Let E be a normed vector space and u be a linear, continuous form on E with kernel H.
We show that: $\forall x \in E \setminus H, \| u \| \le \frac{|u(x)|}{d(x, H)}.$
$\forall \lambda \in \mathbb{R} \setminus \{0\}, \forall y \in H,$
$\frac{|u(\lambda x + y)|}{\|\lambda x + y\|} = \frac{|\lambda||u(x)|}{|\lambda|\|x+\frac{1}{\lambda}y\|} = \frac{|u(x)|}{\|x+\frac{1}{\lambda}y\|} \le \frac{|u(x)|}{d(x,H)}.$
My question is on the last inequality (in particular, the denominator), comparing the norm on E and the distance d(x,H):
So, if I am correct:
$d(x,H) \le d(x,\frac{1}{\lambda}y) = \|x-\frac{1}{\lambda}y\| \le \|x+\frac{1}{\lambda}y\|$
So here, to compare the distance and the norm on E, we assume that $\|w \|:= d(0,w)$, $\forall w \in E$ ; that is the norm on the normed vector space E is the distance from a point w and the origin 0? (My confusion is E is a general normed vector space). Can someone please elaborate, thanks.
You are definitely on the right track, simply adding a minus will solve the problem: $$d(x,H)\leq d(x,-\frac1{\lambda}y)=\|x+\frac1{\lambda}y\|$$