# Seeking help in understanding the proof of the Hahn-Banach extension theorem

by camio   Last Updated July 12, 2019 06:20 AM

I am trying to understand the proof of the extension theorem and have a small difficulty. I would appreciate it very much if you can kindly help me on it. As we know, a standard proof starts in the following way: Suppose $$M$$ is a subspace of a vector space $$X$$. $$x_0 \in X$$ and $$x \notin M$$. $$f$$ is a linear functional defined on $$M$$, satisfying

$$f(x) \leq p(x), \text{ }\text{ }\forall x \in M \text{ }\text{ } (1)$$

where $$p$$ is a sublinear functinonal.

The primary step, before the induction, is to show that there exists a functional (extended from $$f$$), defined on a strictly larger subspace, $$M_1$$, defined by $$M_1\doteq span\{m+\alpha x_0\}, \text{ } m\in M, \alpha \in R \text{ }\text{ } (2)$$

such that the functional is still bounded by $$p$$ on $$M_1$$.

I have no problem in understanding the above. But I have a question in a point in what follows. In some books, what follows is some argument like this:

If $$f$$ can be extended to $$M_1$$, then it must be true that

$$f(m+\alpha x_0)=f(m)+\alpha f(x_0) \text{ }\text{ }(3)$$

Then what we need is to show that there exists such a real constant $$f(x_0)$$ that would enable

$$f(m+\alpha x_0) \leq p(m+\alpha x_0) \text{ }\text{ } (4)$$

to be satisfied. Then what is remained, as we know, will be taking advantage of the RHS of equation (3) above to characterize the desired $$f(x_0)$$.

My issue is: In my opinion, in (3), $$f$$ is no longer the original $$f$$, for otherwise $$f(m+\alpha x_0)$$ would be illegal because $$m+\alpha x_0 \notin M$$ for $$\alpha \ne 0$$. Thus, how can we guarantee that the $$f(m)$$ on the RHS still satisfies (1), which is the base for the remained proof?

Therefore, I am more comfortable with an alternative approach employed by some other authors, like the following: Let

$$g(x)\doteq f(m) + \alpha g(x_0), \text{ }\text{ } x \in M_1, m \in M, \alpha \in R \text{ }\text{ } (5)$$

Then, we will try to find such a $$g(x_0)$$ that will satisfy

$$f(m) + \alpha g(x_0) \leq p(m+\alpha x_0), \text{ } \forall m \in M, \text{ } \forall \alpha \in R \text{ }\text{ } (6)$$

Based on the second approach, here is my understanding about how the "extend" can actually been carried out:

Given a $$f$$ defined on $$M$$ that satisfies (1), we can always extend it to a new linear functional, $$g$$, defined on $$M_1$$, in the form of (7) and by properly selecting a real constant $$c$$, such that $$g(x)$$ is bounded by $$p(x)$$ as well.

$$g(x) \doteq f(x-\alpha x_0) + \alpha c, \text{ } x\in M_1, \text{ } \alpha \in R \text{ }\text{ } (7)$$

Please comment and point out any mistake to help me understand. Thank you in advance!

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