Thread: 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.

Thanks in advance.

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 . 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 , which holds only if p=2.

Hi Opalg

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