Let $X$ be a Banach space and suppose $X^{\prime\prime}=A\oplus B$, where $A$ and $B$ are infinite dimensional and closed. Is $\kappa(X)\cap A$ weak*-dense in $A$? $\kappa\colon X\to X^{\prime\prime}$ is the standard embedding.
Goldstine's theorem
2 Answers
No. Put $X = \ell^{1}$, $A = (\ell_\infty/c_0)'$ and $B = \ell_1$. Then $\ell_1'' = A \oplus B$ but $A \cap \ell^1 = 0$.
In general, if $X = Y'$ then there is a canonical decomposition $X''' = (Y''/Y)' \oplus X$ and typically (but not always) $Y''/Y$ is infinite-dimensional if $Y$ is not reflexive.
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1Did you mean $X'' = (Y''/Y)' \oplus X$? – 2012-10-31
I think the following is a counterexample. Let $X$ be the space of continuous functions on $[0,1]$ endowed with $L^\infty$ norm. Then $X'$ is the space of signed Borel measures endowed with total variation norm. Let $C=\{\mu\in X': |\mu|(\mathbb{Q}\cap[0,1])=0\}$ and $D=\{\mu\in X': |\mu|([0,1]\setminus\mathbb{Q})=0\}$, where $|\mu|$ denotes the variation of $\mu$. By definition, $C$ and $D$ are closed subspace of $X'$ and $X'=C\oplus D$. Let $A=\{f\in X'':f(C)=0\}$ and $B=\{f\in X'':f(D)=0\}$. Then $A$ and $B$ satifsy your assumptions but $\kappa(X)\cap A=0$.
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0@commenter, thank you for your explanation. It is very helpful to me. – 2012-10-31