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 $x_0 \notin M$. $f$ 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 will serve as the basis for us to characterize the proper $f(x_0)$?
In other words, it seems to me that the above approach is deducing the extended functional from the original functional, which I feel is unwarrented.
I am more comfortable with an alternative approach employed by some other authors, like the following: Let
$$g(x)\doteq f(m) + \alpha c, \text{ }\text{ } x \in M_1, m \in M, \alpha \in R \text{ }\text{ } (5)$$
Note that the key difference from the first approach is that here we prescribe the form of the extended functional rather than deduce it from the original one.
Then, we will need to show that there exists a constant $c$ that will enable this prescribed form of the extended functional to do the job, i.e. to satisfy:
$$ f(m) + \alpha c \leq p(m+\alpha x_0), \text{ } \forall m \in M, \text{ } \forall \alpha \in R \text{ }\text{ } (6)$$
Surely to prove the existence of (or equivalently to characterize) such a proper $c$ will have to play with the property (1) and is a major part of the proof.
BTW, I think the extension theorem also provides a specific means to extend a linear functional $f$ on a subspace to a larger subspace:
Given a linear functional, $f$, defined on a subspace $M$ that satisfies (1), we can always extend it to a linear functional, $g$, defined on a strictly larger subspace $M_1$, in the form of (7) and by properly selecting a real constant $c$, such that $g(x)$ is bounded by $p(x)$ as well.
$$g(x) \doteq f(x-\alpha(x;x_0) x_0) + \alpha(x;x_0) c, \text{ } x\in M_1 \text{ }\text{ } (7)$$
Note that, in the above, $\alpha$ is a functional from $M_1$ to $R$, parameterized by the given vector $x_0$.
Please comment and point out any mistake to help me understand. Thank you in advance!
In 3) the symbol $f$ on LHS is the extended function that is being defined using the given function $f$ on $M$ and $f$ on RHS is the original function on $M$. So $f(x) \leq p(x)$ is satisfied.
In your second approach you are just saying that by properly selecting the constant $c$ we can make sure that $g(x) \leq p(x)$ on $M$. But how? That is where a major part of the proof lies.