> [!definitionb] Definition > > Let $\cx$ be a [[Normed Vector Space|normed space]] over $\complex|\real$. Let > $ > \cx^* = L(\cx, \complex|\real) > $ > be the [[Space of Bounded Linear Maps|space of bounded linear maps]] from $\cx$ to the underlying field. $\cx^*$ is the **dual** of $\cx$, with its elements called the [[Bounded Linear Functional|bounded linear functionals]]. > [!theorem] > > Let $\cx$ be a normed space and $f: \cx \to \complex|\real$. $f \in L(\cx, \complex|\real)$ if and only if $\ker(f)$ is closed. > > *Proof*. Assuming that $f \ne 0$. Suppose that $\ker(f)$ is closed, and consider $\cx/\ker(f)$. Define > $ > \ol f: \cx/\ker(f) \to F \quad x + \ker(f) \mapsto f(x) > $ > then $f$ is well-defined as $f(x + y) = f(x)$ for all $y \in \ker(f)$. Moreover, $\ker (\ol{f}) = 0 + \ker(f)$ and $f$ is injective, meaning that $\dim \cx/\ker(f) = 1$. > > Let $x \in \cx$ such that $x + \ker(f) \ne 0$, then $\norm{x + \ker(f)} = \varepsilon > 0$ and $\span(x + \ker(f)) = \cx/\ker(f)$. For any $y + \ker(f) \in \cx/\ker(f)$, write $y + \ker(f) = \lambda x + \ker(f)$, then > $ > \ol f(y + \ker(f)) = \ol f(\lambda(x + \ker(f))) = \lambda \ol f(x + \ker(f)) > $ > and > $ > \begin{align*} > \abs{\ol f(y + \ker(f))} &= \abs{\lambda \cdot \ol f(x + \ker(f))} \\ > &= \abs{\lambda}\cdot \abs{\ol f(x + \ker(f))} > \end{align*} > $ > so $\ol f$ is bounded. Since $\ol{f}$ is continuous in the quotient topology, $f$ is continuous in the original norm topology.