# 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)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)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)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:

- 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)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)Hence $J(\alpha x) = \alpha J(x)$. 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 $x \in X$:

- So $J$ 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 $X^{**}$.