Consider the subset of odd positive integers defined and constructed as follows by these rules :
A) $1$ is in the set.
B) if $x$ is in the set , then $2x + 1$ is in the set.
C) if $x$ and $y$ are in the set then $xy$ is in the set.
I call them extended Mersenne numbers because rule A and B alone ...
Let $X$ be a Banach space with dual space $X^*$, and let $M$ be a closed subspace of $X$. Then $M^\perp=\{x^*\in X^*: x*(m)=0 \text{ for all } m\in M\}$ is a closed subspace in $X^*$. I read the following statement in Garnett's Bounded Analytic Functions:
The Hahn-Banach theorem gives us the ...
Let $X$ be a normed space and $Y$ a closed subspace of $X$ and $Y^0=\{f \in X^*|f(x)=0,\forall y \in Y\}$.
Prove that $Y^0$ is isometricaly isomorphic with $(X/Y)^*$.
I have to find a function specifically between $Y^0 $ and $(X/Y)^*$.
One idea is the function $S: Y^0 \longrightarrow (X/Y)^*$...Consider the subset of odd positive integers defined and constructed as follows by these rules :
A) $1$ is in the set.
B) if $x$ is in the set , then $2x + 1$ is in the set.
C) if $x$ and $y$ are in the set then $xy$ is in the set.
I call them extended Mersenne numbers because rule A and B alone ...
It is mostly correct now, but could be written better.
Consider $l:X'/U^\perp\to U'$, which sends each coset of $U^\perp$ to its restriction to $U$. This map is well-defined because all elements of the coset agree on $U$.
The map is surjective, because every functional $f$ on $U$ can be exten...
