The Natural Embedding J is an Open Map

Recall from The Natural Embedding J page that if $X$ is a normed space, then the natural embedding of $X$ into $X^{**}$ is the map $J : X \to X^{**}$ defined for all $x \in X$ by $J(x) = \hat{x}$, where $\hat{x}(f) = f(x)$ for all $f \in X^*$.

We will now use The Open Mapping Theorem to show that if $X$ is a Banach space, then $J$ is an open map.

Proposition 1: Let $X$ be a Banach space. Then $J$ is an open map.

• Proof: Since $X$ is a Banach space and since $X^{**}$ is always a Banach space, we have that $J : X \to X^{**}$ is a map between Banach spaces. Thus, $J$ is an open mapping if and only if $J(X)$ is (norm) closed subspace in $X^{**}$.

• Since $X^{**}$ is a Banach space, a subspace of $X^{**}$ is norm closed if and only if the subspace is a Banach space. So we will show that $J(X)$ is complete.

• Let $(T_n) \subset J(X)$ be a Cauchy sequence in $J(X)$. Then for every $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $m, n \geq N$ then $\| T_m - T_n \| < \epsilon$. Since $(T_n) \subset J(X)$, for each $n \in \mathbb{N}$ there exists an $x_n \in X$ such that $T_n = J(x_n)$. Thus, if $m, n \geq N$ we have that:

(1)

\begin{align} \quad \| J(x_m - x_n) \| = \| J(x_m) - J(x_n) \| < \epsilon \end{align}

• But $J$ is an isometry, and so $\| J(x_m - x_n) \| = \| x_m - x_n \|$ for all $m, n \in \mathbb{N}$, and in particular, if $m, n \geq N$ then:

(2)

\begin{align} \quad \| x_m - x_n \| < \epsilon \end{align}

• So $(x_n) \subset X$ is a Cauchy sequence in $X$. Since $X$ is a Banach space, $(x_n)$ converges to some $x \in X$. But since $J$ is continuous and $(x_n)$ converges to $x$ we have that $(J(x_n))$ converges to $J(x)$.

• Thus every Cauchy sequence in $J(X)$ converges, so $J(X)$ is a complete subspace of the Banach space $X^{**}$ and so $J(X)$ is closed, and consequentially by the Open Mapping Theorem, $J$ is an open map. $\blacksquare$

Corollary: Let $X$ be a Banach space. Then $J(X)$ is norm dense in $X^{**}$ if and only if $J$ is surjective.

This corollary tells us that a Banach space is reflexive if and only if $J(X)$ is norm dense in $X^{**}$.

• Proof: $\Rightarrow$ Suppose that $J(X)$ is norm dense in $X^{**}$. Then $\overline{J(X)} = X^{**}$. But by the previous proposition, $J(X)$ is norm closed in $X^{**}$ and so $J(X) = \overline{J(X)}$. So $J(X) = X^{**}$, i.e., $J$ is surjective.

• $\Leftarrow$ Suppose that $J$ is surjective. then $J(X) = X^{**}$. Since a set is always contained in its closure, consequentially, $\overline{J(X)} = X^{**}$, i.e., $J(X)$ is norm dense in $X^{**}$. $\blacksquare$