The Operator Algebra of Bounded Linear Operators on X

# The Operator Algebra of Bounded Linear Operators on X

 Definition: Let $X$ be a Banach space and let $\mathcal B(X) = \mathcal B(X, X)$ denote the space of all bounded linear operators on $X$. The Operator Algebra of Bounded Linear Operators on $X$ is the space $\mathcal B(X)$ with the operations of operator addition, scalar multiplication, and composition as operator multiplication, specified for all $S, T \in \mathcal B(X)$ and for all $\alpha \in \mathbb{C}$ by: 1) $(S + T)(x) := S(x) + T(x)$ for all $x \in X$. 2) $(\alpha S)(x) := \alpha S(x)$ for all $x \in X$. 3) $(S \circ T)(x) :=S(T(x))$ for all $x \in X$. and with norm on $B(X)$ as the usual operator norm, $\displaystyle{\| S \| := \sup_{\| x \| = 1} \| S(x) \|}$.

Note that the operator norm on $\mathcal B(X)$ is indeed an algebra norm, since for all $S, T \in \mathcal B(X)$ we have that:

(1)
\begin{align} \quad \| S \circ T \| = \sup_{\| x \| = 1} \| S(T(x)) \| \leq \sup_{\| x \| = 1} \| S \| \| T(x) \| = \| S \| \sup_{\| x \| = 1} \| T(x) \| = \| S \| \| T \| \end{align}

Moreover, recall from the Criterion for B(X, Y) to be a Banach Space page, that if $Y$ is a Banach space then $\mathcal B(X, Y)$ is also a Banach space. In particular, if $X$ is Banach space, then $\mathcal B(X)$ is a Banach space and thus $\mathcal B(X)$ is a Banach algebra.

Note that in general, composition of two functions (when defined) is not commutative. The same applies to compositions of two bounded linear operators, so $\mathcal B(X)$ is a noncommutative Banach algebra.