The Canonical Embedding J is an Isometry

# The Canonical Embedding J is an Isometry

Recall from the Isometries on Normed Linear Spaces page that if $X$ and $Y$ are normed linear spaces then a linear map $T : X \to Y$ is said to be an isometry if:

(1)\begin{align} \quad \| T(x) \| = \| x \|, \quad \forall x \in X \end{align}

Recall that the canonical embedding of a normed linear space $X$ into the topological double dual $X^**$ is the function $J : X \to X^{**}$ defined for all $x \in X$ by:

(2)\begin{align} \quad J(x) = J_x \end{align}

Where $J_x : X^* \to \mathbb{C}$ is defined for all $\varphi \in X^*$ by:

(3)\begin{align} \quad J_x(\varphi) = \varphi(x) \end{align}

Then following theorem tells us that the canonical embedding $J$ is in fact as isometry.

Theorem 1: Let $X$ be a normed linear space. Then the canonical embedding $J : X \to X^{**}$ is an isometry. |

**Proof:**Let $x \in X$ and consider the linear functional $J_x : X^* \to \mathbb{C}$. Then for all $\varphi \in X^*$ we have that:

\begin{align} \quad \| J(x)(\varphi) \| = \| J_x(\varphi) \| = \| \varphi(x) \| \leq \| \varphi \| \| x \| = \| x \| \| \varphi \| \end{align}

- Let $M = \| x \|$ (observe that $\| x \|$ is fixed here). Then:

\begin{align} \quad \| J(x)(\varphi) \| \leq M \| \varphi \| \end{align}

- By choosing $\varphi \in X^*$ such that $\| \varphi \| = 1$ we see that:

\begin{align} \quad \| J(x) \| \leq \| x \| \quad (*) \end{align}

- Now for each $x \in X$ there is a $\varphi \in X^*$ such that $\varphi (x) = \| x \|$ and $\| \varphi \| = 1$. But $\varphi(x) = J(x)(\varphi)$. Hence:

\begin{align} \quad \| x \| \leq \| J(x) \| \quad (**) \end{align}

- From $(*)$ and $(**)$ we conclude that for each $x \in X$ we have that:

\begin{align} \quad \| J(x) \| = \| x \| \end{align}

- So $J$ is an isometry.

Corollary 2: Let $X$ be a normed linear space. Then the canonical embedding $J$ is continuous and injective. |

**Proof:**This follows immediately by the theorems on the page referenced above. $\blacksquare$