Is $L^2(\Omega)$ the only $L^p$ hilbertian space? I've started today studying Hilbertian spaces, and all of the examples seen in class were about the space $L^2(\Omega)$, where $\Omega$ is a limited domain in $\mathbb{R}^N$ $(N \geq 1)$.
Online I didn't manage to find anything related to my doubt regarding other $p \neq 2$ .
So my attempt is to find a function $v : \ \Omega \to \mathbb{C}$   in $L^p(\Omega)$ ($p \neq 2$) for which $$ \langle v, v \rangle := \int_{\Omega} v(x)\overline{v(x)} dx \ \ \notin \mathbb{C} $$
Iìm trying to use the fact that Hölder's Inequality can't be used here. ($p \neq p'$ if $p \neq 2$) and so the integral may diverge.
And the conclusion would be that the "usual" inner product over $L^2$ can't be extended for every $1 \leq p < \infty $.
But i can't manage to find one.
Is my attempt correct?
Can someone provide me a counterexample?
I apologize in advance for not being precise but it's my first time with an hilbert :P

Each $L^p$ space for $p \ge 1$ is what is called a Banach space.  (i.e. a complete normed vector space).  Every Hilbert space is a Banach space.  There is a way to characterize all the Banach spaces that are Hilbert spaces.  Essentially a Banach space is a Hilbert space iff its norm satisfies the parallelogram law: $$2\|f\|^2 + 2\|g\|^2 = \|f-g\|^2 + \|f+g\|^2$$
This works for $L^2$.  You should look for a violation of this for $p\neq 2$.
For the counter example you are looking for, here is one such function for $L^1[0,1]$.  Take the function $v(x) = 1/\sqrt{x}$.  That function is indeed integrable on the interval, but $$\int_{0}^1 \frac{1}{\sqrt{x}} \frac{1}{\sqrt{x}} dx = \int_{0}^1 \frac1x dx = \infty$$
You can fashion similar examples for other spaces.
