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

• Mar 22nd 2010, 01:07 AM
Ase
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.

• Mar 22nd 2010, 11:27 AM
Opalg
Quote:

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 \$\displaystyle \|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 \$\displaystyle 2(2)^{2/p} = 4(1)^{2/p}\$, which holds only if p=2.
• Mar 23rd 2010, 06:29 AM
Ase
Hi Opalg

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