# Norm in a dual space

If $f \in X^*$, with $X^*$ the dual space consisting of all linear bounded functionals on a linear normed space $X$. With the norm defined as $||f||_{X^{*}} = \sup_{||x|| \leqslant 1} |f(x)|$. Why does $|f(x)| \leqslant ||f||_{X^{*}} ||x||_{X}$ hold?

-

Suppose $x\in X$ with $x\neq 0$. Then $y = x/\|x\|$ satisfies $\|y\| = 1$, and hence $|f(y)|\leq \|f\|$. But $f(y) = f(x/\|x\|) = f(x)/\|x\|$, so $|f(x)|/\|x\|\leq \|f\|$. This proves the inequality $|f(x)|\leq \|f\|\|x\|$ when $x\neq 0$. The inequality is trivial with $x = 0$.