# Evaluation Linear Transformation on Normed Vector Space is Linear Isometry

## Theorem

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

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

Let $J : X \to X^{\ast \ast}$ be the evaluation linear transformation for $X$.

Then:

$J$ is a linear isometry.

## Proof

$J$ is a linear transformation.

It remains to show that:

$\norm {\map J x}_{X^{\ast \ast} } = \norm x_X$

for each $x \in X$.

For each $x \in X$, we denote:

$\map J x = x^\wedge$

We then aim to show:

$\norm {x^\wedge}_{X^{\ast \ast} } = \norm x_X$

for each $x \in X$.

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

for each $f \in X^\ast$.

So, we have:

$\norm x_X \in \set {c > 0 : \cmod {\map {x^\wedge} f} \le c \norm f_{X^\ast} \text { for all } f \in X^\ast}$

So, from the definition of the norm on bounded functionals and the definition of infimum, we have:

$\norm {x^\wedge}_{X^{\ast \ast} } \le \norm x_X$

We now aim to show that:

$\norm x_X \le \norm {x^\wedge}_{X^{\ast \ast} }$

From Existence of Support Functional, there exists $f \in X^\ast$ such that:

$\map f x = \norm x_X$

and:

$\norm f_{X^\ast} = 1$

Then we have:

 $\ds \cmod {\map {x^\wedge} f}$ $=$ $\ds \cmod {\map f x}$ $\ds$ $=$ $\ds \norm x_X$ $\ds$ $=$ $\ds \norm f_{X^\ast} \norm x_X$

So we have:

$\norm x_X \in \set {\cmod {\map {x^\wedge} f} : \norm f_{X^\ast} = 1}$

So from the definition of the norm on bounded functionals and the definition of supremum, we have:

$\norm {x^\wedge}_{X^{\ast \ast} } \ge \norm x_X$

So we have:

$\norm {\map {x^\wedge} f} = \norm x_X$

That is:

$\norm {\map J x}_{X^{\ast \ast} } = \norm x_X$

$\blacksquare$