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.
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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 $X'' = (Y''/Y)' \oplus X$? – commenter Oct 31 '12 at 09:13
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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 $\kappa(X)\cap A=0$.
23rd
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You identified the dual space $C[0,1]^{**}$ incorrectly. It is much bigger. – Grebe Oct 30 '12 at 19:42
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@Grebe, I cannot see why $C[0,1]^{**}$ is bigger than the space of bounded Borel functions. Could you please give some explanation? – 23rd Oct 31 '12 at 08:49
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2Every subset $A \subset [0,1]$ defines a linear functional on the subspace $\ell^{1}[0,1] \subset C[0,1]^{\ast}$ of combinations of Dirac measures. Extend it by Hahn-Banach to all of $C[0,1]^{\ast}$. If $A$ is not measurable then this construction gives an element of $C[0,1]^{\ast\ast}$ that cannot be represented by a bounded Borel function. This shows that $C[0,1]^{\ast\ast}$ has cardinality at least $2^{\mathfrak{c}}$ while the bounded Borel functions have cardinality continuum. – commenter Oct 31 '12 at 09:44
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