The Hahn-Banach theorem goes like
If $p:X\to\mathbb R$ is sublinear and $f$ is a linear functional on a subspace $Y\subseteq X$ bounded by $p$ (on $Y$), then there is an extension of $f$ to $X$ that is bounded by $p$ and matches with $f$ on $Y$.
I am trying to do something similar, let $\psi:X\to2^\mathbb R$ be such that
- for any $x\in X$, $\psi(x)$ is a non empty-closed interval
- for any $x\in X$, $a\in \mathbb R$, $\psi(ax)=a\psi(x)$
- for any $x,y\in X$, $\psi(x+y)\subseteq \psi(x)+\psi(y)$
One can check that $x\to \sup \psi(x)$ and $x\to -\inf \psi(x)$ are both sublinear functions (over the extended real line). I want to know if the following holds :
If $f$ is a linear function on a subspace $Y\subseteq X$ such that $f(x)\in\psi(x)$ for any $x\in Y$, then there is an extension of $f$ to $X$ such that $f(x)\in\psi(x)$ for all $x\in X$.
In a sense this is a version of Hahn-Banach where the function is sandwiched between a sublinear and a (forgive the abuse of word) superlinear function.
I have never seen anything like that, but I have this feeling that Hahn Banach is actually already enough to show that, I was thinking of using $p(x)=\sup \psi(x)-\inf\psi(x)$ as a semi-norm but it is hard for me to find a new $f'$ that is less than $p$ that would correspond one to one to $f$ in some way. Any thought is welcome.
For the context I am trying to make progress on this question, however I think this new question has independent interest, the formulation is pleasant to me.
Actually since $-\inf\psi(x)=\sup-\psi(x)=\sup\psi(-x)$, if $g$ is linear such that for all $x\in X$, $g(x)\leq \sup\psi(x)$, then $g(-x)\leq \sup\psi(-x)$ which can be rewritten as $-g(x)\leq -\inf\psi(x)$ and therefore $\inf\psi(x)\leq g(x)\leq \sup\psi(x)$, by closedness of $\psi(x)$, we get $g(x)\in \psi(x)$. Because of that, if $p:x\to\sup\psi(x)$ takes values in the reals then the Hanh-Banach theorem yields exactly the function that I am looking for. If some values of $p$ are infinite, I think that the problem is supposed to be even simpler (even though I am not sure on how to argue in this case).
Any review would be very welcome.