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^*.

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