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
From Evaluation Linear Transformation on Normed Vector Space is Linear Transformation from Space to Second Normed Dual, we have that:
- $J$ 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 $x \in X$.
As shown in Evaluation Linear Transformation on Normed Vector Space is Linear Transformation from Space to Second Normed Dual, we have:
- $\cmod {\map {x^\wedge} f} \le \norm f_{X^\ast} \norm x_X$
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:
- $\norm {\map J x}_{X^{\ast \ast} } = \norm x_X$
$\blacksquare$