Dual spaces

lim Practice= Perfect

Suppose $latex {(A,||cdot||_A)}&fg=000000$ and $latex {(B,||cdot||_B)}&fg=000000$ are Banach space, $latex {A^*}&fg=000000$ and $latex {B^*}&fg=000000$ are their dual spaces. If $latex {Asubset B}&fg=000000$ with $latex {||cdot||_Bleq C||cdot||_A}&fg=000000$, then

$latex displaystyle i:Amapsto B&fg=000000$

$latex displaystyle quad xrightarrow x&fg=000000$

is an embedding. Let us consider the relation of two dual spaces. For any $latex {fin B^*}&fg=000000$

$latex displaystyle |langle f,xrangle|=|f(x)|leq ||f||_{B^*}||x||_Bleq C||f||_{B^*}||x||_Aquad forall, xin A&fg=000000$

Then $latex {f|_{A}}&fg=000000$ will be a bounded linear functional on $latex {A}&fg=000000$

$latex displaystyle i^*:B^*mapsto A^*&fg=000000$

$latex displaystyle qquad frightarrow f|_A&fg=000000$

is a bounded linear operator.

In a very special case that $latex {A}&fg=000000$ is a closed subset of $latex {B}&fg=000000$ under the norm $latex {||cdot||_B}&fg=000000$, one can prove $latex {i^*}&fg=000000$ is surjective. In fact $latex {forall,gin A^*}&fg=000000$ can be extended to $latex {bar{g}}&fg=000000$ on $latex {B}&fg=000000$ by Hahn-Banach thm such that $latex {i^*bar{g}=g}&fg=000000$. Then

$latex displaystyle A^*=B^*/ker i^*.&fg=000000$

Let us take $latex {A=H^1_0(Omega)}&fg=000000$ and $latex displaystyle B=H^1(Omega)&fg=000000$…

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