I am asked the following:
Let $\Omega$ be a measure space such that $\exists A,B \subset \Omega$ where
- $A,B$ are measurable sets
- $0 < |A| < \infty$ and $0 <|B|< \infty$
- $|A \cap B| = 0$
Then $L^{p}(\Omega)$ is a Hilbert space $\iff$ $p=2$
------- Attempt at solution --------------
Showing that $L^{2}(\Omega)$ is a Hilbert space is not difficult. Simply define $$ (f,g) = \int_{\Omega}f \cdot \overline{g} d\mu $$ This is well defined via Hölder's inequality, and the inner product properties are straight forward.
Now, I got stuck trying to show that $L^{p}$ is not a Hilbert space for $p \neq 2$. For $p<1$, $L^{p}$ is not a normed space with such a measure, so I guess the result follows from there. For $p>1$, my idea is to show that $L^{p}$ is not isometrically isomorphic to $l^{2}$, where $l^{2}$ is the space of sequences $(x_{n})$ such that $\sum_{n=0}^{\infty} x_{n}^{2} < \infty$. I was trying to use the fact that simple functions are dense in $L^p$ for $p>1$, but I'm not making much progress. Any help would be aprecciated.