# The Hahn-Banach theorem, and a (related?) result on extension of operators on dense subspaces

I have tried googling for an answer to this question, but have not found anything that I am able to read. I've also tried looking on MSE, but have found nothing.

The Hahn-Banach theorem says the following:

Given a seminorm $$p: V \to \mathbb{K}$$ and any linear subspace $$U \subset V$$ (not necessarily closed), any functional $$f' \in U^*$$ dominated by $$p$$ has a linear extension to $$f \in V^*$$.

There is another result on the extension of linear operators:

For $$E$$ and $$F$$ Banach spaces, and $$E_0$$ a dense subspace of $$E$$, any linear operator $$T' \in \mathcal{L}(E_0, F)$$ extends to a (unique) linear operator $$T \in \mathcal{L}(E, F)$$.

I wonder if there might be some unifying connection between these two results. The Hahn-Banach theorem requires a semi-norm (but in many cases we can use the norm itself), and requires linear functionals (the codomain has to be either the complex or real numbers). On the other hand, the "extension theorem" allows for any codomain that is complete, but requires the domain to be dense for an extension to occur.

The proofs of these results that I am familiar with are not related at all, so there is no immediate reason (to me) for why these should be connected. On the other hand, they are so 'nice' that I can't help but wonder if there might be some connection.

Since I have no idea if such a bridge exists, I've tagged this question as a soft question. Any insights into why these two results may or may not be connected, or a reference to some text that might discuss these in parallel, or anything related to them, would be sufficient as an answer.

Theorem: If $$X,Y$$ are metric spaces and if $$Y$$ is complete, then for any dense subset $$A \subset X$$ and uniformly continuous function $$f:A \to Y$$, there is a unique extension $$\tilde{f}:X \to Y$$ of $$f$$.
Can you see how this result implies the second result you stated? A bounded linear transformation $$T:A\to F$$ where $$A \subset E$$ is a dense subspace, $$E$$ is a normed space(need not be Banach) and $$F$$ is a Banach space, then $$T$$ is always uniformly continuous. This is because
$$$$\|Tx-Ty\| \leq \|T\|\|x-y\|$$$$
By the previous theorem $$T$$ extends to a unique linear transformation $$\tilde{T}:E \to F$$. With a bit of work you can also show that $$\|T\| = \| \tilde{T}\|$$. But the Hahn-Banach theorem is more powerful. It deals with extensions of bounded functions on arbitrary subspaces. In functional analysis, to study a normed space $$X$$, it is very important to study it's dual $$X^*$$ and the Hahn-Banach theorem is very useful in this regard.