characterization of the dual space of the Sobolev space $H_0^1$

Question

I am slightly confused about the properties of the dual space of the Sobolev space $H_0^1$ as outlined on page 299 in Evans.

In particular, following the notation in the book, item 3 says that $\forall u \in H_{0}^{1}(U), v \subset L^2(U) \subset H^{-1}(U)$, $$(v,u)_{L^2(U)}=\langle v,u \rangle.$$

I am not quite sure how to prove this and it might be due to a confusion with notation. Since $v \in H^{-1}(U)$, item 1 in the book states that $\exists \, v^0,v^1, \dots , v^n$ in $L^2(U)$ such that $$\langle v,u\rangle=\int_Uv^0u+\sum_{i=1}^{n}v^{i}u_{x_i} \,dx.$$

In other words, we can identify $v$ with $(v^0,\dots,v^n)$. Since $v\in L^2$ and since this implies that $v$ is "associated" with the above functional $\langle v,u \rangle$ then one of the $v_i$'s have to be $v$ and the rest have to be $0$. Certainly, if $v^0=v$ the above statement follows. But why must this be the case? Why can't $v^1=v$ instead?

Or is this what is meant by $L^2(U) \subset H^{-1}(U)$, i.e. that if $v \in L^2$ then the functional associated with $v$ takes on the form $\int vu\,dx$? I am hoping to clarify this part because certainly $\int v_{x_i}u\,dx$ seems legitimate too.

Answer 1

It appears that my (first) edition of the book does not contain this statement, but I think I understand it. The elements of $H^{-1} $ are bounded linear functionals on $H^1_0$: $$H^{-1}=\{f:H_0^1\to\mathbb R \ ; \ |f(u)|\le C\|u\|_{H^1}\}$$ Then what do we mean by saying that $L^2\subset H^{-1}$? It means that some functionals on $H^1_0$ admit a bound by the $L^2$ norm, and thus can be extended to a functional on $L^2$. $$H^{-1}\supset L^2 = \{f:H_0^1\to\mathbb R \ ; \ |f(u)|\le C'\|u\|_{L^2}\}$$

Now invoke the structure theorem for $H^{-1}$, which identifies $f\in H^{-1}$ with a tuple $(f^0,\dots,f^n)$ of $L^2$ functions, via $$f(u)=\int f^0u+\sum_{i=1}^{n} \int f^{i}u_{x_i} \tag{1}$$ If $f^{i}$ is not a zero function for some $i\ne 0$, the functional (1) is not bounded by the $L^2$ norm of $u$, since the integral norm offers no control of the derivative $u_{x_i}$. Conversely, if $f^1=\dots=f^n=0$, then of course (1) is bounded on $L^2$.

Conclusion: the copy of $L^2$ within $H^{-1}$ can be described as $$\left\{f:H_0^1\to \mathbb R \ ; \ f(u) = \int f^0 u\right\}$$ where $f^0$ is an $L^2$ function.

The statement you quoted identifies $f$ with $f^0$, which is shorter but less precise than identifying it with $(f^0,0,\dots,0)$.

Answer 2

Sorry to bother here but I double the answer by @user127096. Let me quote the result from H. Brezis's book.

This book states that, in Proposition $9.20$, for $v\in H^{-1}(\Omega)$, there $\exists \, v^0,v^1, \dots , v^n$ in $L^2(U)$ such that $$\langle v,u\rangle=\int_Uv^0u+\sum_{i=1}^{n}v^{i}u_{x_i} \,dx.$$ and MOREOVER:

$(1)$: $ \, v^0,v^1, \dots , v^n$ is NOT determined uniquely by $v\in H^{-1}$, that is, $v_0$ can be varied.

$(2)$: If $\Omega$ is bounded, we could take $v^0\equiv 0$.

Hence, not to mention $(1)$, by $(2)$ we see that there has to be some $i\neq 0$ such that $v^i\neq0$, otherwise you functional $v$ will be $0$ all the time.

To explain your question, we have to go back to definition. How we write done $$ H_0^1\subset L^2 \approx (L^2)^*\subset H^{-1} $$ in the first place?

Actually, we use $(L^2)^*$ to discover what is $H^{-1}$ by applying a canonical mapping $T$: $(L^2)^*\to H^{-1}$ that is the restriction to $H_0^1$ of elements in $(L^2)^*$. i.e., $$<Tv,u>_{<H^{-1},\,H^1_0>}=<v,u>_{<(L^2)^*,L^2>} \,\,\,\,\,\,\,\,\,\,\,(*) $$ for all $u\in L^2$.

Hence, we have

$(1)$: $\|Tv\|_{H^{-1}}\leq C\|v\|_{L^2}$

$(2)$: $T$ is injective

$(3)$: $R(T)$ is dense since $H_0^1$ is dense in $L^2$ with respect to $L^2$ norm.

Therefore, we know already a big enough part of $H^{-1}$ by this way. Hence, your question is just the result of $(*)$

For more general explanation, please check out H. Brezis, page 136, Remark $3$.