Let $k$ be an algebraically closed complete non-archimedean field. Let $\mathcal{O}_k$ be its ring of integers. Suppose that $A$ is a $k$-Banach algebra, and $B$ is its closed unitary ball. Note that $B$ is a Banach $\mathcal{O}_k$-algebra.

$\textbf{Question 1:}$ Is it true that $\mathrm{Hom}_{\mathcal{O}_k}(A, \mathcal{O}_k)$ is nonempty? I think this is similar to the result for complex Banach algebras where the space of characters is nonempty. Here, the Hom is taken on continuous $\mathcal{O}_k$-algebra morphisms.

$\textbf{Question 2:}$ Suppose that for all $\sigma \in \mathrm{Hom}_{\mathcal{O}_k}(A, \mathcal{O}_k)$ we have that $\sigma(a) = 0$. Does this imply that $a = 0$?

Let $k$ be an algebraically closed complete non-archimedean field. Let $\mathcal{O}_k$ be its ring of integers. Suppose that $A$ is a $k$-Banach algebra, and $B$ is its closed unitary ball. Note that $B$ is a Banach $\mathcal{O}_k$-algebra.

$\textbf{Question 1:}$ Is it true that $\mathrm{Hom}_{\mathcal{O}_k}(B, \mathcal{O}_k)$ is nonempty? I think this is similar to the result for complex Banach algebras where the space of characters is nonempty. Here, the Hom is taken on continuous $\mathcal{O}_k$-algebra morphisms.

$\textbf{Question 2:}$ Suppose that for all $\sigma \in \mathrm{Hom}_{\mathcal{O}_k}(B, \mathcal{O}_k)$ we have that $\sigma(a) = 0$. Does this imply that $a = 0$?