Modifying Hahn-Banach to prove that any continuous form on a subspace can be extended to the whole space

Question

The usual Hahn-Banach theorem is stated something like this:

Let $E$ be a real vector space, and $f$ a linear form on a subspace of $E$, bounded by some sub-additive, positive-homogeneous function $p$. Then $f$ can be extended to a linear form on all of $E$, also bounded by $p$.

This result can be used to justify that there exist non-trivial continuous forms on certain kinds of vector spaces. But can't the proof of Hahn-Banach be modified slightly to directly prove the following?

Let E be a topological vector space, and $f$ a continuous linear form on a subspace of $E$. Then $f$ can be extended to a continuous linear form on all of $E$.

The proof would go something like this:

1. Say $f$ is defined on a subspace $F$, and let $u\not\in F$. Then we can extend $f$ to $F\oplus\mathbb R u$ by putting $\tilde f(x+\lambda u)=f(x)+\lambda \alpha$, where $\alpha$ is any real number. No matter what constant $\alpha$ is chosen, this will be continuous linear form.
2. Now let $\mathcal F$ be the set of all continuous extensions of $f$, ie all pairs $(g, U)$ where $g$ is a continuous linear form extending $f$ on a subspace $U$ containing $F$. Partially order them by saying one is greater than another if it's an extension of the other.
3. Every totally ordered subset of $\mathcal F$ has an upper bound since we can take the union of all the subspaces involved, which will be a subspace, and define a linear form on it by taking the "union" of the forms. This linear form is "obviously" continuous.
4. Therefore by Zorn's lemma there is a maximal element in $\mathcal F$, and if it weren't defined on all of $E$, we could extend it by using part (1).

I suspect that if there is a hole in the chain, it's the claim that the "union" function is continuous in step (3), but I don't even know where to begin constructing a counter example to show this claim is false.

Is this proof correct?

Answer 1

No. Your proof must be incorrect.

There are non-locally convex topological vector spaces where every continuous functional is zero, for example $L^p$ spaces for $0 < p < 1$. See also this remark.

I think, the problem in your argument is indeed that if you take a totally ordered subset of $\mathcal{F}$ and the union of all its elements, you obtain a linear map on this union, but there is no reason that it is continuous. If the union of these spaces carries the strict inductive limit topology, then it would be. However, the original topology might be completely different.