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.
Why are only Sobolev spaces with certain exponents Hilbert Space?
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functional-analysis
hilbert-spaces
sobolev-spaces
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0@t.b.: Heh, I'd leave that as an exercise for the reader. :-D – 2012-05-24
1 Answers
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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.
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0Okay, thanks for the edit, +1. I removed my comments as they are no longer relevant. An alternative approach and links (see especially the answer on MO by Bill Johnson) is given in [this answer](http://math.stackexchange.com/a/97131/5363). – 2012-05-24