I have been reading in the book Variational Analysis in Sobolev and BV Spaces Applications to PDEs and Optimization by Hedy Attouch, Giuseppe Buttazzo and Gérard Michaille, and in section 3.1 (can be found on Google Books here) they discuss the Riesz identification of $ H $ and $ H^\ast $ for a Hilbert space, and when we may not identify a Hilbert space with its dual. As a particular example they mention the Sobolev space $ H_0^1(\Omega) $ and its dual space $ H^{-1}(\Omega) $, which cannot be identified when already identifying $ L^2(\Omega) $ with its dual thus obtaining the triplet $$ H_0^1(\Omega) \hookrightarrow L^2(\Omega) \hookrightarrow H^{-1}. $$ I can see how this somehow is problematic, but I have a hard time grasping exactly what it is we violate by making the identification?

I have an intuition (most likely completely wrong, please help me correct it) that the underlying abstract linear space should somehow be the same regardless of the representation chosen for the space. So the Riesz isomorphism yields one representation of the dual space of $ H_0^1(\Omega) $, and $ H^{-1}(\Omega) $ is here somehow another but these are apparently not comparable.

Can anyone here help me gain a better understanding of what is going on here, or maybe point me to a book which tackles this issue in more detail than the one I linked above? (Also linkede here again).

Edit: Equivalent to this question


1 Answer 1


This is indeed a delicate situation. Consider the spaces $H^1_0(\Omega)$ and $L^2(\Omega)$. They are Hilbert spaces when endowed with their standard scalar product.

Then using Riesz theorem, we see that $H^1_0(\Omega)$ is isomorphic to its dual space. That is, every continuous functional on $H^1_0(\Omega)$ can be represented as $H^1$-scalar product with a fixed function from $H^1_0(\Omega)$.

Now, there is also a different possibility. This is widely used in PDE theory. Given a function $u\in L^2(\Omega)$, the functional $$ v\mapsto \int_\Omega uv \ dx $$ is a continuous functional on $H^1_0(\Omega)$. This functional is now induced by the $L^2$-scalar product. Now, if we take $u\in H^1_0(\Omega)\subset L^2(\Omega)$, the above construction is an alternative to obtain functionals on $H^1_0(\Omega)$.

So there are two possibilities of constructing functionals on $H^1_0(\Omega)$: using the $H^1_0(\Omega)$ or the $L^2(\Omega)$ scalar product. Both constructions are not equivalent. While the first gives rise to an isomorphism (Riesz), the second construction gives a compact mapping from $H^1_0(\Omega)$ to its dual.


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