Bounded X-Derivations, Z1(A, X)

# Bounded X-Derivations, Z^{1}(A, X)

Definition: Let $\mathfrak{A}$ be a Banach algebra, let $X$ a Banach $\mathfrak{A}$-bimodule. A Bounded $X$-Derivation is a bounded linear map $D : \mathfrak{A} \to X$ such that $D(ab) = D(a)b + aD(b)$ for all $a, b \in \mathfrak{A}$. The Set of All Bounded $X$-Derivations is denoted $Z^1(\mathfrak{A}, X)$. |

The following proposition tells us that $Z^1(\mathfrak{A}, X)$ is a subspace of $\mathrm{BL}(\mathfrak{A}, X)$.

Proposition 1: Let $\mathfrak{A}$ be a Banach algebra and let $X$ be a Banach $\mathfrak{A}$-bimodule. Then the set of all bounded $X$-derivations $Z^1(\mathfrak{A}, X)$ is a subspace of $\mathrm{BL}(\mathfrak{A}, X)$. |

**Proof:**Certainly $Z^1(\mathfrak{A}, X) \subseteq \mathrm{BL}(\mathfrak{A}, X)$ by definition. So all we need to show is that $Z^1(\mathfrak{A}, X)$ is closed under addition and closed under scalar multiplication.

**1. Showing that $Z^1(\mathfrak{A}, X)$ is closed under addition:**Let $D, E \in Z^1(\mathfrak{A}, X)$. Then $D(ab) = D(a)b + aD(b)$ for all $a, b \in \mathfrak{A}$ and $E(ab) = E(a)b + aE(b)$ for all $a, b \in \mathfrak{A}$. So, for all $a, b \in \mathfrak{A}$ we have that:

\begin{align} \quad [D + E](ab) =D(ab) + E(ab) = D(a)b + aD(b) + E(a)b + aE(b) = [D(a) + E(a)]b + a[D(b) + E(b)] = [D + E](a)b + a[D + E](b) \end{align}

- Thus $D + E \in Z^1(\mathfrak{A}, X)$.

**2. Showing that $Z^1(\mathfrak{A}, X)$ is closed under scalar multiplication:**Let $D \in Z^1(\mathfrak{A}, X)$ and let $\alpha \in \mathbf{F}$. Then $D(ab) = D(a)b + aD(b)$ for all $a, b \in \mathfrak{A}$. So, for all $a, b \in \mathfrak{A}$ we have that:

\begin{align} \quad [\alpha D](ab) = \alpha D(ab) = \alpha[D(a)b + aD(b)] = [\alpha D](a)b + a[\alpha D](b) \end{align}

- Thus $\alpha D \in Z^1(\mathfrak{A}, X)$.

- We conclude that $Z^1(\mathfrak{A}, X)$ is a subspace of $\mathrm{BL}(\mathfrak{A}, X)$. $\blacksquare$