Let $\mathbb{K}$ be a field, either $\mathbb{R}$ or $\mathbb{C}$, it is well-known that for any Banach space $(A,||-||_A)$ the injection $J : A \to A^{\ast\ast}$ is an isometry.
I am interesting in the generalisation of this fact to extended seminorms. I am principally interested in the finite-dimensional case so I haven't really thought about completeness, feel free to assume completeness property where needed.
To go from norms to extended seminorms (I will call those esn) we allow for non-zero vectors to have esn 0 and we allow the esn of a vector to be infinite.
So an esn is a function $||-||_A : A \to [0,+\infty]$ such that:
- $||0||_A = 0$
- $||\lambda x||_A = |\lambda|||x||_A$ for any $\lambda \in \mathbb{K}$ and any $x \in A$ with the convention $0 \times \infty = 0$
- $||x + y||_A \leq ||x||_A + ||y||_A$ for any $x,y \in A$
Given two esn vector spaces $(A,||-||_A)$ and $(B,||-||_B)$ we can define the operator esn $||-||_{A \multimap B}$ on $\mathcal{L}(A,B)$ by $||f||_{A \multimap B} := \sup\limits_{x \neq 0} \frac{||f(x)||_B}{||x||_A}$ with the convention that $0 \times \infty = 0$ and $\frac{0}{0} = 0$ (so $\frac{0}{x} = 0$ for all $x \in [0,+\infty]$). In particular we can define the dual norm $||\varphi||_{A^\ast} := \sup\limits_{x \neq 0} \frac{|\varphi(x)|}{||x||_{A}}$.
What I want to prove (or disprove) is that $||J(x)||_{A^{\ast\ast}} = ||x||_A$ for any $x \in A$. Proving that $||J(x)||_{A^{\ast\ast}} \leq ||x||_A$ for any $x \in A$ can be done in the same way that in the case of a norm. The harder part is to prove that $||x||_A \leq ||J(x)||_{A^{\ast\ast}}$. There is a version of the Hahn-Banach theorem for esn that can be found in the article Extended seminorms and extended topological vector spaces by David Salas and Sebastián Tapia-García. For any $x \in A$ such that $||x||_A \neq \infty$ we can define a subspace $M = \{\lambda x | \lambda \in \mathbb{K}\}$ and a functional $\alpha : M^\ast \to \mathbb{K}$ by $\alpha(\lambda x) = \lambda ||x||_A$ and extend it using Hahn-Banach to a functional on $A^\ast$. So we get $||J(x)||_{A^{\ast\ast}} \geq ||x||_A$ following the usual proof. However when $||x||_A = \infty$ we cannot define the functional $\alpha$. So I am not sure how to prove it.
I have tried to unfold the definition: To prove that $||J(x)||_{A^{\ast\ast}} = \infty$ we would need that $\sup\limits_{f \in A^\ast} \frac{|f(x)|}{||f||_{A^\ast}} = \infty$, i.e. that there exists $f \in A^{\ast}$ such that $f(x) \neq 0$ and $||f||_{A^\ast} = 0$. So we need to find $f \in A^\ast$ such that $f(x) \neq 0$ and for any $x' \in A$ if $f(x') \neq 0$ then $||x'||_A = \infty$. I am not sure how to build such an $f$.
Any input would be appreciate, thanks.