The Dual Space of a Normed Linear Space

# The Dual Space of a Normed Linear Space

 Definition: Let $(X, \| \cdot \|)$ be a normed linear space. The Dual Space of $X$ denoted $X^*$ is the set of linear functionals on $X$ with the operator norm $\| \cdot \|_{\mathrm{op}} : X^* \to [0, \infty)$ defined for all $T \in X^*$ by $\| T \|_{\mathrm{op}} = \inf \{ M : |T(x)| \leq M \| x \|, \: \forall x \in X \}$.

Below is a list of a few dual spaces. We will prove some of these laters.

Space Dual Space
$\mathbb{R}^n$ $\mathbb{R}^n$
$c_0$ $\ell^1$
$\ell^1$ $\ell^{\infty}$
$\ell^p$, $1 < p < \infty$ $\ell^q$
$L^1(E)$ $L^{\infty}(E)$
$L^p(E)$, $1 < p < \infty$ $L^q(E)$

The following proposition tells us that $X^*$ with the operator norm is indeed a normed linear space.

 Proposition 1: Let $(X, \| \cdot \|)$ be a normed linear space and let $X^*$ denote the set of all bounded linear functionals on $X$. Let $\| \cdot \|_{\mathrm{op}} : X^* \to [0, \infty)$ be the operator norm, that is, for all $T \in X^*$ we have that $\| T \|_{\mathrm{op}} = \inf \{ M : T(x) \leq M \| x \|, \: \forall x \in X \}$. Then $(X^*, \| \cdot \|_{\mathrm{op}})$ is a normed linear space.
• Proof: The set of all functions defined on $X$ is a linear space and since $X^*$ is a subset of this linear space, it suffices to show that $X^*$ is closed under addition, closed under scalar multiplication, and contains the $0$ function to show that it is a linear space.
• Let $S, T \in X^*$. Then for all $\alpha, \beta \in \mathbb{R}$ and all $x, y \in X$ we have that:
(1)
\begin{align} \quad (S + T)(\alpha x + \beta y) = S(\alpha x + \beta y) + T(\alpha x + \beta y) = \alpha S(x) + \beta S(y) + \alpha T(x) + \beta T(y) = \alpha (S + T)(x) + \beta (S + T)(y) \end{align}
• So $(S + T) \in X^*$.
• Let $t \in \mathbb{R}$ and let $T \in X^*$. Then for all $\alpha, \beta \in \mathbb{R}$ and all $x, y \in X$ we have that:
(2)
\begin{align} \quad (t T)(\alpha x + \beta y) = tT(\alpha x + \beta y) = t\alpha T(x) + t \beta T(y) = \alpha (tT)(x) + \beta (tT)(y) \end{align}
• So $tT \in X^*$.
• Lastly, $0 \in X^*$ trivially. So $X^*$ is a linear space.
• All that remains to show is that $\| \cdot \|_{\mathrm{op}}$ is a norm on $X^*$.
• Showing that $\| T \|_{\mathrm{op}} = 0$ if and only if $T = 0$: Suppose that $\| T \|_{\mathrm{op}} = 0$. Then $\inf \{ M : |T(x)| \leq M\| x \|, \: \forall x \in X \} = 0$. So $|T(x)| \leq 0$ for all $x \in X$ implying that $T = 0$. Conversely, suppose that $T = 0$. Then $\inf \{ M : |0| \leq M \| x \|, \: \forall x \in X \} = 0$. Hence $\| T \|_{\mathrm{op}} = \| 0 \|_{\mathrm{op}} = 0$.
• Showing that $\| \alpha T \|_{\mathrm{op}} = |\alpha| \| T \|_{\mathrm{op}}$: Let $\alpha \in \mathbb{R}$ and let $T \in X^*$. Then:
(3)
\begin{align} \quad \| \alpha T \|_{\mathrm{op}} = \inf \{ M : |\alpha T(x)| \leq M \| x \|, \: \forall x \in X \} = |\alpha| \inf \{ M^* : |T(x)| \leq M^* \| x \|, \: \forall x \in X \} = |\alpha| \| T \|_{\mathrm{op}} \end{align}
• Showing that $\| S + T \|_{\mathrm{op}} \leq \| S \|_{\mathrm{op}} + \| T \|_{\mathrm{op}}$: Let $S, T \in X^*$. Then:
(4)
\begin{align} \quad \| S + T \|_{\mathrm{op}} = \inf \{ M : |S(x) + T(x)| \leq M\| x \|, \: \forall x \in X \} &\leq \inf \{ M : |S(x)| + |T(x)| \leq M\| x \|, \: \forall x \in X \} \\ &\leq \inf \{ M : |S(x)| \leq M_1\| x \|, \: \forall x \in X \} + \inf \{ M_2 : |T(x)| \leq M_2\| x \|, \: \forall x \in X \} \\ & \leq \| S \|_{\mathrm{op}} + \| T \|_{\mathrm{op}} \end{align}
• Hence $(X^*, \| \cdot \|_{\mathrm{op}})$ is a normed linear space. $\blacksquare$