Characterization of Dual Operator
Theorem
Let $\GF \in \set {\R, \C}$.
Let $X$ and $Y$ be normed vector spaces over $\GF$.
Let $X^\ast$ and $Y^\ast$ be the normed dual spaces of $X$ and $Y$ respectively.
Let $X^{\ast \ast}$ and $Y^{\ast \ast}$ be the second normed duals of $X$ and $Y$ respectively.
Let $T : Y^\ast \to X^\ast$ be a bounded linear transformation.
Let $\iota_X : X \to X^{\ast \ast}$ be the evaluation linear transformation for $X$.
Let $\iota_Y : Y \to Y^{\ast \ast}$ be the evaluation linear transformation for $Y$.
The following statements are equivalent:
- $(1) \quad$ $T$ is $\struct {w^\ast, w^\ast}$-continuous
- $(2) \quad$ there exists a bounded linear transformation $S : X \to Y$ such that $S^\ast = T$, where $S^\ast$ denotes the dual operator of $S$
- $(3) \quad$ $T^\ast \sqbrk {\iota_X X} \subseteq \iota_Y Y$
Proof
$(1)$ implies $(2)$
Suppose that $T$ is $\struct {w^\ast, w^\ast}$-continuous.
From Characterization of Continuity of Linear Functional in Weak-* Topology , we have that:
- $x^\wedge : \struct {X^\ast, w^\ast} \to \GF$ is continuous.
From Composite of Continuous Mappings is Continuous:
- $x^\wedge \circ T : \struct {Y^\ast, w^\ast} \to \GF$ is continuous.
Applying Characterization of Continuity of Linear Functional in Weak-* Topology , there exists $S x \in Y$ such that:
- $x^\wedge \circ T = \paren {S x}^\wedge$
We first show that $S : X \to Y$ is linear.
We have:
\(\ds \paren {\map S {\alpha x + \beta y} }^\wedge\) | \(=\) | \(\ds \paren {\alpha x + \beta y}^\wedge \circ T\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \alpha \paren {x^\wedge \circ T} + \beta \paren {y^\wedge \circ T}\) | Evaluation Linear Transformation on Normed Vector Space is Linear Transformation from Space to Second Normed Dual | |||||||||||
\(\ds \) | \(=\) | \(\ds \alpha S x + \beta S y\) |
We now show that $S$ is bounded.
We have, for each $x \in X$:
\(\ds \norm {S x}_Y\) | \(=\) | \(\ds \norm {\paren {S x}^\wedge}_{Y^{\ast \ast} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \norm {x^\wedge \circ T}_{Y^{\ast \ast} }\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \norm {x^\wedge}_{X^{\ast \ast} } \norm T_{\map B {Y^\ast, X^\ast} }\) | Norm on Bounded Linear Transformation is Submultiplicative | |||||||||||
\(\ds \) | \(=\) | \(\ds \norm T_{\map B {Y^\ast, X^\ast} } \norm x_X\) | Evaluation Linear Transformation on Normed Vector Space is Linear Isometry |
So $S$ is bounded.
Finally, for $f \in Y^\ast$ and $x \in X$ we have:
\(\ds \map f {S x}\) | \(=\) | \(\ds \map {\paren {S x}^\wedge} f\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map {\paren {x^\wedge \circ T} } f\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map {x^\wedge} {T f}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map {\paren {T f} } x\) | Definition of Evaluation Linear Transformation on Normed Vector Space |
That is:
- $f \circ S = T f$
for each $f \in Y^\ast$.
From the definition of the dual operator we have:
- $S^\ast f = T f$
for each $f \in Y^\ast$.
So $S^\ast = T$.
$\Box$
$(2)$ implies $(3)$
Suppose that:
- there exists a bounded linear transformation $S : X \to Y$ such that $S^\ast = T$, where $S^\ast$ denotes the dual operator of $S$.
We have $S^{\ast \ast} = T^\ast$, where $S^{\ast \ast}$ denotes the second dual operator.
Let $x \in X$ and $f \in Y^\ast$.
Then:
\(\ds \map {\map {T^\ast} {x^\wedge} } f\) | \(=\) | \(\ds \map {\map {S^{\ast \ast} } {x^\wedge} } f\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map {\paren {x^\wedge \circ S^\ast} } f\) | Definition of Dual Operator | |||||||||||
\(\ds \) | \(=\) | \(\ds \map {x^\wedge} {S^\ast f}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map {x^\wedge} {f \circ S}\) | Definition of Dual Operator | |||||||||||
\(\ds \) | \(=\) | \(\ds \map f {S x}\) | Definition of Evaluation Linear Transformation on Normed Vector Space | |||||||||||
\(\ds \) | \(=\) | \(\ds \map {\paren {S x}^\wedge} f\) | Definition of Evaluation Linear Transformation on Normed Vector Space |
That is:
- $\map {T^\ast} {\iota_X x} = \map {\iota_Y} {S x}$
$\Box$
$(3)$ implies $(1)$
Suppose that $T^\ast \sqbrk {\iota_X X} \subseteq \iota_Y Y$.
For each $x^\wedge \in \iota_X X$, we then have that:
- $\map {T^\ast} {x^\wedge} \in \iota_Y Y$
That is, there exists $y \in Y$ such that:
- $x^\wedge \circ T = y^\wedge$
From Characterization of Continuity of Linear Functional in Weak-* Topology, we then have:
- $x^\wedge \circ T : \struct {Y^\ast, w^\ast} \to \GF$ is continuous
for each $x \in X$.
Conversely from Characterization of Continuity of Linear Functional in Weak-* Topology, we have that every continuous linear functional $\Phi : \struct {Y^\ast, w^\ast} \to \GF$ has the form $\Phi = x^\wedge$.
So from Continuity in Initial Topology, we have that:
- $T : \struct {Y^\ast, w^\ast} \to \struct {X^\ast, w^\ast}$ is continuous.
$\blacksquare$