- #1
SqueeSpleen
- 138
- 5
Let X be an Inner Product Space. If for every closed subspace M, M^{\perp \perp} = M, then X is a Hilbert Space (It's complete).
Hint: Use the following map: T : X \longrightarrow \overset{\sim}{X}: T(y)=(x,y)=f(x) where (x,y) is the inner product of X.
Relevant equations:
S^{\perp} is always closed to every S \subset X
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 \overline{T(M^{\perp})} = M^{\circ} (Where M^{\circ} is the Null Space of M).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: T : X \longrightarrow \overset{\sim}{X}: T(y)=(x,y)=f(x) where (x,y) is the inner product of X.
Relevant equations:
S^{\perp} is always closed to every S \subset X
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 \overline{T(M^{\perp})} = M^{\circ} (Where M^{\circ} is the Null Space of M).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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