Q:

Any idea with this problem of distances???

Let $E$ a normed linear space and $H$ the closed hyperplane $H=\ker f$, where $f\in L(E,\mathbb{R})$, $f\not\equiv 0$. Show that if $a\in E$ then $$d(a,H)=\frac{|f(a)|}{||f||}$$
And the problem have a second part: Let $E$ the space of all $x_n\to 0$. With $\displaystyle||(x_n)||=\sup_{n\in\mathbb{N}}|x_n|$. Let $H$ the closed hyperplane $H=f^{-1}(\{0\})$, with $$f(x)=\sum_{n=0}^{\infty}\frac{x_n}{2^n}$$ if $x=(x_n)$. Show that if $a\notin H$ then there is no $b\in H$ such that $d(a,H)=||a-b||$.
PD: i'm stuck in the two parts, any help. I try to write the definition $d(a,H)=\inf \{||a-x||:x\in H\}$ but no have more ideas! 

A:

Here is a hint for the first part. I'll assume that $a \notin H$. By a corollary of the Hahn-Banach Theorem, there exists $x^* \in L(E,\mathbb{R}) = E^*$ such that:

$H \subset \ker x^*$;
$x^*(a) = d(a,H)$;
$\|x^*\|=1$.

By a well-known result, condition 1. implies that $x^*$ is linearly dependent on $f$, i.e. for some $k \in \mathbb{R}$ we have $x^*=k f$. Now condition 3 implies that $k = \pm 1/\|f\|$, and condition 2 yields $$d(a,H) = x^*(a) = k f(a) = \pm \frac{f(a)}{\|f\|}.$$ It is now easy to conclude.

