# dual of $H^1_0$: $H^{-1}$ or $H_0^1$?

I have a problem related to dual of Sobolev space $H^1_0$. By definition, the dual of $H^1_0$ is $H^{-1}$, which contains $L^2$ as a subspace. However, from Riesz representation theorem, dual of a Hilbert space is itself (in the sense of isomorphism). It seems that this implies dual of $H^1_0$ is $H^1_0$ itself.

From the result, I think the latter argument is not right. But where's the mistake?

This problem may related to the question about Hilbert space and its dual A paradox on Hilbert spaces and their duals

Thanks so much!

Thanks a lot for all the kind solutions!

The following are my conclusions:

1. Dual space strongly depends on action we take. For example, in our case, if we take $H^1$ inner product, the dual space is $H_0^1$ itself; if taking $L^2$ inner product, the dual space is larger than $L^2$; and taking all possible dual actions, the dual is by definition $H^{-1}$.

2. As dual of one space, there are certainly identifications between these spaces. Notice that identification between infinite dimensional spaces can be bijective and ISOMETRIC! Indeed there is a simple example: sequence {1,2,3...} and {2,4,6...}. Just taking metric of the latter to be half.

3. Another basic example is simple elliptic equation in $H_0^1$: $$\int_\Omega \nabla u\cdot \nabla v+\int_\Omega uv=\langle f,v\rangle , \forall v\in H_0^1(\Omega)$$ where $f\in H^{-1}$. This gives isometric identification of $H^{-1}$ and $H_0^1$.

4. Brezis's book reminds that we can't take two different inner products to give both Riesz representation at the same time. The example there is very instructive.

• We do not identify the dual of $H_0^1$ with itself. Brezis gives a nice summary books.google.com/… – Euler....IS_ALIVE Apr 5 '14 at 4:24
• – Jack May 17 '16 at 22:06

As you have said, the dual space of $H_0^1$ is by definition $H^{-1}$. It can be identified with $H_0^1$ via isomorphism, but whether the identification is useful or not, it depends on the problem you are studying.

If, for example, you are studying abstract problems about Hilbert spaces and it's duals then, the identification is recommended and you gain a lot of time by using it. Take a look, for example, in Brezis book chapter 5.

On the other hand, when dealing with some practical problems, like for example, Partial Differential Equations problems, it is more convenient to use just $H^{-1}$. In some further examples its even better when you know a representation for elements of $H^{-1}$ in terms of $L^2$ functions. See, for instance, page 291 of Brezis book.

Let me restate your question as follows. Let $\Omega$ be a nonempty open subset of $\mathbb{R}^d$. On the one hand, we have the sandwich: $$H_0^1(\Omega)\subset L^2(\Omega)\subsetneq H^{-1} (\Omega),\tag{*}$$ where $H^{-1}(\Omega)$ is defined as the dual of $H_0^1(\Omega)$. On the other hand, by Riesz representation theorem, $H^{-1}(\Omega)$ can be identified as $H_0^1(\Omega)$. This suggests that we may write $$H^{-1}(\Omega)=H_0^1(\Omega)\tag{**}$$ which contradicts $(*)$ since $L^2(\Omega)$ is a proper subspace of $H^{-1}(\Omega)$.

The problem is that $(**)$ is not true if one considers "$=$" in $(**)$ as equality of sets rather than an isomorphism of Hilbert space.

As Terry Tao said:

"It is better to think of isomorphic pairs as being equivalent or identifiable rather than identical, as the latter can lead to some confusion if one treats too many of the equivalences as equalities. For instance, $\ell^2(\{0,1,2,\dots\})$ and $\ell^2(\{1,2,\dots\})$ are equivalent (one can simply shift the standard orthonormal basis for the former by one unit to obtain the latter), but one can also identify the latter space as a subspace of the former. It is fairly harmless to treat one of these equivalences as an equality, but of course one cannot do so for both equivalences at the same time."

The dual space $(H_0^1)^*$ is per definition the space of all linear and continuous functionals $(H_0^1)^*\to \mathbb R$.

There are different ways in generating this dual space:

(1) Riesz representation lets you represent each functional $f\in (H_0^1)^*$ by scalar product with an element $u_f\in (H_0^1)$ $$f(v) = \langle u_f,v\rangle \quad\forall v\in H_0^1.$$

(2) Every function $u\in L^2$ also generates a function in $(H_0^1)^*$ via $$f_u(v)=\int_\Omega uv\ dx \quad\forall v\in H_0^1.$$ You can even take $u\in H_0^1$.

If you look at the different functionals, then you see that the functionals in (1) are defined by the $H^1$-scalar product, the functionals in (2) are defined by the $L^2$-scalar product.

Riesz representation theorem is a : "representation" theorem, and not an identification theorem. In the example 3 above, $f = - \Delta u + u$ can be represented by $u$, but cannot be identified by $u$, since $f$ is in $H^{-1}$ whereas $u$ is in $H^1_0$.