Dual Operator is Bounded Linear Transformation

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Theorem

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

Let $X$ and $Y$ be normed vector spaces over $\GF$.

Let $T : X \to Y$ be a bounded linear transformation.

Let $X^\ast$ and $Y^\ast$ be the normed duals of $X$ and $Y$ respectively.


Then the dual operator $T^\ast : Y^\ast \to X^\ast$ is a bounded linear transformation.

Further:

$\norm {T^\ast}_{\map B {Y^\ast, X^\ast} } = \norm T_{\map B {X, Y} }$


Proof

Let $f, g \in Y^\ast$ and $\alpha, \beta \in \GF$.

Then we have, for each $x \in X$:

\(\ds \map {\paren {\map {T^\ast} {\alpha f + \beta g} } } x\) \(=\) \(\ds \map {\paren {\alpha f + \beta g} } {T x}\)
\(\ds \) \(=\) \(\ds \alpha \map f {T x} + \beta \map g {T x}\)
\(\ds \) \(=\) \(\ds \alpha \map {\paren {T^\ast f} } x + \beta \map {\paren {T^\ast g} } x\)

So we have:

$\map {T^\ast} {\alpha f + \beta g} = \alpha T^\ast f + \beta T^\ast g$

In Dual Operator is Well-Defined, it was shown that for each $f \in Y^\ast$ and $x \in X$ we have:

$\cmod {\map {\paren {T^\ast f} } x} \le \norm f_{Y^\ast} \norm T_{\map B {X, Y} } \norm x_X$

So we have:

$\norm {T^\ast f}_{X^\ast} \le \norm T_{\map B {X, Y} } \norm f_{Y^\ast}$

for $f \in Y^\ast$.

Hence:

$\norm {T^\ast}_{\map B {Y^\ast, X^\ast} } \le \norm T_{\map B {X, Y} }$

We finally need to show that $\norm {T^\ast}_{\map B {Y^\ast, X^\ast} } = \norm T_{\map B {X, Y} }$.

Pick a sequence $\sequence {x_n}_{n \in \N}$ such that $\norm {x_n}_X = 1$ and:

$\ds \norm T_{\map B {X, Y} } - \frac 1 n \le \norm {T x_n}_Y \le \norm T_{\map B {X, Y} }$

so that:

$\norm {T x_n}_Y \to \norm T_{\map B {X, Y} }$

For each $n \in \N$, let $f_n \in Y^\ast$ be a support functional for $T x_n$.

Then we have:

$\map {\paren {T^\ast f_n} } {x_n} = \map {f_n} {T x_n} = \norm {T x_n}_Y$

so that:

$\ds \cmod {\map {\paren {T^\ast f_n} } {x_n} } \ge \paren {\norm T_{\map B {X, Y} } - \frac 1 n} \norm {x_n}_X$

We therefore have:

$\ds \paren {\norm T_{\map B {X, Y} } - \frac 1 n} \norm {f_n}_{Y^\ast} \le \norm {T^\ast f_n}_{X^\ast}$

so that:

$\ds \norm {T^\ast}_{\map B {Y^\ast, X^\ast} } \ge \norm T_{\map B {X, Y} } - \frac 1 n$

for each $n \in \N$.

Then:

$\norm {T^\ast}_{\map B {Y^\ast, X^\ast} } \ge \norm T_{\map B {X, Y} }$

With that, we obtain:

$\norm {T^\ast}_{\map B {Y^\ast, X^\ast} } = \norm T_{\map B {X, Y} }$

$\blacksquare$


Sources

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