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!

Updated December 17, 2018 16:20 PM

Updated February 17, 2019 16:20 PM

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