# convex conjugate of $f(x) = (max_{i \leq n}x_i)(-\sum logx_i)$

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?

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

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 $$y>x>0$$.
• 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 Commented Oct 18, 2018 at 20:33
• @MarkoffChainz same story (concave when $y>x>0$, convex when $y>x>0$). It would be convex if $f(x) = \max_i\{x_i \log x_i\}$. Commented Oct 18, 2018 at 22:30