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Let $X$ be a Banach space. Let $X^*$ denote the dual space . Would you help me, How to show that $(X^*)^{**}=(X^{**})^*$?

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Would downvoter explain why all the answers were downvoted? – Norbert Dec 4 '12 at 20:14
up vote 2 down vote accepted

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:


for all normed vector spaces X, so


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No. I find the question from exercise of conway book – beginner Dec 3 '12 at 23:32
@Sean Gomes Please, explain your argument. – MathOverview Dec 4 '12 at 0:14
The 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. – Sean Gomes Dec 4 '12 at 2:38
What is the number of the exercise? – GEdgar Dec 4 '12 at 13:30
@GEdgar: exercise no 2 page 90 – beginner Dec 10 '12 at 20:17

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