Take $(\mathbb{R}^d , \mathfrak{B})$ as measurable space, and denote $\mathcal{M}(\mathbb{R}^d)$ be the linear space of all bounded signed measure. This is known to be a Banach space w.r.t. the total variation norm defined by
$$||\mu|| = \sup_{B \in \mathfrak{B}} (|\mu(B)|+|\mu(B^c)|)$$ Does there exists a closed subspace, i.e. a Banach space $P\subset\mathcal{M}(\mathbb{R}^d)$ that is $\zeta$-convex, in the sense that there exists a biconvex function $\zeta : P \times P \rightarrow \mathbb{R}$ such that
- $\zeta(0,0)>0$
- $\zeta(\mu,\nu)\leq ||\mu+\nu||$ if $||\mu||=||\nu||=1?$
(A function $\zeta$ is said to be biconvex on $P\times P$ if $\zeta(\mu,\cdot)$ and $\zeta(\cdot, \nu)$ are both convex on $P$ for all $\mu,\nu \in P$)
I've tried to do this:
Take $\hat{P}\doteqdot\lbrace \mu \in \mathcal{M}(\mathbb{R}^d) : 0\leq ||\mu||\leq 1\rbrace$ and consider $\hat{\zeta}(\mu,\nu)\doteqdot 1 + \frac{||\mu||}{2}+\frac{||\nu||}{2}$
Indeed this $\hat{\zeta}$ is actually a biconvex function s.t.
- $\zeta(0,0) = 1>0$
However the 2. property is not attained since $1 + \frac{||\mu||}{2}+\frac{||\nu||}{2}$ is not $\leq ||\mu +\nu||$ when $||\mu||=||\nu||=1 $ (??).
I'm looking for an hint or for a converse result.