We proved the following version of the Hahn-Banach extension theorem in a course I'm taking:
Theorem (Hahn-Banach): Let $X$ be a real vector space and $q : X \to \mathbb{R}$ be sublinear. Let $U \subseteq X$ be a linear subspace and $l : U \to \mathbb{R}$ be a linear functional that satisfies $l(u) \leq q(u)$ for all $u \in U$. Then there exists a linear extension $L : X \to \mathbb{R}$ of $l$ that satisfies $L(x) \leq q(x)$ for all $x \in X$.
Is it true that this Hahn-Banach extension $L$ is also continuous, hence a $\mathbb{R}$-linear functional?
The way we use the theorem later in this course seems to rely on getting a functional out of the extension. Additionally, the Wikipedia entry states that Hahn-Banach yields a linear functional.
Some of my thoughts/work so far:
I see that, if $q$ is continuous, then $L$ is also continuous: Let $(x_n) \subseteq X$ with $x_n \to 0$. Then $$ L(x_n) \leq q(x_n) \to 0 $$ $$ -L(x_n) = L(-x_n) \leq q(-x_n) \to 0 $$ So $|L(x_n)| \leq \max\{q(x_n), q(-x_n)\} \to 0$. Hence, $L$ is continuous at $0$, and therefore continuous.
But a sublinear function isn't necessarily continuous... so this doesn't help, right?
From what I showed above, we see that $$ -q(-x) \leq L(x) \leq q(x) \qquad \forall x \in X. $$ Not sure if this inequality might help.
The formulation of Hahn-Banach that you have does not require any topology on $X$. One obvious application is the case where $X$ does have a norm and you use the theorem to extend functionals while preserving the norm.
But the form with the seminorm $q$ is the one that allows one to obtain the geometric form of Hahn-Banach (i.e., the separation theorems) which is often the most useful.