Bounded Real-Valued Linear Functional is Real Part of Unique Bounded Complex-Valued Linear Functional

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Theorem

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

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

Let $f : X_\R \to \R$ be a bounded linear functional.


Then there exists a unique bounded linear functional $g : X \to \C$ such that:

$\map f x = \map \Re {\map g x}$

for each $x \in X$.


Proof

Let $M > 0$ be such that:

$\cmod {\map f x} \le M \norm x$

for each $x \in X$.

From Linear Functional on Complex Vector Space is Uniquely Determined by Real Part, there exists a unique linear functional $g : X \to \C$ such that:

$\map f x = \map \Re {\map g x}$

for each $x \in X$, given by:

$\map g x = \map f x - i \map f {i x}$

for each $x \in X$.

We just need to show that $g$ is bounded.

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

\(\ds \cmod {\map g x}^2\) \(=\) \(\ds \size {\map f x}^2 + \size {\map f {i x} }^2\)
\(\ds \) \(\le\) \(\ds M^2 \norm x^2 + M^2 \norm {i x}^2\)
\(\ds \) \(=\) \(\ds 2 M^2 \norm x^2\) Norm Axiom $\text N 2$: Positive Homogeneity

so that:

$\cmod {\map g x} \le M \sqrt 2 \norm x$

for each $x \in X$.

So $g$ is a bounded linear functional as required.

$\blacksquare$