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A complex linear functional is recovered from its real part by f(x)=u(x)-iu(ix)

Statement

Let $X$ be a complex vector space.

If $f : X \to C$ is complex linear and $u : = Re f$ , then $u$ is real linear on the underlying real vector space and

$f (x) = u (x) - i u (i x) (x \in X) .$

Conversely, if $u : X \to R$ is real linear on the underlying real vector space, then

$g (x) : = u (x) - i u (i x) (x \in X)$

defines a complex linear functional with $Re g = u$ . In particular a complex linear functional is uniquely determined by its real part.

Facts & Assumptions

Given: A complex vector space $X$ , a complex linear functional $f : X \to C$ , and a real linear functional $u$ on the underlying real vector space.

[L1]

A linear functional is additive and homogeneous over the relevant scalar field (Linear functionals and the algebraic dual $V^{*} = L (V, F)$ ).

[L2]

The underlying real vector space is the realification of Realification of a complex vector space by restriction of scalars. Complex coordinate arithmetic gives $i^{2} = - 1$ and $Re (a + i b) = a$ for real $a, b$ ( $C$ is the real coordinate plane, with coordinate arithmetic, Real and imaginary parts, complex conjugation, and modulus).

Proof

technique · direct

1.1L1L2givenalgebra

Let $u : = Re f$ . For $x, y \in X$ and $a \in R$ , $u (x + y) = Re (f (x + y)) = Re f (x) + Re f (y),$ and $u (a x) = Re (f (a x)) = Re (a f (x)) = a u (x) .$ So $u$ is real linear.

1.2L1givenalgebra

Write $f (x) = a + i b$ with $a, b \in R$ . Since $f$ is complex linear, $f (i x) = i f (x) = i a - b,$ so $Re f (x) = a$ and $Re f (i x) = - b$ . Therefore $u (x) - i u (i x) = a - i (- b) = a + i b = f (x) .$

1.3L1L2givenconstructalgebra

Conversely, let $u$ be real linear and define $g (x) : = u (x) - i u (i x)$ . Additivity is immediate from real linearity of $u$ . Also $g (i x) = u (i x) - i u (i^{2} x) = u (i x) + i u (x) = i (u (x) - i u (i x)) = i g (x) .$ Now for $λ = a + i b \in C$ with $a, b \in R$ , $g (λ x) = g (a x + b i x) = a g (x) + b g (i x) = a g (x) + b i g (x) = λ g (x) .$ Hence $g$ is complex linear.

2.1step 1.2step 1.3algebra∎

Taking real parts in the definition of $g$ gives $Re g (x) = u (x)$ for every $x$ . Together with step 1.2, this shows that a complex linear functional is uniquely determined by its real part.

Depends on

Used by

Dependency tree · two levels

11 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Daniel Daners, Introduction to Functional Analysis, Theorem 26.4 (standard reference, not scraped)
Gerald Teschl, Topics in Real and Functional Analysis, Theorem 4.14 (standard reference, not scraped)