Self-concordant barrier for the epigraph of $f(x,y) = x^p y^{1-p}$?

Question

The problem

Assume $p > 1$. Consider the function $$f(x,y) = x^p y^{1-p}, \qquad x,y > 0.$$ Note that $$ f'' = p(p-1)x^{p-2}y^{-1-p} \begin{bmatrix} y \\ & x \end{bmatrix} \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \begin{bmatrix} y \\ & x \end{bmatrix}, $$ which is positive semidefinite. Thus, the following epigraph is convex: $$Q = \{ (x,y,s) \; : \; s > f(x,y), \qquad x,y > 0 \}.$$ What's a self-concordant barrier for $Q$?

Variations and discussion

More generally, I'm interested in the following function: $$ f(x,y) = \|x\|_2^p y^{1-p}, $$ and the corresponding epigraph.

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If you delete variable $y$ (i.e. set $y=1$), a well-known self-concordant barrier is: $$F(x,s) = -\log(s^{2 \over p} - \|x\|_2^2) - 2\log s .$$

I've written a quick and dirty numerical program that computes the relevant third order tensor norm (to see if it's self-concordant). I don't trust this program too much, it might have bugs, but it seems to tell me that the "obvious" modification below, is not self-concordant: $$ F(x,y,s) = -\log(s^{2 \over p} - \|x\|_2^2y^{{2 \over p}-2}) - 2\log s - 2\log y. $$

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An alternative would be to "model" the problem differently by introducing slack variables, etc..., but the epigraph of $f$ has so far eluded my attempts at this approach.

Answer 1

I always ask my questions too fast and then immediately find the answer.

Let $\alpha = 1/p \in (0,1]$. The set $Q$ can be characterized instead by $$ x < s^{\alpha}y^{1-\alpha}. $$

This is a convex cone, the barrier for this has been worked out in a thesis by Chares. Chares gives the following barrier: $$ F(x,y,s) = -\log(s^{2\alpha}y^{2 - 2\alpha} - x^2) - (1-\alpha)\log(s)-\alpha \log (y). $$

The parameter of the barrier is $3$. Several related barriers are conjectured, e.g. if $x$ is a vector.