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$$f(x)=\exp\left\{\max_{i=1,\ldots,n}\left|\log(a_i^Tx)\right| \right\}$$ where $a_i \in \mathbb{R}^m$ and $\operatorname{dom} f = \{x \in \mathbb{R}^m_+ \mid a_i^Tx>0, i = 1,\ldots,n \}$

I know that $|\log(a_i^Tx)|$ is neither convex or concave but is this also the case for $f$? If so how can I show that?

Michael Hardy
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darisoy
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    See my edits for proper MathJax usage. When you write \text{exp} rather than \exp, then instead of $3\exp5$ or $3\exp(5)$ you see $3\text{exp}5$ or $3\text{exp}(5).$ Notice that in $\exp5$ and $\exp(5),$ coded as \exp5 and \exp(5) respectively, there is more space to the right of $\exp$ in one of these expressions than in the other. With \text{exp} you don't get context-dependent spacing. – Michael Hardy Jan 24 '21 at 21:11
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    Notice that with \max_{i=1,\ldots,n} you see $\displaystyle \max_{i=1,\ldots,n}$ (when it's in a "displayed" setting) whereas with \text{max}_{i=1,\ldots,n} you see $\displaystyle \text{max}_{i=1,\ldots,n}. \qquad$ – Michael Hardy Jan 24 '21 at 21:11

1 Answers1

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Since $\exp$ is a monotone transformation, $\exp(\max_i y_i) = \max_i \exp(y_i)$ for any vector $y$. Therefore:

$$f(x) = \exp\left\{\max_{i=1,\ldots,n}\left|\log(a_i^Tx)\right| \right\} = \max_{i=1,\ldots,n} \exp\left\{\left|\log(a_i^Tx)\right| \right\}$$

Since the maximum of convex functions is convex, it suffices to check if $g_i(x) = \exp\left\{\left|\log(x)\right| \right\}$ is convex. You can either plot a graph to convince yourself it's convex, or you can formally prove it.

LinAlg
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  • @aTa perhaps we look at a different plot, what is the base of the logarithm? – LinAlg Jan 25 '21 at 17:26
  • you say you 'cannot use pointwise maximum', but the functions are identical – LinAlg Jan 25 '21 at 17:37
  • i said 'cannot use pointwise maximum' if the inner functions are not convex. but if the log's base is less than or equal to $e$ then pointwise maximum should work – darisoy Jan 25 '21 at 17:39