$(X,m)$ be a measure space ; how to show that for $1 \le p \le \infty$ , $L^p(X,m)$ is a Hilbert space iff $p=2$?

Question

Let $(X,m)$ be a measure space ; how to show that for $1 \le p \le \infty$ , $L^p(X,m)$ is a Hilbert space iff $p=2$ ? I can easily show that $L^2$ is a Hilbert space .

To show $L^p$ is not Hilbert space for $p \ne 2$ , I think showing the failure of parallelogram identity for $p\ne 2$ is not easy , I am not getting anywhere by following that route . Or I was also thinking that $(L^p)^*$ is isometrically isomorphic with $L^q$ for $1<p<\infty , 1/p+1/q=1$ , so if we could show $L^p$ is not isometrically isomorphic with $L^q$ for $1<p<\infty ,p \ne 2, 1/p+1/q=1$ , then we would be done ; but I don't know how to show it . Perhaps there are some other ways . Please help . Thanks in advance

Answer 1

the inner product in $L^2$ is $<f,g>=\int f \bar{g}$ in you take complex valued functions and the norm in this space induced by the inner product is $\sqrt{<f,f>}=\sqrt{\int f \bar{f}}=\sqrt{\int |f|^2}$.Prove that $L^2$ is complete with this norm.

For the parallelogram law the any two measurable sets $A,B$ and the functions $1_A,1_B$.

Then $||1_A+1_B||_p^2+||1_A-1_B||_p^2=2||1_A||_p^2+2||1_B||_p^2$

Take as a counterexample the $L^p(\mathbb{R},\Sigma,m)$ where $m$ is the bebesgue measure and two disjoint subsets of the real line$

If you prove that this law fails in this measure space then it fails in general because you want to prove that $L^p$ is a hilbert space iff $p=2$ for any measure and sigma algebra defined in the space.

But for an arbitrary measure space, $L^2$ is a banach space and keeping that in mind, you can use this to prove that $L^2$ is a hilbert space

A complete inner pproduct space $X$ is a hilbert space iff the parallelogram law holds.

Proof:

Suppose that $X$ is a complete banach space where parallelogram law holds.

Define the function $<.,.>:X \times X \longrightarrow \mathbb{C}$, $\forall x,y \in X$ to be $<x,y>=\frac{1}{4}(||x+y||^2-||x-y||^2)$.

You can easily prove that the inner product's axioms for this function hold using the parallelogram law

Thus you have that any banach space where parallelogram law holds is a hilbert space.

The converse is always true because we know that a hilbert space is a banach space where parallelogram law holds in it.

I hope this helps.