According to wikipedia article on polarisation identity, in a normed space (V,||.||), if the parallelogram law holds, then there is an inner product on V such that ||x||^2=⟨x,x⟩ for all x in V. However, from what I understand, we can prove polarisation identity this way: ||x+y||^2−||x−y||^2=⟨x+y,x+y⟩−⟨x−y,x−y⟩=...=4⟨x,y⟩ However, from this proof, I don't see why we need parallelogram law. On top of that, shouldn't it be the other way round, such that if ||x||^2 =(x,x), then the parallelogram law holds? Can someone explain to me please?

Ebone6v

Ebone6v

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2022-08-17

According to wikipedia article on polarisation identity, in a normed space ( V , | | . | | , if the parallelogram law holds, then there is an inner product on V such that | | x | | 2 = x , x for all x V. However, from what I understand, we can prove polarisation identity this way:
| | x + y | | 2 | | x y | | 2 = x + y , x + y x y , x y = . . . = 4 x , y
However, from this proof, I don't see why we need parallelogram law. On top of that, shouldn't it be the other way round, such that if | | x | | 2 = x , x , then the parallelogram law holds? Can someone explain to me please?

Answer & Explanation

Leroy Cunningham

Leroy Cunningham

Beginner2022-08-18Added 14 answers

The point of the theorem is the existence of the inner product. It is true (and easy to show) that if you are given an inner product , then you can always define a norm via | | x | | 2 = x , x , and this norm satisfies the parallelogram law and the polarisation identity. But that isn't the situation the theorem is considering. It is about the situation where you start with a norm and want to define an inner product such that x , x = | | x | | 2 .
If you are just given a norm to start with, you can't necessarily define an inner product from that. Just defining x , x = | | x | | 2 doesn't even define x , y in general, let alone mean that , has the linearity properties you need for the "=" in your equation. Even defining x , y = ( | | x + y | | 2 | | x y | | 2 ) / 4 doesn't get the linearity properties unless you have the parallelogram law as well.

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