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I wonder if there is any existing result for self-concordant function in the literature about the following question.

Suppose $f$ is a self-concordant barrier function of a proper cone $K$ (pointed, closed,convex and solid), is there any result for a self-concordant barrier function $g$ of the dual of $K$, i.e. $K^*$? Is there a formula to get $g$ from $f$ or any relation between the two?

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    $\begingroup$ I think if $f$ is self-concordant for $K$, then barring some pathology, its Legendre transform $f^*$ should be self-concordant for $K^\circ$ (notice, polar cone not dual-cone)....this sounds like a textbook result. $\endgroup$
    – Suvrit
    Commented Jul 21, 2015 at 11:53
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    $\begingroup$ @Suvrit, that would be very nice. So far what I found is that $f^*$ is also self-concordant but I don't know the domain of it. See stanford.edu/~boyd/cvxbook/bv_cvxbook.pdf,pp517,exercise 9.20. $\endgroup$
    – Brian Ding
    Commented Jul 21, 2015 at 18:11
  • $\begingroup$ See Theorems 2.4.2 and 2.4.4 of the book Interior-point Polynomial Algorithms in Convex Programming of Nesterov and Nemirovskii. $\endgroup$
    – Dan Fox
    Commented Oct 12, 2016 at 18:47

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It is true that the Fenchel dual of a self-concordant function $f$ on a cone $K$ is also self-concordant (with the same parameter) and the domain of $f^*$ will be the image of the gradient of $f$, i.e. the natural domain of $f^*$. This happens to coincide with $K^*$ if $f$ is logarithmically homogeneous or if $f$ is minus the logarithm of a hyperbolic polynomial(the second is only different from the first if the polynomial is not homogeneous), but I don't think it is true in general: if for example $f$ is coercive, the image of the gradient will include non zero points of $K$.

Some authors don't include this but I'm assuming that the definition of self-concordant includes that the interior of $K$ is the natural domain of $f$ (i.e. f goes to infinity in the boundary of $K$). Also, I'm using $K^* = \{ y | \forall x \in K: y^Tx \leq 0\}$, which I guess wikipedia calls the polar cone, but the difference is only a minus sign so it shouldn't be a problem.

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