The Operator Norm on the Set of Bounded Linear Operators

# The Operator Norm on the Set of Bounded Linear Operators

 Definition: Let $(X, \| \cdot \|_X)$ and $(Y, \| \cdot \|_Y)$ be normed linear spaces and let $T$ be a linear operator from $X$ to $Y$. The Operator Norm of $T$ is defined to be $\| T \|_{\mathrm{op}} = \inf \{ M : \| T(x) \|_Y \leq M \| x \|_X, \: \forall x \in X \}$.

The notion of the operator norm for a bounded linear operator is analogous to the operator norm for a bounded linear functional.

Later on The Space of Bounded Linear Operators page we will see that $\mathcal B(X, Y)$ with the operator norm $\| \cdot \|_{\mathrm{op}}$ is a normed linear space.

 Proposition 1: Let $(X, \| \cdot \|_X)$ and $(Y, \| \cdot \|_Y)$ be normed linear spaces and let $T$ be a linear operator from $X$ to $Y$. If there exists an $M > 0$ such that $\| T(x) \|_Y \leq M$ for all $x \in X$ then $T$ is bounded.
• Proof: Suppose that $\| T(x) \|_Y \leq M$ for all $x \in X$. Then for all $x \in X$ with $x \neq 0$ we have that:
(1)
\begin{align} \quad \left \| T \left ( \frac{x}{\| x \|_X} \right ) \right \|_Y \leq M \end{align}
• That is:
(2)
\begin{align} \quad \| T(x) \|_X \leq M \| x \|_X \end{align}
• So $T$ is bounded. $\blacksquare$

The following proposition gives us various ways to define the operator norm of a bounded linear operator.

 Proposition 2: Let $(X, \| \cdot \|_X)$ and $(Y, \| \cdot \|_Y)$ be normed linear spaces and let $T$ be a bounded linear operator from $X$ to $Y$. Then: a) $\displaystyle{\| T \|_{\mathrm{op}} = \sup_{x \in X, x \neq 0} \left \{ \frac{\| T(x) \|_Y}{\| x \|_X} \right \}}$. b) $\displaystyle{\| T \|_{\mathrm{op}} = \sup_{x \in X, \| x \| = 1} \left \{ \| T(x) \|_Y \right \}}$.