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I want to find convex conjugate of $f(x) = (\max_{i \leq n}x_i)(-\sum \log x_i)$ where $x \in \mathbb{R}^n_{++}$. This function looks like the negative entropy function but there is $(\max_{i \leq n}x_i)$ instead of $x_i$.

It seems like $f^*(y) = \sup_x \{xy - f(x)\}$ is unbounded above for any $y$. I think so because I first tried to find conjugate of $g(x) = -\sum x_ilogx_i$. Then $x_iy_i+x_ilogx_i$ is unbounded for every $i$ even when all $y_i < 0$. The function in the title is even greater then negative entropy, so I thought that it is unbounded too. So what is conjugate of this function if the domain is empty?

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  • $\begingroup$ Are you sure that the supremum is unbounded? Do you use $f(x)=\infty$ if $x\not> 0$? $\endgroup$
    – LinAlg
    Commented Oct 18, 2018 at 19:18
  • $\begingroup$ @LinAlg, I edited and tried to explain why i think it is unbounded. Yes I use that $f(x) = \infty$ outside of domain $\endgroup$
    – Markoff Chainz
    Commented Oct 18, 2018 at 19:34

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The problem is that your function is not convex. It is not concave either. This can be seen for $n=2$ by changing just one variable: $g(x) = f(x,y) = -\max\{x,y\}\cdot(\log x + \log y)$. The function $g$ is concave when $x>y>0$ and convex when $y>x>0$.

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  • $\begingroup$ Thanks! What if there wasn't a minus sign in the function, so that $f(x) = (max_{i \leq n}x_i)(\sum logx_i)$. I thought the function is convex, because it seems to be pointwise maximum of convex functions $\endgroup$
    – Markoff Chainz
    Commented Oct 18, 2018 at 20:33
  • $\begingroup$ @MarkoffChainz same story (concave when $y>x>0$, convex when $y>x>0$). It would be convex if $f(x) = \max_i\{x_i \log x_i\}$. $\endgroup$
    – LinAlg
    Commented Oct 18, 2018 at 22:30

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