Talk:Convex conjugate

It is stated that ${\displaystyle f^{**}}$  is always convex and also always dominated by ${\displaystyle f}$ . On the other hand, it seems to me that a concave function going to ${\displaystyle -\infty }$  cannot dominate any convex function, unless the dominated function is the constant function of value ${\displaystyle -\infty }$ . But this is not allowed by the definition of ${\displaystyle {\overline {R}}}$  and, in turn, of ${\displaystyle f^{*}}$  and ${\displaystyle f^{**}}$ . (Notably, this definition does not agree with the definition in the entry of fr.wikipedia, where $f^*$ and $f^{**}$ are allowed to take value ${\displaystyle -\infty }$ .) I don't see a way out of this paradox. Delio.mugnolo (talk) 21:24, 2 December 2015 (UTC)Reply

The lead text says that this is a generalisation of the Legendre transform, but looking at the Legendre transformation's page it's difficult to see in what the generalisation actually is; modulo differences in notation they seem pretty much the same, but I'm probably missing something. It would help a lot if someone with the required knowledge would write a brief paragraph about the differences between the two. Nathaniel Virgo (talk) 07:22, 10 February 2015 (UTC)Reply

It appears to me as if the Legendre transform page is incorrect. The Legendre transform is for convex differentiable functions only and is defined by ${\displaystyle f^{*}(x^{*})=\langle x^{*},x\rangle -f($x)}   where ${\displaystyle x}$  is such that ${\displaystyle \nabla f($x)=x^{*}}  . The convex conjugate of a convex differentiable function coincides with the Legendre transformation, but as the convex conjugate can be defined for all functions it is more general. See page 94 of Boyd and Vandenbergheor[1]. Zfeinst (talk) 08:00, 10 February 2015 (UTC)Reply