User:Elroch/Notes

Assuming the well-ordering theorem, pick a well-ordered basis $$ \left\{\ v_i: i \in I \right\}\ $$ extending $$U\;$$ to $$V\;$$. If the theorem is false, there is a minimum element $$v_j\;$$ of this basis such that the functional $$\phi\;$$ may not be extended to the subspace generated by $$U\;$$ and $$ \left\{\ v_i: i \leq\ j \right\}\ $$

Hence it suffices to extend the functional to a space generated by a subspace $$W\;$$ (on which the condition holds) and a single extra basis element $$e\;$$.

Let $$y_1, y_2 \in W$$

By assumption,

$$ \phi(y_1-y_2)\leq p(y_1-y_2)\ $$

$$ \Rightarrow \phi(y_1-y_2)\leq p(y_1-e+(e-y_2)) $$

$$ \Rightarrow \phi(y_1-y_2)\leq p(y_1-e)+p(e-y_2) $$

$$ \phi(y_1)-p(y_1-e) \leq p(e-y_2)+\phi(y_2) $$

Since this is true for all $$y_1,y_2 \in W\;$$ we can pick $$c\in R\;$$ satisfying:

$$ \sup_{y_1 \in W}\left\{\phi(y_1)-p(y_1-e) \right\} \leq c \leq \inf_{y_2\in W}\left\{p(e-y_2)+\phi(y_2) \right\} $$

We set $$ \phi(e) = c \;$$

To complete the proof, we need to show that $$y\in W, \lambda\in R \Rightarrow \phi(y+\lambda e) \leq p(y+\lambda e)$$. We already know this if $$\lambda=0\;$$

Case 1: $$\lambda < 0\;$$

$$ \phi(y+\lambda e) = \phi(y) + \lambda c\; $$

$$\leq \phi(y) - \lambda p(y_1-e) + \lambda \phi(y_1) $$ for any $$y_1 \in W$$, by definition of $$c\;$$.

Set $$y_1 = -\frac{y}{\lambda}$$

$$\Rightarrow \phi(y+\lambda e) \leq -\lambda p(-\frac{y}{\lambda}-e) = p(y+\lambda e)$$

Case 2: $$\lambda > 0\;$$

$$ \phi(y+\lambda e) = \phi(y) + \lambda c \;$$

$$\leq \phi(y) + \lambda p(e-y_2)+\lambda \phi(y_2) $$ for any $$y_2 \in W$$, by definition of $$c\;$$.

Set $$ y_2 = -\frac{y}{\lambda}$$

$$ \Rightarrow \phi(y+\lambda e) \leq \lambda p(e+\frac{y}{\lambda}) = p(y + \lambda e )$$

QED