Goldstine's Theorem

# Goldstine's Theorem

Lemma 1: Let $X$ be a normed linear space and let $B_{X^{**}}$ be the closed unit ball of $X^{**}$. Then $B_{X^{**}}$ is weak-* closed. |

**Proof:**By definition we have that:

\begin{align} \quad B_{X^{**}} = \{ \varphi \in X^{**} : \| \varphi \| \leq 1 \} \end{align}

- Let $(\varphi_n)$ be a sequence in $B_{X^{**}}$ that weak-* converges to $\varphi$. Then for all $f \in X^*$ we have that:

\begin{align} \quad \lim_{n \to \infty} \varphi_n(f) = \varphi(f) \end{align}

- So for each $f \in X^*$ we have that:

\begin{align} \quad \| \varphi (f) \| = \| \lim_{n \to \infty} \varphi_n(f) \| = \lim_{n \to \infty} \| \varphi_n(f) \| \leq \| \varphi_n \| \| f \| \leq 1 \| f \| = \| f \| \end{align}

- Thus $\| \varphi \| \leq 1$ which shows that $\varphi \in B_{X^{**}}$. Thus $B_{X^{**}}$ is weak-* closed. $\blacksquare$

Theorem 2 (Goldstine's Theorem): Let $X$ be a normed linear space. Let $B_X$ be the closed unit ball of $X$ and let $B_{X^{**}}$ be the closed unit ball of $X^{**}$. If $J : X \to X^{**}$ is the natural embedding of $X$ into $X^{**}$ then the weak-* closure of $J(B_X)$ is $B_{X^{**}}$. |

*Goldstine's theorem tells us that $J(B_X)$ is weak-* dense in $B_{X^{**}}$.*

**Proof:**Recall that the natural embedding $J$ is an isometry. Thus, if $x \in B_X$ then $\| x \| \leq 1$, and so:

\begin{align} \quad \| J(x) \| = \| \hat{x} \| = \| x \| \leq 1 \end{align}

- Thus $J(x) \in B_{X^{**}}$. Therefore $J(B_X) \subseteq B_{X^{**}}$. From Lemma 1 we have that $B_{X^{**}}$ is weak-* closed. Therefore:

\begin{align} \quad \mathrm{cl}_{\mathrm{wk}-*} (J(B_X)) \subseteq B_{X^{**}} \end{align}

- Now since $B_X$ is convex and $J$ is linear we have that $J(B_X)$ is convex. Since $X^{**}$ with the weak-* topology is a locally convex topological vector space we have from the theorem on The Closure of a Convex Set is Closed in a LCTVS page that $\mathrm{cl}_{\mathrm{wk}-*} (J(B_X))$ is convex.

- Suppose that $\mathrm{cl}_{\mathrm{wk}-*} (J(B_X)) \neq B_{X^{**}}$. Then there exists a $\varphi \in B_{X^{**}} \setminus \mathrm{cl}_{\mathrm{wk}-*} (J(B_X))$. Since $\varphi \in B_{X^{**}}$ there exists an $x \in B_X$ such that $\varphi = \hat{x}$.

- Since $\mathrm{cl}_{\mathrm{wk}-*} (J(B_X))$ is nonempty, weak-* closed, and convex, and $\varphi \in B_{X^{**}} \setminus \mathrm{cl}_{\mathrm{wk}-*} (J(B_X))$, by one of The Separation Theorems there exists a continuous linear functional $T$ on $\mathrm{cl}_{\mathrm{wk}-*} (J(B_X))$ with $\| T \| = 1$ such that:

\begin{align} \quad T(\varphi) < \inf_{\psi \in \mathrm{cl}_{\mathrm{wk}-*} (J(B_X))} T(\psi) \end{align}

- We have proven that $\mathrm{cl}_{\mathrm{wk}-*} (J(B_X)) \subseteq B_{X^{**}}$. In general, if $A \subseteq B$ then $\inf_{a \in A} a \leq \inf_{b \in B} b$. Therefore:

\begin{align} \quad T(\varphi) < \inf_{\psi \in \mathrm{cl}_{\mathrm{wk}-*} (J(B_X))} T(\psi) \leq \inf_{\psi \in J(B_X)} T(\psi) = -1 \end{align}

- But then $\| T \| > 1$, a contradiction. Therefore:

\begin{align} \quad \mathrm{cl}_{\mathrm{wk}-*} (J(B_X)) = B_{X^{**}} \quad \blacksquare \end{align}