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

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 $$W^{-1,q}(a,b) := (W_0^{1,p}(a,b))^{\ast}$$.

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.)