For my quetion in MO is $\forall X$, $X^{**}$=X$\oplus Y$ for a $Y$ another set I am not really sure in Thomas answer why the first assumption saying that such a $Y$ exist iff the sequence $0 \to X \to \varphi(X^∗)^∗ \to \eta \mathrm{coker}\varphi \to 0$ splits?
Double Dual of $ \ell^\infty$
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functional-analysis
duality-theorems
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0No. This question has been asked and answered on [MathOverflow](http://mathoverflow.net/questions/106695/double-duals-characteristic-closed). – 2012-09-16
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3...in fact, it looks like *you* asked this question there, and I answered it. – 2012-09-16
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2What is true is that every dual space is complemented in its bidual. All the spaces you listed are dual spaces. The space $c_0$ is not complemented in its bidual [Phillips's lemma](http://math.stackexchange.com/q/132520): $(c_0)^{\ast\ast} = \ell^{\infty} \ncong c_0 \oplus Y$ for any $Y$. – 2012-09-16
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1If there are some points that are unclear in Alex's answer from MO, perhaps inquire about these specific points instead of asking the same question. – 2012-09-16
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0I personally would be interested to know a further characterization of when the sequence does and does not split, based on properties of $X$. To add, it seems that the splitting condition is just a restatement of the original issue but in different terms. – 2012-09-16
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0would you please give me a reference why such a Y exist iff the sequence 0→X→φ(X∗)∗→ηcokerφ→0 splits? – 2012-11-13
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0Would you please give me a reference or a more detailed explanation on why such a Y exist iff the sequence 0→X→φ(X∗)∗→ηcokerφ→0 splits? – 2012-11-13