Normed linear space and non-zero linear bounded functional 
Let $V$ be a normed linear space and $l$ be a non-zero linear bounded functional on $V$.
If $d=\inf\{\|v\|:l(v)=1\}$, show that $\|l\|=\frac{1}{d}$.

 A: Recall that the norm of a bounded linear functional $l$ is given by
$$ \| l \| = \sup_{\| u \| = 1} |l(u)|.$$
For any unit vector $u$ such that $l(u)\not=0$, it follows
$$l\left(\frac{u}{l(u)}\right)=\frac{l(u)}{l(u)}=1\implies \left\|\frac{u}{l(u)}\right\|\geq d\implies |l(u)|\leq \frac{1}{d}\implies \|l\|\leq \dfrac{1}{d}.$$
On the other hand, the definition of $d$ implies that for $\epsilon>0$ there is $v\in V$ such that $l(v)=1$ and $\|v\|\leq d+\epsilon$. Consider the unit vector $u:=v/\|v\|$ then
$$\|l\|\geq |l(u)|=\frac{|l(v)|}{\|v\|}\geq \frac{1}{d+\epsilon}.$$
Hence for all $\epsilon>0$,
$$\frac{1}{d+\epsilon}\leq \|l\|\leq \dfrac{1}{d}$$
and we may conclude that $\|l\|=1/d$.
A: You must have $\forall v \in V,|l(v)| \leq ||l||.||v||$. Without loss of generality you can assume that $l(v)=1$ (using the linearity of $l$ on the LHS and the absolute homogeneity of the norm on the RHS; of course in this case $||v|| \neq 0$). 
Your equation becomes $\forall v \in V, \frac{1}{||v||} \leq ||l||$ meaning that $||l|| = sup(\frac{1}{||v||}) = \frac{1}{inf(||v||)}$
