Why Are Hilbert Space Embeddings Used in FEM Eigenvalue Approximation?

[SqueeSpleen]

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 beggining 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.

 

[akeleti8]

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 .

 

[Xiuh]

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
$$\langle\mathcal{l},u\rangle_{H^{-1},H^1_0} = (v_o,u)_{L^2} + \sum_{i=1}^d (v_i, u_{x_i})_{L^2}$$