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Elementary Identities in a Real Inner Product Space

Bilinearity in the second argument, behaviour of the zero vector, vanishing and homogeneity of the norm, the expansions of |x±y|^2, the parallelogram law and the polarisation identity.

Statement

Let R\mathbb{R} be the ordered field of real numbers, with the notation of that item, and let EE be a real inner product space with inner product ⟨⋅,⋅⟩\langle\cdot,\cdot\rangle, norm ∣⋅∣|\cdot| and zero vector 0E0_{E}; for x,y∈Ex,y\in E, −x-x is the additive inverse of xx and x−y=x+(−y)x-y=x+(-y) as in Elementary Identities in a Vector Space, and for λ∈R\lambda\in\mathbb{R}, ∣λ∣|\lambda| is its absolute value. Then for all x,y,z∈Ex,y,z\in E and all λ∈R\lambda\in\mathbb{R} the following hold.

1. (Bilinearity) ⟨x,y+z⟩=⟨x,y⟩+⟨x,z⟩\langle x,y+z\rangle=\langle x,y\rangle+\langle x,z\rangle, ⟨x,λy⟩=λ⟨x,y⟩\langle x,\lambda y\rangle=\lambda\langle x,y\rangle, ⟨x−y,z⟩=⟨x,z⟩−⟨y,z⟩\langle x-y,z\rangle=\langle x,z\rangle-\langle y,z\rangle, ⟨x,y−z⟩=⟨x,y⟩−⟨x,z⟩\langle x,y-z\rangle=\langle x,y\rangle-\langle x,z\rangle, and ⟨−x,y⟩=⟨x,−y⟩=−⟨x,y⟩\langle -x,y\rangle=\langle x,-y\rangle=-\langle x,y\rangle.

2. (Zero vector) ⟨x,0E⟩=⟨0E,x⟩=0\langle x,0_{E}\rangle=\langle 0_{E},x\rangle=0 and ∣0E∣=0|0_{E}|=0.

3. (Vanishing) ∣x∣=0|x|=0 if and only if x=0Ex=0_{E}; consequently 0<∣x∣0<|x| whenever x≠0Ex\ne 0_{E}.

4. (Homogeneity) ∣λx∣=∣λ∣ ∣x∣|\lambda x|=|\lambda|\,|x|; in particular ∣−x∣=∣x∣|-x|=|x|.

5. (Expansion) ∣x+y∣2=∣x∣2+2⟨x,y⟩+∣y∣2|x+y|^{2}=|x|^{2}+2\langle x,y\rangle+|y|^{2} and ∣x−y∣2=∣x∣2−2⟨x,y⟩+∣y∣2|x-y|^{2}=|x|^{2}-2\langle x,y\rangle+|y|^{2}.

6. (Parallelogram law) ∣x+y∣2+∣x−y∣2=2∣x∣2+2∣y∣2|x+y|^{2}+|x-y|^{2}=2|x|^{2}+2|y|^{2}.

7. (Polarisation) 4⟨x,y⟩=∣x+y∣2−∣x−y∣24\langle x,y\rangle=|x+y|^{2}-|x-y|^{2}.

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