Linear Functionals and Bounded Linear Functionals

Linear Functionals and Bounded Linear Functionals

Definition: Let $X$ be a linear space. A Linear Functional on $X$ is a map $T : X \to \mathbb{R}$ which satisfies the following properties:
a) $T(x + y) = T(x) + T(y)$ for all $x, y \in X$.
b) $T(\alpha x) = \alpha T(x)$ for all $\alpha \in \mathbb{R}$ and for all $x \in X$.
Definition: Let $(X, \| \cdot \|)$ be a normed linear space. A Bounded Linear Functional on $X$ is a linear functional $T$ such that there exists an $M > 0$ where $|T(x)| \leq M \| x \|$ for every $x \in X$.

The notion of a linear functional being bounded is not a new concept. The following proposition tells us that a linear functional is bounded if and only if it is continuous, and moreover, is continuous at $0$.

Proposition 1: Let $(X, \| \cdot \|)$ be a normed linear space. Then the following are equivalent:
a) $T$ is a bounded linear operator.
b) $T$ is continuous on $X$.
c) $T$ is continuous at the origin, $0$.
  • Proof:
  • $(a) \Rightarrow (b)$ Suppose that $T$ is a bounded linear operator. Let $\epsilon > 0$ be given and let $y \in X$.. Let $\delta = \frac{\epsilon}{M} > 0$. Then if $\| x - y \| < \delta$ and by the boundedness of $T$ we have that:
(1)
\begin{align} \quad T(x - y) \leq M \| x - y \| < M \delta = M \cdot \frac{\epsilon}{M} = \epsilon \end{align}
  • So $T$ is continuous at $y$ and since $y$ is an arbitrary element of $X$ we have that $T$ is continuous on $X$.
  • $(b) \Rightarrow (c)$ Suppose that $T$ is continuous on $X$. Then it is clearly continuous at $0$ by definition.
  • $(c) \Rightarrow (a)$ Suppose that $T$ is continuous at $0$. Then for $\epsilon = 1 > 0$ there exists a $\delta > 0$ such that if $\| x \| \leq \delta$ then:
(2)
\begin{align} |T(x)| < \epsilon = 1 \end{align}
  • Take $x \in X$. Then $\frac{\delta x}{\| x \|}$ is such that:
(3)
\begin{align} \biggr \| \frac{\delta x}{\| x \|} \biggr \| = \delta \end{align}
  • Therefore:
(4)
\begin{align} \quad \biggr | T \left ( \frac{\delta x}{\| x \|} \right ) \biggr | < 1 \end{align}
  • Hence, for every $x \in X$ we have that:
(5)
\begin{align} \quad |T(x)| < \frac{1}{\delta} \| x \| \end{align}
  • So $T$ is a bounded linear operator. $\blacksquare$
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