## Dual spaces

Suppose ${(A,||\cdot||_A)}$ and ${(B,||\cdot||_B)}$ are Banach space, ${A^*}$ and ${B^*}$ are their dual spaces. If ${A\subset B}$ with ${||\cdot||_B\leq C||\cdot||_A}$, then

$\displaystyle i:A\mapsto B$

$\displaystyle \quad x\rightarrow x$

is an embedding. Let us consider the relation of two dual spaces. For any ${f\in B^*}$

$\displaystyle |\langle f,x\rangle|=|f(x)|\leq ||f||_{B^*}||x||_B\leq C||f||_{B^*}||x||_A\quad \forall\, x\in A$

Then ${f|_{A}}$ will be a bounded linear functional on ${A}$

$\displaystyle i^*:B^*\mapsto A^*$

$\displaystyle \qquad f\rightarrow f|_A$

is a bounded linear operator.

In a very special case that ${A}$ is a closed subset of ${B}$ under the norm ${||\cdot||_B}$, one can prove ${i^*}$ is surjective. In fact ${\forall\,g\in A^*}$ can be extended to ${\bar{g}}$ on ${B}$ by Hahn-Banach thm such that ${i^*\bar{g}=g}$. Then

$\displaystyle A^*=B^*/\ker i^*.$

Let us take ${A=H^1_0(\Omega)}$ and $\displaystyle B=H^1(\Omega)$ for example. Define the inner product to be

$\displaystyle \langle u,v\rangle=\int_\Omega uv+D_iuD_iv$

When ${\Omega}$ is a bounded subset of ${\mathbb{R}^n}$, ${H^1_0(\Omega)}$ is a proper closed subset of ${H^1(\Omega)}$. From the above explanation,

$\displaystyle H^{-1}(\Omega)=(H^1_0(\Omega))^*=(H^1(\Omega))^*/\ker i^*$

Define the continuous linear functional ${f(u)= \int_{\partial \Omega}u}$ for ${u\in H^1(\Omega)}$. One can see that ${i^*f=0}$.

$\textbf{Remark:}$ Should attribute this to Lun Zhang. I am always confused that $B^*\subset A^*$.