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Sobolev spaces of order 2 are known to form a Hilbert space. Consider such a Sobolev space of (order 2) functions on the domain $f:\mathbb{R}\rightarrow \mathbb{R}$. What is an example for the basis of such a Sobolev space.

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    $\begingroup$ First of all, there are infinitely many orthonormal bases for any Hilbert space of dimension $>1$. More importantly, finding a basis depends very much on the domain in question and about which Sobolev space you are looking at, i.e., how many derivatives you consider. $\endgroup$ – Lukas Geyer Oct 16 '12 at 4:35
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As mentioned above, the question is not well-posed in this form since the answer depends not only on the precise space you are considering but also on the norm you are using. Perhaps the following comment might nevertheless be useful. If your space can be identified with (or is defined as) the domain of definition of an unbounded, self-adjoint operator (as many of the useful ones are) and if the latter has discrete spectrum, then its eigenfunctions (suitably normed) form an ONB for the Sobolev space with the corresponding norm. Simple examples are the Laplace operator on the circle or the standard one-dimensional Schrödinger operator on the line. More sophisticated examples are provided by the Laplacian on a compact Riemann manifold or general Schrödinger operators under suitable conditions on the potential function.

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My guess for (a nice!) basis: use the Hermite functions and the Gram-Schmidt process. It is may be rather lengthy, but the procedure and ingredients are as follows.

1) $h_n' = \sqrt{n/2}h_{n-1} + \sqrt{(n+1)/2)}h_{n+1}, n \geq 0, h_{-1}=0$

2) $h_n$ are what is called the 'physicists polynomials', denoted by $\psi_n$ there.

3) Sobolev product would be $(f,g)_S=(f,g)_0 + (f',g')_0$. Could be also the product containing the Fourier transform, but this seems to me more lenghty ($Fh_n$ is nice, due to Wiener thorem, and equals $(-i)^nh_n$, but the factor $(1+|\xi|^2)^r)$ in the product in H^r is also a bit disturbing the smoothness of the computation.

4) The space ($\Omega$ for PDE-researchres) would be R.

5) Advantage of the Hermite fucntions is due to their easy dimensional properties $h_{a_1...a_n}(x^1,...,x^n)=h_{a_1}(x^1)....h_{a_n}(x^n)$, where $a=(a_1,...,a_n) \in N_0^n.$

One should compute the first three and then to see the system... The derivative property of the Hermits enables to speak about the neighbour basis terms only. But this must be done properly. [Formally by an induction, after "guessing".]

Note that this gives you the orthogonal basis only, one should normalize than! (In the Sobolev norm!)

It is quite strange - I could not found this results in the literature (even not as excercises for students). (Chebyshev polynomials should be orthonormal for $H^1([-1,1])$ but changing the domain to $\mathbb R$ change the polynomials substantially even if we take the correct domain diffeomorphic to $\mathbb R$, i.e., $(-1,1)$.)

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