Dual Operator is Well-Defined

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 duals of $X$ and $Y$ respectively.

Then the dual operator $T : Y^\ast \to X^\ast$ is well-defined.

We first show that $T^\ast$ is well-defined as a mapping $Y^\ast \to X^\ast$.

That is, we want to show that $T^\ast f \in X^\ast$ for each $f \in Y^\ast$.

For $x, y \in X$ and $\lambda, \mu \in \GF$, we have:

$\ds \map {\paren {T^\ast f} } {\alpha x + \beta y}$

$=$

$\ds \map f {\map T {\alpha x + \beta y} }$

$\ds $

$=$

$\ds \map f {\alpha T x + \beta T y}$

$\ds $

$=$

$\ds \alpha \map f {T x} + \beta \map f {T y}$

$\ds $

$=$

$\ds \alpha \map {\paren {T^\ast f} } x + \beta \map {\paren {T^\ast f} } y$

So $T^\ast f$ is a linear functional.

We show that $T^\ast f$ is bounded.

For each $x \in X$, we have:

$\ds \cmod {\map {\paren {T^\ast f} } x}$

$=$

$\ds \cmod {\map f {T x} }$

$\ds $

$\le$

$\ds \norm f_{Y^\ast} \norm {T x}_Y$

$\ds $

$\le$

$\ds \norm f_{Y^\ast} \norm T_{\map B {X, Y} } \norm x_X$

So $T^\ast f$ is bounded.

$\blacksquare$