# 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:

$\map J x \in X^{\ast \ast}$

$\blacksquare$