PDA

View Full Version : Parallelogram Equality+Inner Product Spaces


gravenewworld
Dec21-04, 01:03 PM
It is true that if a norm satisfies the parallelogram equality then it must come from an inner product right (i.e. < , > is an inner product).? How in the world could you go about proving/showing this?

fourier jr
Dec21-04, 02:03 PM
ya that's true; here's how it's done:
a) you need to know the parallelogram law (duh) but also the polarization identity: \| x+y\|^2 - \|x-y\|^2 = 4<x,y>

b) let V be a normed linear space in which the parallelgram law holds. Define <x,y> by the polarisation identity & prove that V with that inner product is an inner product space, and that \|x\| = \sqrt{<x,x>}

b*) see that spaces like l^\infty, l^1, C[a,b] with the uniform norm, c_0, don't satisfy the parallelogram law, and that there's no inner product (by a) ) that gives the norms for those spaces

matt grime
Dec21-04, 02:11 PM
You are of course not allowing fields of characteristic 2.