Graduate Why Are Hilbert Space Embeddings Used in FEM Eigenvalue Approximation?

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SUMMARY

Hilbert Space embeddings are crucial in the Finite Element Method (FEM) for eigenvalue approximation, particularly in the context of mixed forms as discussed in Danielle Boffi's work. The embeddings H_{\Phi} and H_{\Xi} must maintain dense and continuous relationships with their respective spaces, ensuring compatibility. Identifying spaces like \Phi with their duals can lead to significant mathematical inaccuracies, particularly when dealing with variational formulations and partial differential equations. The discussion emphasizes the importance of maintaining distinct identities for these spaces to avoid misinterpretations in functional analysis.

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  • Understanding of Hilbert Spaces and their properties
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Hi. I'm studying Finite Elements Method, I was readding a paper written by Danielle Boffi and in a part dedicated to the approximation of eigenvalues in mixed form, it's about approximating eigenvalues in the Hilbert Spaces \Phi and \Xi
Then it says:
"If we suppose that there exists Hilbert Spaces H_{\Phi} and H_{\Xi} such that the following dense and continuous embeddings hold in a compatible way
\Phi \subset H_{\Phi} \simeq H_{\Phi} ' \subset \Phi'
\Xi \subset H_{\Xi} \simeq H_{\Xi} ' \subset \Xi'
Here \subset means dense and continously embedded.

I'm not sure why it does that, I have read more than one time that they work with a space identified with it's dual space, as they're Hilbert Spaces I know you can do it using Riesz Representation theorem, but I don't exacly see why they're doing this.

I didn't know why they don't simply identify \Phi with \Phi'.

I have found an advice against identifying a space which is not L^{2} with it's dual, because otherwise in constructions like this, if H_{\Phi}=L^{2} and \Phi=H^{1} we would end up identifying the four spaces \Phi \equiv H_{\Phi} \equiv H_{\Phi} ' \equiv \Phi' and it would mean that you're identifying a function with it's laplacian which is (and here I'm quoting a book) ""the beginning of the end""

I'm new in this of variational formulations and I don't have a strong background on partial differential equations, I have more background on functional and real analysis, I'm studying this subject to make my Licenciatura's thesis in this but I'm really lost sometimes, this is as clear as I could make the question so feel free to ask me to clarify something if I wasn't clear enough.

PD: I don't know how to write in latex without making a new line.
 
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Hi there.
Φ≡HΦ≡H′Φ≡Φ'
does not make sense. i do not think i am negligent with my work in saying a function can't be equal to it's derivative. unless it's e^x. albeit Danielle boffi might have a different finding.

This comes from a guy who would enjoys going to a Lollapalooza wearing a rippling red scirt. This site helps me with your problem http://math.stackexchange.com/questions/644879/function-is-equal-to-its-own-derivative .
 
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When working with PDEs and variational methods it is not advised to identify H^1_0 with its dual H^{-1}, because you lose some subtleties. The space L^2 is sometimes called the pivot space and is sort of in the middle in terms of regularity, i.e. functions in L^2 are first derivatives of functions in H^1_0 and the elements of H^{-1} are first derivatives of functions in L^2. We can use this to characterize the dual space H^{-1} in the following way:

For every \mathcal{l}\in H^{-1} there exist v_0,v_1,\ldots,v_d such that
\langle\mathcal{l},u\rangle_{H^{-1},H^1_0} = (v_o,u)_{L^2} + \sum_{i=1}^d (v_i, u_{x_i})_{L^2}
 
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