Let \(V\) be the set of all complex-valued functions \(f\) on the real line such that (for all \(t\) in \(\RR\)) \[ f(-t) = \overline{f(t)}. \] The bar denotes complex conjugation. Show that \(V\), with the operations \[ (f + g)(t) = f(t) + g(t), \qquad (cf)(t) = cf(t) \] is a vector space over the field of real numbers. Give an example of a function in \(V\) which is not real-valued.
Solution
The zero function belongs to \(V\). If \(f,g\in V\), then \[ (f+g)(-t)=f(-t)+g(-t)=\overline{f(t)}+\overline{g(t)} =\overline{(f+g)(t)}, \] so \(f+g\in V\). If \(c\in\RR\), then \[ (cf)(-t)=c f(-t)=c\overline{f(t)}=\overline{c f(t)}, \] because \(c\) is real. Thus \(cf\in V\). Hence \(V\) is a real vector subspace of the vector space of all complex-valued functions on \(\RR\).
A non-real-valued example is \[ f(t)=it. \] Indeed, \[ f(-t)=-it=\overline{it}=\overline{f(t)}, \] and \(f(t)\) is nonreal whenever \(t\ne0\).