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Let $X$ be a Hilbert space and $M\subset X$. We know that the following is true: $$(M^{\perp})^{\perp}=\overline{\operatorname{span} M}.$$ But I want to know is it true if $X$ is an inner product space . Can someone help me prove it or give a counterexample? Thank you in advance.

I have tried some examples such as $X=C[0,1] $ and $M=\{f\in C[0,1]:f \text{ is constant}\}$,but all of them show the positive answer.

Martin Sleziak
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C. Ding
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1 Answers1

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This identity holds in all inner product spaces. The usual proof uses Hahn-Banach and no completeness is needed.

gerw
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  • Thank you! But I have found a counterexample here. http://math.stackexchange.com/questions/859092/closed-subspace-m-m-perp-perp-in-pre-hilbert-spaces/859115#859115.@gerw – C. Ding Apr 17 '16 at 13:12
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    Oh, I see. It depends on the definition of $(\cdot)^\perp$. I am used to use $M^\perp := { x^* \in X^* : x^*(x) = 0 \forall x \in M}$, which is different for the (maybe more common) $M^\perp := {y \in X : (x,y) = 0 \forall x \in M}$. – gerw Apr 17 '16 at 20:15
  • Sorry,I should add the definition. But I'm also interested in the situation that $M^\perp := { x^* \in X^* : x^(x) = 0 \forall x \in M}$.Do you mean prove like this http://math.stackexchange.com/questions/1135883/showing-m-perp-perp-overlinem .What I don't understannd in the proof is why $x_0\in E$ , is it necessary to add E is a reflexive space(e.g. ,$E^{*}=E$)? – C. Ding Apr 17 '16 at 23:52
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    No, for $N \subset X^$, you define $N$ to live in $X$, i.e. $N^\perp = {x \in X : x^(x) = 0 \forall x^* \in N}$. (So actually you have two definitions of $(\cdot)^\perp$ in order to work with the dual pairing $(X, X^*)$) – gerw Apr 18 '16 at 06:28