> [!definition] > > Let $\cx$ be a [[Normed Vector Space|normed space]]. An element of its [[Topological Dual|dual]] $\cx^*$ is known as a **(bounded) linear functional**. > [!theorem] > > Let $\cx$ be a vector space over $\complex$. > - If $f$ is a complex linear functional on $\cx$ and $u = \re{f}$, then $u$ is a real linear functional, and $f(x) = u(x) - iu(ix)$ for all $x \in \cx$. > - If $u$ is a real linear functional on $\cx$ and $f: \cx \to \complex$ is defined by $f(x) = u(x) - iu(ix)$, then $f$ is complex linear. > - If $\cx$ is normed, then $\norm{u} = \norm{f}$. > > *Proof*. For any $x, y \in \cx$, $\lambda \in \real$, > $ > \begin{align*} > u(\lambda x + y) &= \re{f(\lambda x + y)} \\ > &= \lambda\re{f( x)} + \re{f(y)} \\ > &= \lambda u(x) + u(y) > \end{align*} > $ > so $u$ is linear, and > $ > \begin{align*} > u(x) - iu(ix) &= \re{f(x)} - i\re{f(ix)} \\ > &= \re{f(x)} - i\re{if(x)} \\ > &= \re{f(x)} + \im{f(x)} > \end{align*} > $ > Now take, > $ > f(ix) = u(ix) - iu(i^2x) = u(ix) + iu(x) = if(x) > $ > and for any $\lambda \in \real$, > $ > f(\lambda x) = u(\lambda x) - iu(i\lambda x) = \lambda u(x) - \lambda iu(ix) = \lambda f(x) > $ > Therefore for any $a + bi \in \complex$, > $ > \begin{align*} > f((a + bi)x) &= f(ax) + f(bix) \\ > &= af(x) + bif(x) \\ > &= (a + bi)f(x) > \end{align*} > $ > > Lastly, let $x \in \cx: \norm{x} = 1$, then > $ > \begin{align*} > \norm{f(x)} = |f(x)| &= \sqrt{\re{f(x)}^2 + \im{f(x)}^2} \\ > &\ge |\re{f}| = |u(x)| = \norm{u(x)} > \end{align*} > $ > and $\norm{f} \ge \norm{u}$. Let $\alpha = \ol{\sgn f(x)}$, then by rotating $x$ such that $f(x)$ is real, > $ > \begin{align*} > \abs{f(x)} &= \abs{\alpha} \cdot \abs{f(x)} \\ > &= \abs{\alpha f(x)} = \abs{\re{f(x)}} \\ > &= |f(\alpha x)| \\ > &= |u(\alpha x)| \\ > &\le \norm{u} \norm{\alpha x} = \norm{u}\norm{x} > \end{align*} > $ > therefore $\norm{f} = \norm{u}$.