TITLE: Norm of a bounded linear functional as the reciprocal of the distance from zero to an hyperplane
QUESTION [5 upvotes]: Let $(X,\|\;\|)$ be a normed vector space over $K$ and let $f\in B(X,K)$ (bounded linear functional) $(f\neq0$). Let $L=\{x\in X: f(x)=1\}$. I want to prove that:
$$
\|f\|=\frac{1}{d(0,L)}
$$
My attempt goes like this:
Since $f$ is bounded, is clear that $|f(x)|\le\|f\|\|x\|\;\;\forall x\in X$. So,
$$
1\le\|f\|\|x\|\;\forall x\in L
$$
and,
$$
\frac{1}{\|f\|}\le\|x\|\;\;\forall x\in L
$$
thus
$$
\frac{1}{\|f\|}\le\inf_{x\in L}(\|x\|)=\inf_{x\in L}(d(0,x))=d(0,L)
$$
so
$$
\|f\|\ge\frac{1}{d(0,L)}
$$
Any ideas for the other part would be appreciated since I couldn't figure out how to show that $\|f\|\|x\|\le 1\;\;\forall x\in L$

REPLY [5 votes]: By definition of $\Vert f \Vert$, for all $\epsilon >0$, one can find $x$ with $\Vert x \Vert=1$ and $$\vert f(x) \vert \ge \frac{\Vert f \Vert}{1+\epsilon}$$
Now take $y=\frac{x}{f(x)}$. You have $f(y)=1$, hence by definition of $d(0,L)$, $\Vert y \Vert \ge d(0,L)$ or $$1=\Vert x \Vert \ge \vert f(x) \vert d(0,L) \ge \frac{\Vert f \Vert}{1+\epsilon} d(0,L)$$
As this is true for all $\epsilon >0$, you finally get $$1 \ge \Vert f\Vert d(0,L)$$ as desired.