Dual Spaces

Why

Take a vector space, and consider the set of continuous linear functionals on that space. Given a suitable norm, this space is a complete normed space.

Defining result

Let $(V, \norm{\cdot })$ be a normed space. The set $\dual{V}$ of all continuous linear functionals on $V$ is a complete normed space with respect to pointwise algebraic operations and norm $\dnorm{\cdot }: V \to \R $ defined by

\[ \dnorm{F} = \underset{x \in V, \;\norm{x} \leq 1}{\sup} \abs{F(x)}. \]

We argue (1) $\dual{V}$ is a vector space, (2) $\dnorm{\cdot }$ is a norm, and (3) $(V, \dnorm{\cdot })$ is complete.¹

We call $(\dual{V}, \dnorm{\cdot })$ the dual space (or conjugate space, conjugate dual, or Banach dual of $V$). Notice that $(\dual{V}, \dnorm{\cdot })$ is complete regardless of whether the original normed space $(V,\norm{\cdot })$ is complete.

Basic dual norm property

Notice that the dual norm satisfies a familiar property.

For any vector $x$ in a normed space $(V, \norm{\cdot })$ and any continuous linear functional $F$ on $E$,

\[ \abs{F(x)} \leq \dnorm{F}\norm{x}. \]

If $x = 0$, then $\norm{x} = 0$ and $F(x) = 0$ ($F$ is linear). Otherwise, $x/\norm{x}$ is a unit vector and so

\[ \dnorm{F} \geq \abs{F(x/\norm{x})} = \frac{\abs{F(x)}}{\norm{x}}. \]

where the inequality is from the definition of $\dnorm{\cdot }$ (as a supremum) and the equality follows from the linearity of $F$.

Future editions will include an account. ↩︎