$H^{-1}(\Omega)$ given an inner product involving inverse Laplacian, explanation required

Question

Let $\Omega$ be a bounded domain and define $V=L^2(\Omega)$ and $H=H^{-1}(\Omega)$.

Endow $H$ with the inner product $$(f,g)_{H} = \langle f, (-\Delta)^{-1}g \rangle_{H^{-1}, H^1}$$ where $(-\Delta)^{-1}g = \tilde g$ is the solution of $-\Delta \tilde g = g$ on $\Omega$, $\tilde g= 0$ on $\Gamma$.

In Lions' Quelques methodes... on page 192, he uses this set up to deal with a PDE of the form $$u_t - \nabla \cdot (|u|\nabla u) = f.$$

Now, we have a Hilbert triple $V \subset H \subset V^*$. I remember that we identify $H$ with its dual, but not $V$ with its dual. But $V^*= (L^2)^* = L^2$. So how should I think of the dual of $V$? I ask because Lions then says the following:

...standard theory from before tells us there is a unique $u \in L^2(0,T;V)$ with $u_t \in L^2(0,T;V^*)$ such that $$(u'(t), v)_H + a(u(t),v) = L(t)(v)$$ where $a(u,v) = \frac{1}{2}\int_\Omega |u|uv$

(and $L(t)$ is given but it's not relevant here). My question is, why does Lions write $(u_t,v)_H$ and not $\langle u_t, v \rangle_{V^*,V}$? We only know that $u_t \in L^2(0,T;V^*)$, don't know that it's in $L^2(0,T;H)$.

But if we did identify $V^* = L^2$, then we get $L^2 \subset H^{-1} \subset L^2$ (but this would mean $L^2=H^{-1}$!!!), and the duality pairing becomes the $H^{-1}$ inner product as defined above so it kinda makes sense.

Finally, any references to other work where PDEs are tackled with the pivot space equal to $H^{-1}$ with the inverse Laplacian are hugely appreciated. Thanks.

Answer 1

Most of this is repeated from my comments following the answer to this question, but the basic problem is here:

I remember that we identify $H$ with its dual, but not $V$ with its dual.

Actually, it's the other way around; it's ok to identify $V = L^2$ with its dual, but if we also insist on identifying $H$ with its dual, we have exactly the problem you observe: this then forces us to identify $H$ with $V$.

Now, on one level this is fine, because all separable Hilbert spaces can be identified (they're all isometrically isomorphic to $\ell^2(\mathbb Z)$), but whether or not one should do this is a different matter. In fact, you've demonstrated exactly why one should refrain from doing this: the identifications aren't compatible with the natural inclusions.

So really, we should be 'pivoting' around $V$, i.e. identifying $V$ with it's dual so that we have the inclusions

$$ H^* \subset V \subset H $$

Then, this will solve the concern you raise when you say

We only know that $u_t \in L^2(0,T;V^∗)$, don't know that it's in $L^2(0,T;H)$.

i.e. we will have $u_t \in H$ (to state it briefly) as well. Then, as you say

But if we did identify $V^∗=L^2$, then we get $L^2\subset H^{−1}\subset L^2$ (but this would mean $L^=H^{−1}$!!!this is no longer a concern), and the duality pairing becomes the $H^{−1}$ inner product as defined above so it kinda makes sense.

and our worries are taken care of.

If you have further concerns, let me know, and I'll try and address them.