# Thread: How to prove that LP(R) does not form a Hilbert space for p different from 2.

1. ## How to prove that LP(R) does not form a Hilbert space for p different from 2.

Hi

I am trying to prove that LP(R) does not form a Hilbert space for p different from 2.

My thoughts are to try and disprove either Cauchy-Schwarz' inequality, the parallelogram law or similar. I am however not quite sure how to do this.

2. Originally Posted by Ase
Hi

I am trying to prove that LP(R) does not form a Hilbert space for p different from 2.

My thoughts are to try and disprove either Cauchy-Schwarz' inequality, the parallelogram law or similar. I am however not quite sure how to do this.
Cauchy–Schwarz won't work, because it involves the inner product, and if the space is not a Hilbert space you don't have an inner product to work with. But the parallelogram law approach is a good idea. You want to know whether or not $\|f+g\|_p^2 + \|f-g\|_p^2 = 2(\|f\|_p^2 + \|g\|_p^2)$. Try taking f to be the function that is equal to 1 on the interval [0,1] and zero elsewhere, and g to be the function that is equal to 1 on the interval [1,2] and zero elsewhere. The parallelogram identity applied to these two functions reduces to the equation $2(2)^{2/p} = 4(1)^{2/p}$, which holds only if p=2.

3. Hi Opalg

Thanks a lot for the brilliant answer. Just what I needed.

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# Lp space is not Hilbert space

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