0
I’m interested in using laplacian (−Δ) eigenfunction as a basis for H1(Rn) . I know that in H1(Ω) , Ω bounded this can be done so I was wandering about H1(Rn) .
Now let eλ be an eigenfunction corresponding to λ so that: −Δeλ=λeλ
In addition, consider that ⟨∗,∗⟩ is the scalar product on H1(Rn)
Here is what I know:
1.The spectrum of −Δ is (0,∞)
2.We have the representation: ∀u∈H1(Rn) u(x)=∫∞0⟨u,eλ⟩eλdλ∀u∈H1(Rn)
3.In general, if H is a hilbert space ( with a countable or uncountable basis) any element of H can be represented as an infinite series. Namely, if M si a countable or uncountable set and Tμ|μ∈M is a base for H then: ∀u∈H ∃N⊂M,N countable such that: u=∑n∈NcnTn
First thing first. Do we actually have eigenfunctions of Laplacian in H1(Rn) ?. It is clear that sin(x1) satisfies the PDE but is not square integrable. And if we don’t have then I guess that spectral theory can’t be of use with Δ and H1(Rn) right ?
II Assuming that we actualiyactually have eλ∈H1(Rn) Are the statement 1, 2, 3 correct ? I’m confident in 1 and 3 but I want to check.
III Does 2 imply some type of density of spaneλ|λ∈span eλ|λ∈(0,∞) in H1(Rn) ?
IV. can we combine 2 and 3 to get ∀u∈H ∃N⊂(0,∞),N countable such that: u=∑n∈Ncnen ? Basically, for a specific element of H can we chose to work with the infinite sum and not the integral ?