Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question

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.

share|cite|improve this answer
Did you mean $X'' = (Y''/Y)' \oplus X$? – commenter Oct 31 '12 at 9:13

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 $\kappa(X)\cap A=0$.

share|cite|improve this answer
You identified the dual space $C[0,1]^{**}$ incorrectly. It is much bigger. – Grebe Oct 30 '12 at 19:42
@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 8:49
Every 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 9:44
@commenter, thank you for your explanation. It is very helpful to me. – 23rd Oct 31 '12 at 11:57

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.