Inner Bounded X-Derivations, B^1(A, X)

# Inner Bounded X-Derivations, B^1(A, X)

Recall from the Bounded X-Derivations, Z^1(A, X) page that if $A$ is a Banach algebra with unit and $X$ is a Banach $A$-bimodule then a bounded linear operator $D : A \to X$ is said to be a bounded $X$-derivation if for all $a, b \in A$ we have that:

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

We said that the set of all bounded $X$-derivations is denoted $Z^1(A, X)$ and proved that $Z^1(A, X)$ is a linear subspace of $\mathrm{BL}(A, X)$.

We will now look at a subset $B^1(A, X)$ of $Z^1(A, X)$ called the set of inner bounded $X$-derivations, which we define below.

Definition: Let $A$ be a Banach algebra with unit and let $X$ be a Banach $A$-bimodule. Let $x \in X$. The Inner Bounded $X$-Derivation (Associated with $x$) is the bounded $X$-derivation $\delta_x : A \to X$ defined for all $a \in A$ by $\delta_x(a) = ax -xa$. The Set of All Inner Bounded $X$-Derivations is denoted by $B^1(A, X)$. |

We first need to verify that for each $x \in X$, $\delta_x$ is indeed a bounded $X$-derivation.

Proposition 1: Let $A$ be a Banach algebra with unit and let $X$ be a Banach $A$-bimodule. Then for every $x \in X$, $\delta_x$ is a bounded $X$-derivation. |

**Proof:**Let $x \in X$. There are a few things to check.

**Showing that $\delta_x : A \to X$ is a bounded linear operator:**Since $X$ is a normed $A$-bimodule, there exists a $K > 0$ such that $\| ax \| \leq K \| a \| \| x \|$ and $\| xa \| \leq K \| a \| \| x \|$ for all $a \in A$ and all $x \in X$.

- Then for all $a \in A$ we have that:

\begin{align} \quad \| \delta_x(a) \| = \| ax - xa \| \leq \| ax \| + \| xa \| \leq K \| a \| \| x \| + K \| a \| \| x \| = [2K \| x \|] \| a \| \end{align}

- So $\delta_x : A \to X$ is a bounded linear operator.

**2. Showing that $\delta_x$ is a bounded $X$-derivation:**From above we have proven that $\delta_x$ is a bounded operator. Now let $a, b \in A$. Then:

\begin{align} \quad \delta_x(ab) = (ab)x - x(ab) = abx - xab = (abx - axb) + (axb - xab) = a[bx - xb] + [ax - xa]b = \delta_x(a)b + a\delta_x(b) \end{align}

- Thus $\delta_x$ is a bounded $X$-derivation. $\blacksquare$

Proposition 2: Let $A$ be a Banach algebra with unit and let $X$ be a Banach $A$-bimodule. Then $B^1(A, X)$ is a subspace of $Z^1(A, X)$. |

**Proof:**There are three things to show.

**1. Showing that $B^1(A, X)$ is closed under addition:**Let $x, y \in X$, and let $\delta_x$ and $\delta_y$ be the corresponding inner $X$-derivations. Then for all $a, b \in A$ we have that:

\begin{align} \quad [\delta_{x} + \delta_{y}](ab) = \delta_x(ab) + \delta_y(ab) = [(ab)x - x(ab)] + [(ab)y - y(ab)] = (ab)[x + y] - [x + y](ab) = \delta_{x + y}(ab) \end{align}

- Thus $\delta_{x} + \delta_{y} = \delta_{x + y} \in B^1(A, X)$, showing that $B^1(A, X)$ is closed under addition.

**2. Showing that $B^1(A, X)$ is closed under scalar multiplication:**Let $x \in X$, and let $\delta_x$ be the corresponding inner $X$-derivation. Then for all $a, b \in A$ we have that:

\begin{align} \quad \alpha \delta_x(ab) = \alpha[(ab)x - x(ab)] = (ab)[\alpha x] - [\alpha x](ab) = \delta_{\alpha x}(ab) \end{align}

- Thus $\alpha \delta_x = \delta_{\alpha x} \in B^1(A, X)$, showing that $B^1(A, X)$ is closed under scalar multiplication.

**3. Showing that $0 \in B^1(A, X)$:**Let $0_X$ be the additive identity in $X$. Then for all $a, b \in A$ we have that:

\begin{align} \quad \delta_{0_X}(ab) = (ab)0_X - 0_X(ab) = 0 \end{align}

- So $0 \in B^1(A, X)$. Thus we conclude that $B^1(A, X)$ is a subspace of $Z^1(A, X)$. $\blacksquare$