Bounded X-Derivations, Z^1(A, X)

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

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

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

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

**Proof:**Certainly $Z^1(A, X) \subseteq \mathrm{BL}(A, X)$ by definition. So all we need to show is that $Z^1(A, X)$ is closed under addition, closed under scalar multiplication, and contains the $0$ operator.

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

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

**3. Showing that $0 \in Z^1(A, X)$:**Let $a, b \in A$. Note that the zero operator $0 : A \to X$ is bounded and is such that $0(a) = 0(b) = 0(ab) = 0$ for all $a, b \in A$. Thus:

\begin{align} \quad 0(ab) = 0 = 0(a)b + a0(b) \end{align}

- Thus $0 \in Z^1(A, X)$. We conclude that $Z^1(A, X)$ is a subspace of $\mathrm{BL}(A, X)$. $\blacksquare$