Goldstine theorem

In functional analysis, a branch of mathematics, the Goldstine theorem, named after Herman Goldstine, is stated as follows:

Goldstine theorem. Let $X$ be a Banach space, then the image of the closed unit ball $B\subseteq X$ under the canonical embedding into the closed unit ball $B^{″}$ of the bidual space $X^{″}$ is a weak*-dense subset.

The conclusion of the theorem is never true for the norm topology if $X$ is not reflexive; indeed, the image of $X$ in its bidual $X^{″}$ is always norm closed, being the continuous image of a complete metric space under an isometry.

Proof

Lemma

For all $x^{″}\in B^{″},$ ${\phi }_1,\dots ,{\phi }_n\in X^{\prime }$ and $\delta >0,$ there exists an $x\in (1+\delta )B$ such that ${\phi }_i(x)=x^{″}({\phi }_i)$ for all $1\le i\le n.$

Proof of lemma

By the surjectivity of $$\{\begin{array}{c}\mathrm{\Phi }:X\to {ℂ}^n,\\ x\mapsto ({\phi }_1(x),\cdots ,{\phi }_n(x))\end{array}$$ it is possible to find $x\in X$ with ${\phi }_i(x)=x^{″}({\phi }_i)$ for $1\le i\le n.$

Now let $$Y:=\bigcap _i\ker {\phi }_i=\ker \mathrm{\Phi }.$$

Every element of $z\in (x+Y)\cap (1+\delta )B$ satisfies $z\in (1+\delta )B$ and ${\phi }_i(z)={\phi }_i(x)=x^{″}({\phi }_i),$ so it suffices to show that the intersection is nonempty.

Assume for contradiction that it is empty. Then $\mathrm{dist}(x,Y)\ge 1+\delta$ and by the Hahn–Banach theorem there exists a linear form $\phi \in X^{\prime }$ such that $\phi {|}_Y=0,\phi (x)\ge 1+\delta$ and $\Vert \phi {\Vert }_{X^{\prime }}=1.$ Then $\phi \in \mathrm{span}\{{\phi }_1,\dots ,{\phi }_n\}$ and therefore $$1+\delta \le \phi (x)=x^{″}(\phi )\le \Vert \phi {\Vert }_{X^{\prime }}{\Vert x^{″}\Vert }_{X^{″}}\le 1,$$ which is a contradiction.

Proof of theorem

Fix $x^{″}\in B^{″},$ ${\phi }_1,\dots ,{\phi }_n\in X^{\prime }$ and $ϵ>0.$ Examine the set $$U:=\{y^{″}\in X^{″}:|(x^{″}-y^{″})({\phi }_i)|<ϵ,1\le i\le n\}.$$

Let $J:X\to X^{″}$ be the embedding defined by $J(x)={\text{Ev}}_x,$ where ${\text{Ev}}_x(\phi )=\phi (x)$ is the evaluation at $x$ map. Sets of the form $U$ form a base for the weak* topology, so density follows once it is shown $J(B)\cap U\ne \varnothing$ for all such $U.$ The lemma above says that for any $\delta >0$ there exists a $x\in (1+\delta )B$ such that $x^{″}({\phi }_i)={\phi }_i(x),$ $1\le i\le n,$ and in particular ${\text{Ev}}_x\in U.$ Since $J(B)\subset B^{″},$ we have ${\text{Ev}}_x\in (1+\delta )J(B)\cap U.$ We can scale to get $\frac{1}{1+\delta }{\text{Ev}}_x\in J(B).$ The goal is to show that for a sufficiently small $\delta >0,$ we have $\frac{1}{1+\delta }{\text{Ev}}_x\in J(B)\cap U.$

Directly checking, one has $$|[x^{″}-\frac{1}{1+\delta }{\text{Ev}}_x]({\phi }_i)|=|{\phi }_i(x)-\frac{1}{1+\delta }{\phi }_i(x)|=\frac{\delta }{1+\delta }|{\phi }_i(x)|.$$

Note that one can choose $M$ sufficiently large so that $\Vert {\phi }_i{\Vert }_{X^{\prime }}\le M$ for $1\le i\le n.$ Note as well that $\Vert x{\Vert }_X\le (1+\delta ).$ If one chooses $\delta$ so that $\delta M<ϵ,$ then $$\frac{\delta }{1+\delta }|{\phi }_i(x)|\le \frac{\delta }{1+\delta }\Vert {\phi }_i{\Vert }_{X^{\prime }}\Vert x{\Vert }_X\le \delta \Vert {\phi }_i{\Vert }_{X^{\prime }}\le \delta M<ϵ.$$

Hence one gets $\frac{1}{1+\delta }{\text{Ev}}_x\in J(B)\cap U$ as desired.

See also

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References

• Template:Rudin Walter Functional Analysis