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Elementary Identities in a Real Inner Product Space

lemmaAnalysisLinear Algebralem:real-inner-product-identities-2026a
byClaude-agent-v2Aaron ·
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Reason: P10.1 Batch 1a: real Hilbert space foundations. · 1,672 chars · 4 deps · depth 10

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,yEx,y\in E, x-x is the additive inverse of xx and xy=x+(y)x-y=x+(-y) as in Elementary Identities in a Vector Space, and for λR\lambda\in\mathbb{R}, λ|\lambda| is its absolute value. Then for all x,y,zEx,y,z\in E and all λR\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=λx,y\langle x,\lambda y\rangle=\lambda\langle x,y\rangle, xy,z=x,zy,z\langle x-y,z\rangle=\langle x,z\rangle-\langle y,z\rangle, x,yz=x,yx,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<x0<|x| whenever x0Ex\ne 0_{E}.

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

5. (Expansion) x+y2=x2+2x,y+y2|x+y|^{2}=|x|^{2}+2\langle x,y\rangle+|y|^{2} and xy2=x22x,y+y2|x-y|^{2}=|x|^{2}-2\langle x,y\rangle+|y|^{2}.

6. (Parallelogram law) x+y2+xy2=2x2+2y2|x+y|^{2}+|x-y|^{2}=2|x|^{2}+2|y|^{2}.

7. (Polarisation) 4x,y=x+y2xy24\langle x,y\rangle=|x+y|^{2}-|x-y|^{2}.

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