5
$\begingroup$

Let $M$ be a Hadamard manifold and let $F:M\to\mathbb{R}$ be a real-valued convex function on $M$. What would be the Fenchel-Young conjugate of $F$?

In general for a real locally convex vector space $V$ the Fenchel-Young conjugate (sometimes referred to as the Legendre transform) $F^*:V^*\to\overline{\mathbb{R}}$ of a functional (function, often convex) $F:V\to\overline{\mathbb{R}}$ (where $\overline{\mathbb{R}}:=\mathbb{R}\cup\{+\infty\}$) is defined to be $$F^*(v^*):=\sup_{v\in V}\{(v^*,v)-F(v)\}, v^*\in V^*$$ Here $V^*$ is the dual space of $V$. However the last definition depends on the linear structure of $V$ and when $V=M$ we are not guaranteed such a structure.

Related is also the question of how to define a dual problem on a Hadamard manifold $M$ given an optimization problem (the primal problem) to solve. While there is extensive research on optimization problems on manifolds, I haven't been successful in finding works which deal particularly with the dual problems on manifolds. Is there current/previous research in this direction?

Added: Following Ferreira and Oliveira. Let $M$ be a Hadamard manifold and $F:M\to\mathbb{R}$ a convex real valued function on $M$. Let $x\in M$ and $TM_x$ be the tangent space at $x$. A vector $u\in TX_x$ is a subgradient of $F$ at $x$ whenever $$F(y)\geqslant F(x)+\langle u,\exp_x^{-1}y\rangle,\forall y\in M$$ where $\exp_x(\cdot):TM_x\to M$ is the exponential map and this is a global diffeomorphism (well known result due to Hadamard). The above definition coincides with the usual one when $M=\mathbb{R}^n$ since then $\exp_x^{-1}y=y-x$. Folowing the lines of the definition of the subgradient I tried to define "some" Fenchel conjuguate for our $F$ as follows. Since we need to work in a space with some linear structure a good candidate would be to use tangent spaces. For that we need to fix some $z\in M$ and its corresponding tangent space $TM_z$. Then define $$F^*_z(u):=\sup_{y\in M}\{\langle u,\exp^{-1}_zy\rangle-F(y)\}$$ In the special case $M=\mathbb{R}^n$ it coincides with the shifted usual Fenchel conjugate in $\mathbb{R}^n$ i.e. $$F^*_z(u)=F^*(u)-\langle u,z\rangle=(F(y+z))^*$$ Obviously the disadvantage is that the definition depends on $z$ hence pointwise. It is desirable of course to get a definition independent of $z$ so that the special case $\mathbb{R}^n$ is also recovered.

$\endgroup$
10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.