[Math] Complex linear functionals and real linear functionals

functional-analysisreal-analysis

Proposition: Let $X$ be a vector space on $\mathbb{C}$. If $f$ is a complex linear functional on $X$ and $u=\text{Re} f$, then $u$ is a
real linear functional and $f(x) = u(x) – i u(ix)$ for all $ x \in X$.

I am using Folland, and the proof he gives of this is very brief:

"If $f$ is complex linear and $u=\text{Re} f$, $u$ is clearly real
linear and $\text{Im} f(x) = – \text{Re}[if(x)] = -u(ix)$"

I am not very good with complex numbers, and only just started functional analysis this week. Where does folland get that $\text{Im} f(x) = – \text{Re}[if(x)] = -u(ix)$? I.e. why is this true?

Best Answer

Let $z \in \def\C{\mathbf C}\C$ be any complex number, then $\def\Re{\operatorname{Re}}\def\Im{\operatorname{Im}}$ $$ z = \Re z + i\Im z $$ multiplying by $i$, we have $$ iz = -\Im z + i\Re z$$ that is the real part of $iz$ is $\Re(iz) = -\Im z$, so we get $$ \Im z = -\Re (iz) $$ Folland applies this to $z = f(x)$, giving $$ \Im f(x) = -\Re [if(x)] $$ As $f$ is complex linear, $if(x) = f(ix)$, giving $$ \Im f(x) = -\Re f(ix) $$ As $u$ was defined to be $\Re f$, we have $$ \Im f(x) = -u(ix) $$ Hence \begin{align*} f(x) &= \Re f(x) + i\Im f(x)\\ &= u(x) + i \cdot \bigl(-u(ix)\bigr)\\ &= u(x) -iu(ix) \end{align*}