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Dual spaces for Quotients and Subspaces

Idea: to describe, for $ M \leq X$, $ (X/M)^*$ and $ M^*$ in terms of $ X^*$.

Definition 10.3.1   $ M^\perp = \{ f \in X^* : f(M) = \{0\} \}$

Remark:
If $ X$ is a Hilbert space, then $ f = \left\langle{ \cdot, h }\right\rangle \in M^\perp$ iff $ h \perp g$ for all $ g \in M$. That is, $ h \in M^\perp$ (Hilbert space definition).
So, this definition can be viewed a generalization of orthogonality for Hilbert spaces.

We can reformulate the 5th corollary of the Hahn-Banach theorem, $ x \in \bar M$ if and only if $ f(x) = 0$ for all $ f \in M^\perp$.

Notice that $ M^\perp \leq X^*$ so $ X^*/M^\perp$ is a Banach space. We have $ \rho : X^* \rightarrow M^*$ defined by $ \rho(f) = f\vert _M$. Then,

$\displaystyle \ker \rho = M^\perp $

We have:

$\displaystyle \tilde \rho : X^*/M^\perp \rightarrow M^* $

with $ \rho(f + M^\perp) = \rho(f)$. Also, $ \Vert\tilde \rho\Vert = \Vert\rho\Vert$ (using proposition at end of section on quotients). We also have:

$\displaystyle \sigma : (X/M)^* \rightarrow X^* $

given by $ \sigma(f) = f \circ Q$ where $ Q:X \rightarrow X/M$ is the quotient map.

Theorem 10.3.2  
  1. $ \tilde \rho$ gives an isometric isomorphism from $ X^*/M^\perp$ onto $ M^*$
  2. $ \sigma$ gives an isometric isomorphism from $ (X/M)^*$ onto $ M^\perp$

Remark:

  1. Dual of a subspace is a quotient of the dual space.
  2. Dual of a quotient is a subspace of the dual space.

Proof. To see that $ \tilde \rho$ is isometric, given $ \phi \in M^*$ by Hahn-Banach, there exists $ f \in X^*$ such that $ f\vert _M = \phi$ and $ \Vert f\Vert = \Vert\phi\Vert$. Then, $ \tilde \rho(Q(f)) = \rho(f) = \phi$, so $ \tilde \phi$ is onto and $ \Vert\tilde \rho(Qf) \Vert = \Vert\rho(f)\Vert = \Vert f\Vert \geq \Vert Qf\Vert$.
Since $ \Vert\tilde f\Vert = 1$, $ \Vert Qf\Vert \geq \Vert\tilde \rho(Qf)\Vert$. Hence, $ \Vert\tilde \rho(Qf)\Vert = \Vert Qf\Vert$. To see that % latex2html id marker 17808
$ \textrm{ran }\sigma = M^\perp$, notice that $ Q(M) = \{0\}$ and $ f(0) = 0$, so

$\displaystyle \sigma(f) = f \circ Q $

maps $ M$ to $ \{0\}$. So, % latex2html id marker 17820
$ \textrm{ran }\sigma \subset M^\perp$
Given $ g \in M^\perp$, $ M \subset \ker g$. By the proposition, there is $ \tilde g : X / M \rightarrow \mathbb{F}$ such that

$\displaystyle \tilde g \circ Q = g $

$\displaystyle \Vert\tilde g\Vert = \Vert g\Vert $

So, % latex2html id marker 17832
$ \textrm{ran }\sigma \supset M^\perp$.
To show $ \sigma$ is isometric, we need $ \Vert\sigma(f) \Vert = \Vert f\Vert$ for all $ f \in (X/M)^*$. Given $ f \in X/M$, there is $ x_n \in X$ such that $ \Vert Qx_n\Vert < 1$ and $ \Vert f(Qx_n)\Vert \rightarrow \Vert f\Vert$ Pick $ y_n \in M$ such that $ \Vert x_n + y_n\Vert < 1$. Then,

$\displaystyle \Vert f \circ Q(x_n + y_n)\vert = \vert f(Q(x_n))\vert \rightarrow \Vert f\Vert $

and $ \Vert x_n + y_n\Vert < 1$ So, $ \Vert f \circ Q\Vert \geq \Vert f\Vert$. Since we already have $ \Vert f \circ Q\Vert \leq \Vert f\Vert$ already, we're done. $ \qedsymbol$


next up previous
Next: Reflexivity Up: Linear Functionals Previous: Banach Limits
Brian Bockelman 2005-12-12