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Suppose $f(x_1,x_2,...,x_n)$ is an affine function, $\mathbb{R}^n\rightarrow\mathbb{R}^m$. Let $$ g(x_1,x_2,...x_n) = \max\{f(x_1,x_2,...,x_n)\} - \min\{f(x_1,x_2,...,x_n)\} $$ The max and min functions find the m-dimensional component's corresponding maximum/minimum value. The question is whether $g(x_1,x_2,...x_n)$ is a convex function.

I taught myself some simple knowledge of convex optimization, including convex sets, convex functions, convex preserving properties, etc. I'd appreciate it if you could answer this question with what I know. Thank you for your time!

OvinaSun
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    Welcome to [math.se] SE. Take a [tour]. You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an [edit]): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. – Martin R Sep 10 '23 at 13:20
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    It seems there's something missing from your question. The set ${f(x_1, x_2, \ldots, x_n)}$ contains only a single element of $\Bbb{R}^m$, which isn't ordered like $\Bbb{R}$. What does the maximum and minimum mean here? Are you referring to the maximum/minimum coordinate (with respect to the standard coordinate system)? – Theo Bendit Sep 10 '23 at 13:31

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Let $f_1, \ldots, f_n : \Bbb{R}^n \to \Bbb{R}$ be the components of $f$, i.e. $$f(\vec{x}) = (f_1(\vec{x}), \ldots, f_m(\vec{x}))$$ for all $\vec{x} \in \Bbb{R}^n$. Since $f$ is affine, then so are $f_1, \ldots, f_n$ (apply the definition of $f$ being affine, then equate the components of each side). This means that, using your notation, $$\max\{f(x_1, \ldots, x_n)\} = \max_{1 \le i \le m} f_i(x_1, \ldots, x_n)$$ is a maximum of $m$ affine, hence convex functions, and hence is convex.

Applying the same argument to $-f$, which is also affine, we get $$-\min\{f(x_1, \ldots, x_n)\} = -\min_{1 \le i \le m} f_i(x_1, \ldots, x_n) = \max_{1 \le i \le m} -f_i(x_1, \ldots, x_n)$$ is also affine, and hence convex. Note that $g$ is the sum of these two convex functions, and hence is convex itself.

Theo Bendit
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