Linear Operators and Bounded Linear Operators

# Linear Operators and Bounded Linear Operators

Recall from the Linear Functionals and Bounded Linear Functionals page that if $X$ is a linear space then a linear functional on $X$ is a function $T : X \to \mathbb{R}$ such that for all $x, y \in X$ and for all $\alpha \in \mathbb{R}$ we have that:

• $T(x + y) = T(x) + T(y)$.
• $T(\alpha x) = \alpha T(x)$.

Furthermore, if $(X, \| \cdot \|)$ is a normed linear space then a linear functional $T$ on $X$ was said to be bounded if there exists an $M > 0$ such that $|T(x)| \leq M \| x \|$ for all $x \in X$.

The concept of a linear functional and bounded linear functional can be extended if the codomain is changed from $\mathbb{R}$ to another linear space, $Y$. Such functions are called linear operators from $X$ to $Y$.

 Definition: Let $X$ and $Y$ be linear spaces. A Linear Operator from $X$ to $Y$ is a function $T : X \to Y$ such that: a) $T(x_1 + x_2) =T(x_1) + T(x_2)$ for all $x_1, x_2 \in X$. b) $T(\alpha x) = \alpha T(x)$ for all $x \in X$. The Set of All Linear Operators from $X$ to $Y$ is denoted by $\mathcal L (X, Y)$.
 Definition: Let $(X, \| \cdot \|_X)$ and $(Y, \| \cdot \|_Y)$ be normed linear spaces and let $T$ be a linear operator from $X$ to $Y$. $T$ is said to be Bounded if there exists an $M > 0$ such that $\| T(x) \|_Y \leq M \| x \|_X$ for every $x \in X$. The Set of All Bounded Linear Operators from $X$ to $Y$ is denoted by $\mathcal B(X, Y)$.

Sometimes the subscripts on the norms for $X$ and $Y$ will be omitted if no ambiguity arises in which norm we are talking of.

Note: If $Y = \mathbb{R}$ with the usual Euclidean norm then $\mathcal B(X, Y)$ is the set of bounded linear functionals from $X$ to $Y$, that is, it is the dual space, $X^*$.

Recall that a linear functional $T$ on $X$ is bounded if and only if $T$ is continuous on $X$ and if and only if $T$ is continuous at $0$. The same result is true for linear operators from $X$ to $Y$.

 Proposition 1: Let $(X, \| \cdot \|_X)$ and $(Y, \| \cdot \|_Y)$ be normed linear spaces and let $T$ be a linear operator from $X$ to $Y$. Then the following statements are equivalent: a) $T$ is a bounded linear operator. b) $T$ is continuous on $X$. c) $T$ is continuous at $0 \in X$.
• Proof:
• $(a) \Rightarrow (b)$ Suppose that $T$ is a bounded linear operator. Then there exists an $M > 0$ such that $\| T(x) \|_Y \leq M \| x \|_X$ for all $x \in X$. Let $x_0 \in X$. Let $\epsilon > 0$ be given and choose $\delta = \frac{\epsilon}{M} > 0$. Then if $\| x - x_0 \|_X < \delta$ we have that:
(1)
\begin{align} \quad \| T(x) - T(x_0) \|_Y = \| T(x - x_0) \|_Y \leq M \| x - x_0 \|_X < M \cdot \delta = M \cdot \frac{\epsilon}{M} = \epsilon \end{align}
• Therefore $T$ is continuous at $x_0$. Since $x_0$ is arbitrary, $T$ is continuous on $X$.
• $(b) \Rightarrow (c)$ Suppose that $T$ is continuous on $X$. Then $T$ is trivially continuous at $0 \in X$.
• $(c) \Rightarrow (a)$ Suppose that $T$ is continuous at $0 \in X$. Then for $\epsilon = 1 > 0$ there exists a $\delta > 0$ such that if $\| x \|_X \leq \delta$ then:
(2)
\begin{align} \quad \| T(x) - T(0) \|_Y = \| T(x) \|_Y < \epsilon = 1 \end{align}
• If $x \neq 0$ then $\frac{\delta x}{\| x \|_X}$ has norm $\delta$. Therefore:
(3)
\begin{align} \quad \left \| T \left ( \frac{\delta x}{\| x \|_X} \right ) \right \|_Y < 1 \end{align}
• Which implies that for every $x \in X$ we have that:
(4)
\begin{align} \quad \| T(x) \|_Y < \frac{1}{\delta} \| x \|_X \end{align}
• So $T: X \to Y$ is a bounded linear operator. $\blacksquare$