Let $X$ be a Banach space. Let $X^*$ denote the dual space . Would you help me, How to show that $(X^*)^{**}=(X^{**})^*$?
Showing that $(X^*)^{**}=(X^{**})^{*}$, where $X$ is a Banach space
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functional-analysis
banach-spaces
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0Would dow$n$voter explai$n$ why all the answers were downvoted? – 2012-12-04
2 Answers
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This is just playing with symbols. By definition $ Y^*=\mathcal{B}(Y,\mathbb{C}) $ for any normed space $Y$. So $ X^{**}=(X^*)^*=\mathcal{B}(X^*,\mathbb{C})=\mathcal{B}(\mathcal{B}(X,\mathbb{C}),\mathbb{C}) $ $ (X^{**})^*=\mathcal{B}(X^{**},\mathbb{C})=\mathcal{B}(\mathcal{B}(\mathcal{B}(X,\mathbb{C}),\mathbb{C}),\mathbb{C}) $ and on the other hand $ (X^*)^{**}=\mathcal{B}(\mathcal{B}(X^*,\mathbb{C}),\mathbb{C})=\mathcal{B}(\mathcal{B}(\mathcal{B}(X,\mathbb{C}),\mathbb{C}),\mathbb{C}) $ Hence $ (X^{**})^*=(X^*)^{**} $
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It's just a matter of notation:
$X^{**}:=(X^{*})^{*}$
for all normed vector spaces X, so
$(X^*)^{**}=((X^*)^*)^*=(X^{**})^*.$
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0@GEdgar: exercise no 2 page 90 – 2012-12-10