Hahn Banach Theorem: Let $X$ be a normed Vector space and $U$ a subspace. For every linear continuous function $u':U \rightarrow \mathbb{K}$, there exists a continuous linear functional $x': X \rightarrow$ such that: $x'_{|_U}=u'$ and $\lVert x' \rVert=\lVert u' \rVert$
$\lVert T \rVert=sup_{\lVert x \rVert \leq 1} \lVert Tx \rVert$
Now I want to try this out on an example:
Consider the Subspace $S={(x_1,x_2,x_3): x_1+2x_2=0,x_3=0}$ of the Banach space $(\mathbb{R^3},\lVert.\rVert_1)$. With the functional $\phi((x_1,x_2,x_3)):=x_1$.
As far as I understood the proof, first I have to define $p(x)=\lVert u' \rVert \lVert x \rVert for x \in \mathbb{R}$ Then there exists a linear mapping $x':\mathbb{R}^3 \rightarrow \mathbb{R}$ such that $x'_{|_U}=u'$ and $x'(x)\leq p(x)$ forall $x \in \mathbb{R}$. Since we also have $x'(-x)\leq p(x)$ we get that $|x'(x)|\leq \lVert u' \rVert \lVert x \rVert forall x \in \mathbb{R}^3$, this means $\lVert x' \rVert \leq \lVert u' \rVert$ On the other hand, $\lVert u' \rVert=sup_{u \in B}|u'(u)|=sup_{u \in B}{|x'(u)|} \leq sup_{x \in B}|x'(x)|=\lVert x' \rVert$
First, I determine $p$ $p(x)=\lVert u' \rVert \lVert x \rVert_1 =sup_{\lVert x \rVert \leq 1}\lVert u'x \rVert |x|=|x_1|(|x_1|+|x_2|+|x_3|)$
Now we need to choose (I think) $x'$ such that $\lVert x' \rVert = \lVert u' \rVert$
$\lVert x' \rVert=sup_{\lVert x \rVert_1 \leq 1}\lVert x'(x) \rVert=|x_1|(|x_1|+|x_2|+|x_3|)=\lVert u' \rVert$
Now my question is, how do I exactly extend $u'$ to $x'$?
Isn't x' for example $x'(x_1,x_2,x_3)=x_1 x_2$
But shouldn't also $x'(x_1,x_2,x_3)=x_1 x_3$ work? Why are there two extensions? Is my calculation/understanding wrong?