Realification of Normed Dual is Isometrically Isomorphic to the Normed Dual of Realification

Theorem

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

Let $\struct {X_\R, \norm {\, \cdot \,} }$ be the realification of $\struct {X, \norm {\, \cdot \,} }$.

Let $\struct {X^\ast, \norm {\, \cdot \,}_{X^\ast} }$ be the normed dual of $\struct {X, \norm {\, \cdot \,} }$.

Let $\struct {X^\ast_\R, \norm {\, \cdot \,}_{X^\ast_\R} }$ be the normed dual of $\struct {X_\R, \norm {\, \cdot \,} }$.

Let $\struct {X^\ast, \norm {\, \cdot \,}_{X^\ast} }_\R$ be the realification of $\struct {X^\ast, \norm {\, \cdot \,}_{X^\ast} }$.

Then $\struct {X^\ast, \norm {\, \cdot \,}_{X^\ast} }_\R$ and $\struct {X^\ast_\R, \norm {\, \cdot \,}_{X^\ast} }$ are isometrically isomorphic.

Further $T : \struct {X^\ast, \norm {\, \cdot \,}_{X^\ast} }_\R \to \struct {X^\ast_\R, \norm {\, \cdot \,}_{X^\ast} }$ defined by:

$T f = \map \Re f$

is an isometric isomorphism.

Proof

We first need to show that if $f \in X^\ast$, then $T f = \map \Re f \in X^\ast_\R$.

From Real Part of Linear Functional is Linear Functional, we have that $\map \Re f : X \to \R$ is $\R$-linear.

We just need to show that $\map \Re f$ is bounded.

We have for each $x \in X$:

\(\ds \cmod {\map \Re {\map f x} }\)

\(=\)

\(\ds \sqrt {\paren {\map \Re {\map f x} }^2}\)

\(\ds \)

\(\le\)

\(\ds \sqrt {\paren {\map \Re {\map f x} }^2 + \paren {\map \Im {\map f x} }^2}\)

\(\ds \)

\(=\)

\(\ds \cmod {\map f x}\)

\(\ds \)

\(\le\)

\(\ds \norm f_{X^\ast} \norm x\)

Fundamental Property of Norm on Bounded Linear Functional

So we have that:

$\map \Re f$ is bounded

with:

$\norm {\map \Re f}_{X^\ast_\R} \le \norm f_{X^\ast}$

We want to show that:

$\norm {\map \Re f}_{X^\ast_\R} = \norm f_{X^\ast}$

for each $f \in X^\ast$.

Let $f \in X^\ast$.

Let $\sequence {u_n}_{n \in \N}$ be a sequence with $\norm {u_n} = 1$ and:

$\ds \norm f_{X^\ast} - \frac 1 n \le \cmod {\map f {u_n} } \le \norm f_{X^\ast}$

so that $\cmod {\map f {u_n} } \to \norm f_{X^\ast}$.

Pick $\alpha_n \in \C$ with $\cmod {\alpha_n} = 1$ such that $\cmod {\map f {u_n} } = \alpha_n \map f {u_n}$.

Since $f : X \to \C$ is $\C$-linear, we have $\alpha_n \map f {u_n} = \map f {\alpha_n u_n}$.

So we have $\map \Re {\map f {\alpha_n u_n} } = \cmod {\map f {u_n} }$.

Then:

$\map \Re {\map f {\alpha_n u_n} } \to \norm f_{X^\ast}$

while:

$\norm {\alpha_n u_n} = 1$ for each $n \in \N$.

This gives:

$\norm f_{X^\ast} \le \norm {\map \Re f}_{X^\ast_\R}$

so that:

$\norm {\map \Re f}_{X^\ast_\R} = \norm f_{X^\ast}$

Finally, we show that $T$ is surjective.

This is given by Bounded Real-Valued Linear Functional is Real Part of Unique Bounded Complex-Valued Linear Functional.

$\blacksquare$