Fundamental Property of Norm on Bounded Linear Functional

Theorem

Let $\Bbb F \in \set {\R, \C}$.

Let $\struct {X, \norm \cdot_X}$ be a normed vector space over $\Bbb F$.

Let $\struct {X^\ast, \norm \cdot_{X^\ast} }$ be the normed dual of $\struct {X, \norm \cdot_X}$.

Then for all $x \in X$ and $f \in X^\ast$:

$\cmod {\map f x} \le \norm f_{X^\ast} \norm x_X$

Proof

Let $x \in X$ and $f \in X^\ast$.

If $x = 0$, the claim is trivial, since by the linearity of $f$:

$\cmod {\map f x} = \cmod {\map f 0} = 0$

Let $x \ne 0$.

Recall by Definition 3 of Norm of Bounded Linear Functional:

$\norm f_{X^\ast} = \sup \set {\dfrac {\cmod {\map f x} } {\norm x_X} : \forall x \in X, x \ne 0}$

Thus, by $\paren 1$ of Definition of Supremum:

$\cmod {\map f x} = \dfrac {\cmod {\map f x} }{\norm x_X} {\norm x_X} \le \norm f_{X^\ast} \norm x_X$

$\blacksquare$