Normed Linear Space ,$p \neq 2$ is $\left \| f\right \|_{p}= \sqrt{f,f}$ for each $ f \in L^P([0,1])$?

Question

For $p \neq 2$, is there an inner product $< ., .>$ on $L^P([0,1])$ such that $\left \| f\right \|_{p}= \sqrt{f,f}$ for each $ f \in L^P([0,1])$?

is it true?for P=2 norm is induced by inner product which is not Euclidean.if its true how to prove it?

Answer 1

Hint: take a look at the parallelogram law.

Answer 2

Hint: if a norm is derived from an inner product in that way, then the inner product is uniquely determined by the norm and there is an explicit algebraic expression for (f,g) in terms of ||f||, ||g||, ||f+g|| and ||f-g||.

Answer 3

No. If the norm were induced by an inner product it would satisfy the parallelogram law $$||f+g||^2+||f-g||^2=2(||f||^2+||g||^2).$$Simple examples show this is not so (for example, the characteristic functions of two disjoint sets).