LemmaStatement: Literature-sourcedProof: AI-generated

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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.

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

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources