We could identity $H^{-1}$ with $H_0^1$ but we don’t. Why

partial differential equationssobolev-spaces

Today in our lecture on partial differential equations while discussing dual spaces of Sobolev spaces:

We could identify $H^{-1}$ with $H_0^{-1}$ by the $H_0^1$ inner product (Riesz) But won't and rather identify
$$
H_0^1(a,b) \hookrightarrow L^2(a,b) \cong (L^2(a,b))^* \hookrightarrow H^{-1}(a,b)
$$
and therefore regard $H_0^1$ as subspace of $H^{-1}$ via the $L^2$ inner product.

I'm interested on why one would do this.

Notation: $H^k = W^{k,2}$, where $W$ is the Sobolev space.
$W_0^{1,p}$ is the closure of $\mathcal{C}_0^{\infty}$ with respect to the Sobolev norm $\| \cdot \|_{1,p}$ and define $W^{-1,q}(a,b) := (W_0^{1,p}(a,b))^{\ast}$.

Best Answer

I think the situation is maybe more clear in the case $(H^1)^*$ and $H^1$, which are also isomorphic Hilbert spaces by the Riesz isomorphism w.r.t. the $H^1$ scalar product.

In short

The problem is that the following diagram does not commute! $$ \begin{array}{ccccccc} H^1 & \hookrightarrow & L^2 \\ \downarrow & \unicode{x21af} & \downarrow \\ (H^1)^* & \hookleftarrow & (L^{2})^* \end{array} $$ If we use multiple Riesz isomorphisms, it is ambiguous to write $f \in (H^1)^*$ for a function $f \in H^{1}$. Since it is not clear which path we took in the diagram above.

Therefore we should stick to one isomorphism and not use multiple (non-commuting) ones!

An example

We consider $\Omega = [0,1]$ and $H^1([0,1])$. This space contains a function defined via $f(x) = x$.

  • using the embedding into $L^2$, we get $f_{L^2}(x) = x$
  • using the $L^2$-Riesz isomorphism, we find $f_{(L^{2})^*}[ g ] = \int_0^1 x \cdot g(x) \, \mathrm{d} x $, for $g \in L^2$
  • and finally, $f_{(H^1)^*}[g] = \int_0^1 x \cdot g(x)$, for $g \in H^1$

Now, what if we use the $H^1$-Riesz isomorphism directly?

The scalarproduct is given by $\langle f , g \rangle_{H^1} = \int f(x) g(x) + f'(x) g'(x)\, \mathrm d x$

Therefore, for $g \in H^1$, we get $$ \widetilde{f}_{(H^1)^*}[g] = \int x g(x) + g'(x)\, \mathrm d x. $$

Hence $$\widetilde{f}_{(H^1)^*} \neq f_{(H^1)^*}! $$

As we see in the example, if we apply the isomorphisms implied by the $L^2$ and the $H^1$ scalar products, we would need to use different symbols for the same function, in order to keep track of the way we took from $H^1$ to $(H^1)^*$.

(Notice: Of course, one should be careful with point evaluations, expecially in $L^2$. But to keep the formulas short, I commited this crime here. Feel free to correct the formulas above into terms like $f_{(L^2)^*}[g] = \int_{0}^1 f_{L^2} g \, \mathrm d \lambda$ instead.)

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