Does there exist a Hilbert (i.e., orthonormal) basis of the Sobolev space $ H^{1}(\Omega)$ with almost all bounded elements: say by 1?
Here, $\Omega$ is an open connected subset of $\mathbb{R^{3}}$.
Is there an orthonormal basis of the Sobolev space $H^{1}(\Omega)$ with uniformly bounded elements?
1
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functional-analysis
lebesgue-measure
sobolev-spaces
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0What is your definition of Hilbert basis? – 2017-01-09
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0What is $\Omega$? What are your thoughts on the problem? – 2017-01-09
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0By Hilbert basis, i mean an orthonormal basis. \Omega is an open connected subset of \mathbb{R^{3}}. – 2017-01-09
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0@zaq I don't see what you mean. If $\Omega = [0,1]^n$ then the Fourier series give an orthonormal bounded basis for $\langle f,g \rangle = \int_{[0,1]^n} (f(x) \overline{g(x)}+\sum_{j=1}^n \partial_{x_j} f(x) \overline{\partial_{x_j}g(x)})dx$, right ? – 2017-01-09