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In my homework I am trying to understand the concept of Hilbert spaces with the following simple problem:

Let H be a Hilbert space and A and B be closed subspaces of H so that $A^{\perp} = B^{\perp}$

Determine if A=B.

My idea is no because A and B could be disjoint and still meet the condition. However I am not sure and any input would be appreciated

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Since $B$ is a closed subspace, we have a decomposition $H=B \oplus B^{\perp}$. We will show that $A\subset B$. Let $a\in A$, $a=b_1+b_2$ with $b_1\in B,b_2\in\ B^{\perp}$. Now because $A^{\perp}=B^{\perp}$, projection of $a$ onto $B^{\perp}$ is the same as projection of a onto $A^{\perp}$, which is $0$, since $a\in A$. Therefore $b_2=0$ and thus $a=b_1\in B$. By symmetry $B\subset A$.

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Since $A$ and $B$ are closed, the answer must be YES. Here's the proof:

Let $C=A^{\perp }=B^{\perp }$ and consider the space $C^{\perp } $. This is a closed subspace of $H$ and we can show that $C^{\perp }=A=B$. For any element $a\in A$ and $b\in B$, it follows that for every $c\in C$ we have $\langle c,a\rangle =\langle c,b\rangle =0$ and hence $A\subset C^{\perp } $ and $B\subset C^{\perp } $. If $A\subsetneq C^{\perp } $ then by Hahn-Banach theorem there's an element $c_0\in C$ so that $\langle a,c_0 \rangle\neq 0$ for some $a\in A$, but $A^{\perp } =C$, a contradiction. Therefore $A=C^{\perp } $ and similarly $B=C^{\perp } $.

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  • $\begingroup$ Can you explain why there is a $c_0 \in C$? I don't know Hahn-Banach and it is not in our book so it would be nice to understand the existence of $c_0$ without H-B $\endgroup$
    – Daniel
    Oct 6 '19 at 12:30

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