Problem in dual space of $H_0^1$

Question

This problem is discussed many times. Let $H^{-1}(U)$ be the dual of $H_0^1$.

Further i know this thing also that $$ H_{0}^{1}(U) \subset L^2(U) \subset H^{-1}(U)$$ Clearly as 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 $u_f\in (H_0^1)$ $$ f(v) = \langle u_f,v\rangle = \int_{\Omega} u_fv+Du_f.Dv\ dx \quad\forall v\in H_0^1. $$

(2) Every function $u\in L^2$ also generates a function in $(H_0^1)^*$ via $$ f_u(v)=\int_\Omega uv\ dx \quad\forall v\in H_0^1. $$ These things are fine. Now what is written that

1. We donot identify the space $H_0^1$ With its dual.

2. Further while making characterization of $H^{-1}$ in the evans book what is written that for any $f \in H^{-1}(U)$ by Reisz representation theorem $$<f, v> = \int_{U} Du. Dv +uv dx$$ Why it is so? Since on one side we are saying that we donot identify the space $H_0^1(U)$ with its dual and on other side we are saying (2) holds?

3. So there must be some functional which must be not in $L^2(U)$ and $H_0^1(U)$?

I have read all the posts regarding this question but i don't get my answer which is the main thing which i am missing or not getting?

Answer 1

The reason we do not identify $H^{-1}$ with $H^1_0$ is that we want $H^{-1}$ to contain the derivatives of $L^2$ functions. For example, for any $f \in L^2(U)$, it turns out that $f_{x_i}$ belongs to $H^{-1}$ where $f_{x_i}$ is interpreted in the distributional sense, so

$$\langle f_{x_i},v\rangle := -\int_\Omega f\, v_{x_i} \, dx$$

for all $v \in H^1_0(\Omega)$. Similarly, second derivatives of $H^1_0$ functions belong to $H^{-1}$, since

$$\langle f_{x_ix_j},v\rangle:= -\int_{\Omega} f_{x_i} v_{x_j} \, dx.$$

You can verify both of the above are bounded linear functionals on $H^1_0(\Omega)$, and they agree with the $L^2$ innerproduct when everything is smooth.

Even though we don't identify $H^1_0$ with its dual, we can still use the Riesz-representation theorem to characterize $H^{-1}$. Notice in Evans he does not write what you wrote in (2) in the statement of the theorem. Instead he writes

$$\langle f,v\rangle = \int_\Omega f^0 v + \sum_{i=1}^n f^iv_{x_i}\, dx$$

for $f^0,f^1,\dots,f^n \in L^2(\Omega)$. This basically says that everything in $H^{-1}$ is the sum of $L^2$ functions and first derivatives of $L^2$ functions. Yes it is true that we can take $f^0=u$ and $(f^1,\dots,f^n)=Du$ for $u$ found by the Riesz-representation theorem, as you indicated, but we do not identify $f$ with $u$.

There are indeed elements of $H^{-1}(\Omega)$ that are not in $L^2(\Omega)$ (and hence not in $H^1_0(\Omega)$) using these identifications. As an example, consider $\Omega=(-1,1)$ and let $f$ be the Heaviside function

$$f(x) = \begin{cases} 1,&\text{if } 0 \leq x < 1\\ 0,&\text{if } -1 < x < 0.\end{cases}$$

Then $f'(x)$ defined by

$$\langle f',v\rangle :=-\int_{-1}^1 f(x)v'(x)\,dx$$

is an element of $H^{-1}(\Omega)$ that does not belong to $L^2(\Omega)$. In fact, $f'$ is the Delta function, since

$$\langle f',v\rangle = -\int_0^1 v'(x) \, dx = v(0)-v(1) = v(0).$$

Note that we used the fact that $v \in H^1_0(\Omega)$ so $v$ is continuous and $v(-1)=0=v(1)$.

In this case, if we were to identify $f'\in H^{-1}(\Omega)$ with an element $u \in H^1_0(\Omega)$, $u$ should satisfy

$$\int_{-1}^1 uv + u'v' \,dx = v(0)$$

for all $v \in H^1_0(\Omega)$. This is the weak form of asking that

$$u - u'' = f'.$$

The solution (you can check by plugging in) is

$$u(x) = \frac{1}{2\cosh(1)}\begin{cases} \sinh(1-x),&\text{if } 0 \leq x \leq 1\\ \sinh(1+x),&\text{if } -1 \leq x < 0.\end{cases}$$

This also the Green's function for $u-u'' = 0$ centered at $x=0$. So we are not identifying $f'$ with $u$.