Deane Jun 24 '23 at 15:26
  • @Deane My question is not about just convexity. Please see the above comments when it may not preserve the strict convexity. – Jinrae Kim Jun 25 '23 at 23:31
  • First, if $A$ is injective, it easy to show that $g$ is strictly convex if $f$ is. Therefore, if you restrict the domain of $g$ to any affine subspace transversal to $\ker A$, it will also be strictly convex. In general, $g$ will be constant along any subspace parallel to $\ker A$ and strictly convex on any affine subspace transversal to the kernel. – Deane Jun 25 '23 at 23:47

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