I am not quite sure whether I understand the statement of the theorem or not. The following is my understanding so far.
Let $p: V \to \mathbb{R} $ is a sub-linear function. If $\phi : U \to \mathbb{R}$ is a linear functional on a linear subspace $U \subset V$ which is dominated by p on U, then there exists a linear extension $\psi : V \to \mathbb{R}$ of $\phi$ to the whole space $V.$
The theorem essentially answers the question: "When we can extend a functional?" We simply need to impose a restriction/condition on our baby functional (functional on the subspace), and the condition is: this baby functional should be dominated by a sub-linear function on the subspace $\mathbb{U}.$
For a $\mathbb{C}$– vector space, the theorem needs a little bit of modification on the hypothesis, and that is: the sub-linear function is replaced by a semi-norm on the vector space $V.$ In fact, this is a natural choice for the replacement because $\mathbf{C}$ is not an ordered field like $\mathbb{R}.$
We also note that the Hahn-Banach Theorem is really general. It does not require that $V$ is a normed linear space. I can appreciate this much.
I have heard that this is one of the three main pillars of analysis including the open mapping theorem and the uniform bound theorem. Once my professor told that the Hahn Banach theorem has many many consequences.
What are some obvious consequences that can be derived from the theorem without much work? By obvious I mean it just requires a two to four sentences' argument to derive such a consequence/corollary.
Best Answer
A simple corollary which I studied, for example,is that for every $x\in V$ you have that $\left\lVert x\right\rVert_V = \sup_{f\in B_{V'}} |f(x)|=\max_{f\in B_{V'}} |f(x)|$, which is nice since in general the "$\sup$" is not achieved.
Anyway I'll let more experienced mathematicians answering you better, for instance you could try to prove this ;)