Normed Dual Space of Normed Quotient Vector Space is Isometrically Isomorphic to Annihilator

Theorem
Let $\GF \in \set {\R, \C}$.

Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space over $\GF$.

Let $X^\ast$ be the normed dual space of $X$.

Let $Y$ be a closed linear subspace of $X$.

Let $\paren {X/Y}^\ast$ be the normed dual space of the normed quotient vector space $X/Y$.

Then:


 * $\paren {X/Y}^\ast$ is isometrically isomorphic to $Y^\bot$

where $Y^\bot$ is the annihilator of $Y$.

Proof
Let $\pi$ be the quotient mapping associated with $X/Y$.

Note that for each $f \in \paren {X/Y}^\ast$ we have:


 * $f \circ \pi \in X^\ast$

by Composition of Bounded Linear Transformations is Bounded Linear Transformation.

Further, for $y \in Y$ we have $\map \pi y = 0_{X/Y}$ from Kernel of Quotient Mapping, and so:


 * $\map {\paren {f \circ \pi} } y = \map f {0_{X/Y} } = 0$ for all $y \in Y$.

So we have $f \circ \pi \in Y^\bot$.

Now define $\phi : \paren {X/Y}^\ast \to Y^\bot$ by:


 * $\map \phi f = f \circ \pi$

From Linear Isometry is Injective: Corollary, it suffices to show that $\phi$ is a linear isometry and surjective.

First, take $f, g \in \paren {X/Y}^\ast$ and $\lambda, \mu \in \GF$.

We have:

so $\phi$ is linear.

Now, we have for $f \in \paren {X/Y}^\ast$:

so $\phi$ is a linear isometry.

It remains to show that $\phi$ is surjective.

Since we have shown that:


 * $\ds \sup_{\mathbf x \in \map \pi {B_X} } \cmod {\map f {\mathbf x} } = \sup_{\mathbf x \in B_{X/Y} } \cmod {\map f {\mathbf x} }$

we know that if $f \circ \pi \in X^\ast$, then $f \in \paren {X/Y}^\ast$.

From Condition for Mapping from Quotient Vector Space to be Well-Defined, we have that for each $g \in X^\ast$, there exists $f : X/Y \to \GF$ such that:


 * $g = f \circ \pi$

From the above, it follows that $f \in \paren {X/Y}^\ast$ and so $g = \map \phi f$.

So $\phi$ is surjective, and we are done from Linear Isometry is Injective: Corollary.