The Natural Embedding J

# The Natural Embedding J

Let $(X, \| \cdot \|_X)$ be a normed linear space. Recall that the dual of $X$ is denoted $X^*$ and is the set of all bounded linear functionals from $X$ to $\mathbb{R}$ equipped with the operator norm. We noted that $X^* = \mathcal B(X, \mathbb{R})$ and since $\mathbb{R}$ is a Banach space, so is $X^*$. We can treat $X^*$ as its own normed linear space, and consider its dual, $(X^*)^* = X^{**}$, which is the set of bounded linear functionals from $X^*$ to $\mathbb{R}$.

 Definition: Let $(X, \| \cdot \|_X)$ be a normed linear space. The Second Dual or Bidual of $X$ is denoted $X^{**}$ and is the set of bounded linear functionals from $X^*$ to $\mathbb{R}$.

Let $(X, \| \cdot \|_X)$ be a normed linear space. For each $x \in X$ define $\hat{x} : X^* \to \mathbb{R}$ for all $f \in X^*$ by:

(1)
\begin{align} \quad \hat{x}(f) = f(x) \end{align}

That is, $\hat{x}$ evaluates each bounded linear functional $f \in X^*$ at $x$. For each $x \in X$ we have that $\hat{x}$ is linear since for all $f, g \in X^*$ and all $\alpha \in \mathbb{R}$:

(2)
\begin{align} \quad \hat{x}(f + g) &= (f + g)(x) = f(x) + g(x) = \hat{x}(f) + \hat{x}(g) \quad \hat{x}(\alpha f) &= (\alpha f)(x) = \alpha f(x) = \alpha \hat{x}(f) \end{align}

So each $\hat{x}$ is a linear functional from $X^*$ to $\mathbb{R}$. Furthermore, for each $x \in X$, let $M = \| x \|$. Then for all $f \in X^*$:

(3)
\begin{align} \quad |\hat{x}(f)| = |f(x)| \leq \| f \| \| x \| = M \| f \| \end{align}

So each $\hat{x}$ is a bounded linear functional from $X^*$ to $\mathbb{R}$. In other words, for all $x \in X$ we have that $\hat{x} \in X^{**}$. The following proposition tells us what the operator norm of $\hat{x}$ is.

 Proposition 1: Let $(X, \| \cdot \|_X)$ be a normed linear space. Then for all $x \in X$, $\| \hat{x} \| = \| x \|_X$.
• Proof: As we saw above, $\| \hat{x} \| \leq \| x \|_X$.
• For the opposite inequality, we will use one of the Corollaries to the Hahn-Banach Theorem. Recall that if $(X, \| \cdot \|_X)$ is a normed linear space and $x_0 \in X$ then there exists an $F \in X^*$ with $F(x_0) = \| x_0 \|_X$ and $\| F \| = 1$. Fix $x \in X$. Then there exists an $F \in X^*$ with $F(x) = \| x \|_X$ and $\| F \| = 1$. Hence:
(4)
\begin{align} \quad \| \hat{x} \| = \sup_{f \in X^*, \| f \| = 1} |\hat{x}(f)| \geq |\hat{x}(F)| = |F(x)| = \| x \|_X \end{align}
• Hence $\|\hat{x} \| = \| x \|_X$. $\blacksquare$
 Definition: Let $(X, \| \cdot \|_X)$ be a normed linear space. The Natural Embedding of $X$ into $X^{**}$ is the map $J : X \to X^{**}$ defined for all $x \in X$ by $J(x) = \hat{x}$.

Note that $J$ is linear operator from $X$ to $X^{**}$. To see this, let $x_1, x_2 \in X$. Then $J(x_1 + x_2) = \hat{x_1 + x_2}$ and $J(x_1) + J(x_2) = \hat{x_1} + \hat{x_2}$. We want to show that these functionals agree on $X^*$. Let $f \in X^*$. Then:

(5)
\begin{align} \quad \hat{x_1 + x_2}(f) = f(x_1 + x_2) = f(x_1) + f(x_2) = \hat{x_1}(f) + \hat{x_2}(f) \end{align}

Hence $J(x_1 + x_2) = J(x_1) + J(x_2)$.

Now let $x \in X$ and $\alpha \in \mathbb{R}$. Then $J(\alpha x) = \hat{\alpha x}$ and $\alpha J(x) = \alpha \hat{x}$. Again, we want to show that these functionals agree on $X^*$. Let [[f \in X^*]]. Then: (6) \begin{align} \quad \hat{\alpha x}(f) = f(\alpha x) = \alpha f(x) = \alpha \hat{x}(f) \end{align} HenceJ(\alpha x) = \alpha J(x)$. So$J$is linear. The following corollary tells us that$J : X \to X^{**}$is an isometry.  Corollary 2: Let$(X, \| \cdot \|_X)$be a normed linear space. Then the natural embedding$J : X \to X^{**}$is an isometry. • Proof:$J$is a linear operator from$X$to$X^{**}$from above, and by proposition$1$that for all$x \in X: (7) \begin{align} \quad \| J(x) \| = \| \hat{x} \| = \| x \|_X \end{align} • SoJ$preserves the norm on$x$, that is,$J$is a isometry.$\blacksquare$ Corollary 3: Let$(X, \| \cdot \|_X)$be a normed linear space. Then the natural embedding$J : X \to X^{**}$is injective. • Proof: Isometries are injective and so by corollary 2,$J$is injective. Alternatively, let$x_1, x_2 \in X$and suppose that$J(x_1) = J(x_2)$. Then$J(x_1 - x_2) = 0$. So$\| J(x_1 - x_2) \| = 0$. But then$\| x_1 - x_2 \|_X = 0$which happens if and only if$x_1 = x_2$, so$J$is injective. Corollary 3 tells us that$J$embeds$X$into$X^{**}$in some sense. So in some sense,$X$is a subspace of$X^{**}\$.