Evaluation Linear Transformation on Normed Vector Space is Linear Transformation from Space to Second Normed Dual
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 $\struct {X, \norm \cdot_X}$.
Let $\map L {X^\ast, \Bbb F}$ be the set of linear functionals on $X^\ast$.
For each $x \in X$, define $x^\wedge : X^\ast \to \Bbb F$ by:
- $\map {x^\wedge} f = \map f x$
Define $J : X \to \map L {X^\ast, \Bbb F}$ by:
- $\map J x = x^\wedge$
for each $x \in X$.
Then:
- $J$ is a linear transformation $X \to X^{\ast \ast}$.
where $X^{\ast \ast}$ denotes the second normed dual.
Proof
We first show that $J$ is a linear transformation.
Let $x, y \in X$ and $\alpha, \beta \in \Bbb F$.
Then, we have, for each $f \in X^\ast$:
\(\ds \map {\paren {\map J {\alpha x + \beta y} } } f\) | \(=\) | \(\ds \map {\paren {\alpha x + \beta y}^\wedge} f\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map f {\alpha x + \beta y}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \alpha \map f x + \beta \map f y\) | Definition of Linear Functional | |||||||||||
\(\ds \) | \(=\) | \(\ds \alpha \map {x^\wedge} f + \beta \map {y^\wedge} f\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \alpha \map {\paren {\map J x} } f + \beta \map {\paren {\map J y} } f\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map {\paren {\alpha \map J x + \beta \map J y} } f\) |
So:
- $\map J {\alpha x + \beta y} = \alpha \map J x + \beta \map J y$
So $J$ is linear.
We now show that, for each $x \in X$, we have:
- $\map J x = x^\wedge \in X^{\ast \ast}$
where $X^{\ast \ast}$ is the second normed dual.
We first show that $x^\wedge$ is a linear functional for each $x \in X$.
Let $x \in X$, $f, g \in X^\ast$ and $\alpha, \beta \in \Bbb F$.
Then:
\(\ds \map {x^\wedge} {\alpha f + \beta g}\) | \(=\) | \(\ds \map {\paren {\alpha f + \beta g} } x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \alpha \map f x + \beta \map g x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \alpha \map {x^\wedge} f + \beta \map {x^\wedge} g\) |
so $x^\wedge$ is a linear functional for each $x \in X$.
We now show that $x^\wedge$ is bounded for each $x \in X$.
Let $x \in X$.
Then, for each $f \in X^\ast$, we have:
\(\ds \cmod {\map {x^\wedge} f}\) | \(=\) | \(\ds \cmod {\map f x}\) | Definition of Evaluation Linear Transformation on Normed Vector Space | |||||||||||
\(\ds \) | \(\le\) | \(\ds \norm f_{X^\ast} \norm x_X\) | Fundamental Property of Norm on Bounded Linear Functional |
So:
- $x^\wedge$ is a bounded linear functional.
That is:
- $\map J x \in X^{\ast \ast}$
$\blacksquare$
Sources
- 2020: James C. Robinson: Introduction to Functional Analysis ... (previous) ... (next) $26.1$: The Second Dual