- #1

SqueeSpleen

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Let [itex]X[/itex] be an Inner Product Space. If for every closed subspace [itex]M[/itex], [itex]M^{\perp \perp} = M[/itex], then [itex]X[/itex] is a Hilbert Space (It's complete).

Hint: Use the following map: [itex]T : X \longrightarrow \overset{\sim}{X}: T(y)=(x,y)=f(x)[/itex] where [itex](x,y)[/itex] is the inner product of [itex]X[/itex].

Relevant equations:

[itex]S^{\perp}[/itex] is always closed to every [itex]S \subset X[/itex]

Attemp to solution.

I don't really know how to solve it, most theorems I have read, have Hilbert Space as a hyphotesis.

The only idea I had was trying to prove that [itex]\overline{T(M^{\perp})} = M^{\circ}[/itex] (Where [itex]M^{\circ}[/itex] is the Null Space of [itex]M[/itex]).Edit: How do I do to have latex without jumping lines? I uploaded a pdf with a itext more closely formated to what I tried to write.

Edit2: I read other posts to find out how to do it, I only had to add an i to the "tex".

Hint: Use the following map: [itex]T : X \longrightarrow \overset{\sim}{X}: T(y)=(x,y)=f(x)[/itex] where [itex](x,y)[/itex] is the inner product of [itex]X[/itex].

Relevant equations:

[itex]S^{\perp}[/itex] is always closed to every [itex]S \subset X[/itex]

Attemp to solution.

I don't really know how to solve it, most theorems I have read, have Hilbert Space as a hyphotesis.

The only idea I had was trying to prove that [itex]\overline{T(M^{\perp})} = M^{\circ}[/itex] (Where [itex]M^{\circ}[/itex] is the Null Space of [itex]M[/itex]).Edit: How do I do to have latex without jumping lines? I uploaded a pdf with a itext more closely formated to what I tried to write.

Edit2: I read other posts to find out how to do it, I only had to add an i to the "tex".

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