Why are only Sobolev spaces with certain exponents Hilbert Space?

Question

I would like to know why $W^{k,2} (\Omega) $ is a Hilbert space , why is it impossible to define inner product in other Sobolev spaces, ie exponent $\ge2$ . Here $||u||_{W^{k,2} (\Omega)} $ = $(\sum_{|\alpha|\le k}||D^\alpha u||^2_{L^2(\Omega)})^{1/2}$. I would want to know it technically as well.

Answer 1

As $1\leq p \leq \infty$, the only Hilbert space among $L^p$ spaces is $L^2$. You can use the parallelogram law to prove this. See here.

Edit: A Banach space $X$ admits an equivalent norm $\|\cdot \|$ such that $\| \cdot \|^2$ is twice Fréchet differentiable on $X$ if, and only if, $X$ is isomorphic to a Hilbert space. This is stated here. But, as I wrote in the comments, Sobolev spaces are interesting when they are endowed with their natural norms.