Linear Operators Are Bounded If and Only If They're Continuous

# Linear Operators Are Bounded If and Only If They're Continuous

Let $X$ and $Y$ be normed linear operators. So far, we have defined bounded linear operators from $X$ to $Y$. We are about to see that a linear operator is bounded if and only if it is continuous.

 Theorem 1: Let $X$ and $Y$ be normed linear spaces and let $T : X \to Y$ be a linear operator. Then $T$ is a bounded linear operator if and only if $T$ is continuous on $X$.
• Proof: $\Rightarrow$ Suppose that $T : X \to Y$ is a bounded linear operator. Then there exists an $M > 0$ such that for all $x \in X$ we have that:
(1)
\begin{align} \quad \| T(x) \| \leq M \| x \| \end{align}
• Let $\epsilon > 0$ be given and let $\delta = \frac{\epsilon}{M} > 0$. Then if $\| x - y \| < \delta$ we have that:
(2)
\begin{align} \quad \| T(x) - T(y) \| = \| T(x - y) \| \leq \| T \| \| x - y \| \leq M \| x - y \| < M\delta = M \cdot \frac{\epsilon}{M} = \epsilon \end{align}
• Therefore $T$ is continuous. $\blacksquare$
• $\Leftarrow$ Suppose that $T : X \to Y$ is continuous on $X$. Then in particular, $T$ is continuous at $0 \in X$. So for $\epsilon_0 = 1 > 0$ there exists a $\delta_0 > 0$ such that if $\| x \| = \| x - 0 \| \leq \delta$ then:
(3)
\begin{align} \quad \| T(x) \| = \| T(x) - T(0) \| \leq 1 \end{align}
• Note that if $x \neq 0$ then:
(4)
\begin{align} \quad \biggr \| \frac{\delta x}{\| x \|} \biggr \| = \delta \frac{\| x \|}{\| x \|} = \delta \end{align}
• Therefore:
(5)
\begin{align} \quad \biggr \| T \left ( \frac{\delta x}{\| x \|} \right ) \biggr \| \leq 1 \end{align}
• Hence, for all $x \in X$ with $x \neq 0$ we have that:
(6)
\begin{align} \quad \delta \frac{\| T(x) \|}{\| x \|} \leq 1 \quad \Leftrightarrow \quad \| T(x) \| \leq \frac{1}{\delta} \| x \| \end{align}
• Hence $T$ is a bounded linear operator. $\blacksquare$