Talk:Convex conjugate

Latest comment: 3 months ago by Konuts in topic Incomplete statement

Clarification for a notation

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What is the definition of the term used in the table of convex conjugates (cases 6 and 8) in the column of the ? Kellertuer (talk) 08:57, 31 January 2014 (UTC)Reply

Typo in rule for translation

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What is y in the rule for the conjugate of f(x+b)? Is it x*? — Preceding unsigned comment added by 193.157.253.66 (talk) 08:21, 23 April 2012 (UTC)Reply

Fixed. Zfeinst (talk) 15:49, 23 April 2012 (UTC)Reply

Scaling properties

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What does this (from the "Scaling properties" section) mean?

In case of an additional parameter (α, say) moreover

where is chosen to be the maximizing argument.

JadeNB (talk) 19:17, 23 March 2011 (UTC)Reply

Biconjugate?

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

Relationship to Legendre transform

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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   where   is such that  . 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 Vandenberghe or [1]. Zfeinst (talk) 08:00, 10 February 2015 (UTC)Reply

Incomplete statement

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The article contains the statement "It allows in particular for a far reaching generalization of Lagrangian duality." However, it is nowhere explained what is this generalization about and what does "far reaching" really means here. More details are needed, possibly with a reference to a book/article. — Preceding unsigned comment added by Konuts (talkcontribs) 11:17, 28 August 2024 (UTC)Reply