Let $H$ be a Banach space and $C\subset H$ be a convex cone (I'm thinking of $H=L^2(U)$ or $C(U)$ and $C=\{\text{positive functions on }U\}$ for some open $U\subset\mathbb R^d$).
Moreover, let $H'\subset H$ a subspace and $\phi:H'\to\mathbb R$ be a functional which is
- continuous in the norm of $H$
- nonnegative on $H'\cap C$.
Does there exist an extension $\tilde\phi:H\to\mathbb R$ which is both bounded on $H$ and nonnegative on $C$?
Best Answer
This is not true in every infinite-dimensional Banach space.
Let $x^*$ be a linear functional on $H$ which is not continuous. We choose $C=\{x\in H : x^*(x)\ge 0\}$. Let $x\in C\subset H$ be such that $x^*(x)=1$. For $H'$ we choose a one-dimensional subspace generated by $x$. We then choose $\phi$ as the restriction of $x^*$ to $H'$.
It is clear that $\phi$ is continuous on $H'$ and nonnegative on $H'\cap C$.
Suppose there is a bounded extension $\bar\phi:H\to\mathbb R$ of $\phi$ such that $\bar\phi$ is nonnegative on $C$. Since $x^*$ is not continuous, there exists a sequence $x_n$ such that $x_n\to 0$ but $x^*(x_n)=1$. It follows that $x^*(x_n-x)=0$ and thus $x_n-x\in C$. Therefore, $\bar\phi(x_n-x)\geq0$. By continuity of $\bar\phi$ it follows that $\bar\phi(-x)\geq0$.
However, we also have $\bar\phi(-x)=-\phi(x)=-1$, which is a contradiction to $\bar\phi(-x)\geq0$.