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 $f \in X^\ast$:

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:

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:

So:


 * $x^\wedge$ is a bounded linear functional.

That is:


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