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 downvoter explain 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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0No. I find the question from exercise of conway book – 2012-12-03
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2@Sean Gomes Please, explain your argument. – 2012-12-04
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0The double dual of a normed vector space $X$ is the vector space that arises from taking the dual of the dual of $X$. If we explicitly write this by using brackets, both the left hand side and the right hand side reduce to the middle...the space that arises from taking the dual of X three times recursively. – 2012-12-04
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1What is the number of the exercise? – 2012-12-04
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0@GEdgar: exercise no 2 page 90 – 2012-12-10