This is what we need for the proof of the theorem:
There is the theorem:
Let $M$ be a linear subspace of a normed linear space $X$, and let $x_0$ $\in$ $X$. Then $x_0$ is in the closure $\bar M$ of $M$ if and only if there is no bounded linear functional $f$ on $X$ such that $f(x)$ $=$ $0$ for all $x$ $\in$ $M$ but $f(x_0)$ $\neq$ $0$.
There is the proof: If $x_0 \in \bar M$, $f$ is a bounded linear functional on $X$, and $f(x) = 0 $ for all $x \in M$, the continuity of $f$ shows that we also have $f(x_0)$ $=$ $0$.
Conversely, suppose $x_0$ $\notin$ $\bar M$. Then there exists a $\delta$ $\gt$ $0$ such that $||x-x_0||$ $\gt$ $\delta$ for all $x$ $\in$ $M$. Let $M'$ be the subspace generated by $M$ and $x_0$, and defined $f(x + \lambda x_0)=\lambda$ if $x\in$ $M$ and $\lambda$ is a scalar. Since
$\delta$$|\lambda|$ $\leq$ $|\lambda|$ $||x_0 + \lambda^{-1} x||$ $=$ ||$\lambda x_0 + x $||,
we see that $f$ is a linear functional on $M'$ whose norm is at most $\delta^{-1}$. Also $f(x)$ $=$ $0$ on $M$, $f(x_0) =1$. The Hahn-Banach theorem allows us to extend this $f$ from $M'$ to $X$.
I don't understand how do we have the inequality $\delta$$|\lambda|$ $\leq$ $|\lambda|$ $||x_0 + \lambda^{-1} x||$ and how we see that $f$ is a linear functional on $M'$ whose norm is at most $\delta^{-1}$.
I also don't understand how does the Hahn-Banach Theorem allows us to extend this $f$ from $M'$ to $X$.
Any help would be appreciated.
We're given $f(x+\lambda x_0) = \lambda$, a natural extension for $f$ from $M$ to $M'$. Using linearity of $f$, we have that $f(x+\lambda x_0) = f(x) + \lambda f(x_0) = \lambda \implies f(x_0) = 1$ (since $f(x) = 0$ on $M$). Moreover, $|f(x+\lambda x_0)| = |\lambda| \leq \delta^{-1}\|x + \lambda x_0\|$. Note that $x + \lambda x_0$ is the general form of a vector in $M'$, so we could equivalently say $|f(x')| \leq \delta^{-1} \|x'\|$ for any $x' \in M'$. Now using the definition of the norm of $f$ which is given by: $$\|f\| = \sup\{|f(x)|: x' \in M', \|x'\|\leq1\}$$ We get that $|f| \leq \delta^{-1}$ whenever $\|x'\| \leq 1$ so $\|f\|$ is at most $\delta^{-1}$, i.e $\|f\| \leq \delta^{-1}$. Now we've established an upper bound on the norm of $f$ on $M'$, therefore by the Hahn Banach theorem there is an extension of $f$ on $X$