This is a long comment rather than an answer.

In this paper of Klartag and Milman, the following operation on functions is defined and called the Asplund sum (as far as I can tell, that name has not been used elsewhere):

$$ f \star g (x) = \sup_y f(y)g(x-y). $$

So if $w^-(x) = w(-x)$, then your function is $w^* = w\star (1/w^-)$. Some properties of the $\star$ operation, including its relationship with the Legendre transform, are discussed in the paper. However, the class of functions of interest there is not closed under taking reciprocals, so the usefulness in your setting may be limited.

This is a long comment rather than an answer.

In this paper of Klartag and Milman, the following operation on functions is defined and called the Asplund sum (as far as I can tell, that name has not been used elsewhere):

$$ f \star g (x) = \sup_y f(y)g(x-y). $$

So if $w^-(x) = w(-x)$, then your function is $w^* = w\star (1/w^-)$. Some properties of the $\star$ operation, including its relationship with the Legendre transform, are discussed in the paper. However, the class of functions of interest there is not closed under taking reciprocals, so the usefulness in your setting may be limited.

Trivial update: since Nicolò has clarified that he is interested in even functions $w$, of course the above simplifies to $w^* = w\star (1/w)$.