There is the theorem:
If $M$ is a subspace of a normed linear space $X$ and if $f$ is a bounded linear functional on $M$, then $f$ can be extended to a bounded linear functional $F$ on $X$ so that $||F||$ $=$ $||f||$.
There is the proof:
We first assume that $X$ is a real normed linear space and, consequently, that $f$ is a real-linear bounded functional on $M$ . If
$||f||$ $=$ $0$, the desired extension is $F$ $=$ $0$. Omitting this case, there is no loss of generality in assuming that $||f||$ $=$ $1$.
Choose $x_0$ $\in$ $X$, $x_0$ $\notin$ $M$, and let $M_1$ be the vector space spanned by $M$ and $x_0$.
Then $M_1$ consists of all vectors of the form $x$ $+$ $\lambda x_0$, where $x$ $\in$ $M$ and $\lambda$ is a real scalar. If
we define $f_1(x + \lambda x_0)$ $=$ $f(x)$ $+$ $\lambda \alpha$, where $\alpha$ is any fixed real number, it is trivial to verify that an extension of $f$ to a linear functional on $M_1$ is obtained.
The problem is to choose $\alpha$ so that the extended functional still has norm $1$.
This will be the case provided that
$|f(x) + \lambda \alpha |$ $\leq$ $|| x + \lambda x_0||$ ($x$ $\in$ $M$, $\lambda$ real).
Replace $x$ by $-\lambda x$ and divide both sides of $|f(x) + \lambda \alpha |$ $\leq$ $|| x + \lambda x_0||$ by $|\lambda|$. The requirement is then that
$|f(x) - \alpha|$ $\leq$ $||x-x_0||$ ($x$ $\in$ $M$),
i.e., that $A_x$ $\leq$ $\alpha$ $\leq$ $B_x$ for all $x$ $\in$ $M$, where
$A_x$ $=$ $f(x)$ $-$ $||x-x_0||$ and $B_x$ $=$ $f(x)$ $+$ $||x-x_0||$.
There exists such an $\alpha$ if and only if all the intervals $[A_x,B_x]$ have a common point, i.e., if and only if
$A_x$ $\leq$ $B_y$
for all $x$ and $y$ $\in$ $M$.
I have two questions :
$(1)$ - How is discussing the case where $||f||$ $=$ $1$ enough for the proof of the theorem?
$(2)$ - How do we conclude that There exists such an $\alpha$ if and only if all the intervals $[A_x,B_x]$ have a common point, i.e., if and only if
$A_x$ $\leq$ $B_y$ ?
Any help would be appreciated.
For (1), suppose we have solved the problem for functionals of norm $1$. Then we can apply that fact to $g := \frac{1}{\Vert f \Vert}f$ to get an extension $G$ of $g$, with $\Vert G \Vert = \Vert g \Vert = 1$. Then $F = \Vert f \Vert G$ is the desired extension of $f$.
For (2), there are two statements here.
First, the earlier statement is that we want $\alpha$ such that $|f(x)-\alpha|\leq\Vert x - x_0\Vert$. If $f(x) \geq \alpha$ that becomes $f(x) - \alpha \leq \Vert x - x_0 \Vert$, i.e., $\alpha \geq f(x) - \Vert x - x_0\Vert = A_x$. If $f(x) < \alpha$ then the original inequality is instead $\alpha - f(x) \leq \Vert x - x_0 \Vert$, which is $\alpha \leq f(x) + \Vert x - x_0 \Vert = B_x$. Thus the permissible $\alpha$'s are exactly those that are in $[A_x, B_x]$ for every $x$.
Next, why is it true that having a common point for the $[A_x, B_x]$'s is equivalent to $A_x \leq B_y$ for all $x, y$? In one direction, if $\alpha$ is in every interval $[A_x, B_x]$ then given any $x, y$ we have $A_x \leq \alpha$ (because $\alpha \in [A_x, B_x]$) and $\alpha \leq B_y$ (because $\alpha \in [A_y, B_y]$) so $A_x \leq B_y$.
For the other direction, suppose that $A_x \leq B_y$ for all $x, y$. I claim that the intersection of any finitely many intervals of the form $[A_x, B_x]$ is non-empty. To see this, suppose we are given $x_1, x_2, \ldots, x_n$. Let $y$ be such that $A_y = \max\{A_{x_k} : k=1,2, \ldots, n\}$ and let $z$ be such that $B_z = \min\{B_{x_k} : k=1, 2, \ldots, n\}$. Then $A_{x_k} \leq A_y$ for all $k$ and $B_{x_k} \geq B_z$ for all $k$, so for each $k$ we have $[A_y, B_z] \subseteq [A_{x_k}, B_{x_k}]$. Thus $[A_y, B_z] \subseteq \bigcap_{k=1}^n [A_{x_k}, B_{x_k}]$. Also, $A_y \leq B_z$ so $[A_y, B_z] \neq \emptyset$, meaning $\bigcap_{k=1}^n[A_{x_k}, B_{x_k}] \neq \emptyset$ as well. Now we know that $\{[A_x, B_x] : x \in X\}$ is a family of compact sets where every finitely many of them have non-empty intersection, and it follows from standard facts about compactness that this implies the whole intersection $\bigcap_{x \in X}[A_x, B_x]$ is non-empty.