Micromechanics of composites/Proof 12

Question
Let $$\boldsymbol{A}$$ and $$\boldsymbol{B}$$ be two second-order tensor fields. Let the average of any second-order tensor field ($$\boldsymbol{S}$$) over the region $$\Omega$$ (of volume $$V$$) be defined as

\langle \boldsymbol{S} \rangle := \cfrac{1}{V}\int_{\Omega} \boldsymbol{S}~\text{dV} ~. $$ Show that

\langle \boldsymbol{A}\cdot\boldsymbol{B} \rangle - \langle \boldsymbol{A} \rangle\cdot\langle \boldsymbol{B} \rangle = \langle [\boldsymbol{A} - \langle\boldsymbol{A} \rangle]\cdot[\boldsymbol{B} - \langle \boldsymbol{B} \rangle] \rangle~. $$

Proof
Expanding out the right hand side, we have

\begin{align} \langle [\boldsymbol{A} - \langle\boldsymbol{A}\rangle]\cdot[\boldsymbol{B} - \langle \boldsymbol{B} \rangle] \rangle& = \cfrac{1}{V}\int_{\Omega}[\boldsymbol{A} - \langle \boldsymbol{A} \rangle]\cdot[\boldsymbol{B} - \langle \boldsymbol{B} \rangle]~\text{dV} \\ & = \cfrac{1}{V}\int_{\Omega}[ \boldsymbol{A}\cdot\boldsymbol{B} - \langle \boldsymbol{A} \rangle\cdot\boldsymbol{B} - \boldsymbol{A}\cdot\langle \boldsymbol{B} \rangle + \langle \boldsymbol{A} \rangle\cdot\langle \boldsymbol{B} \rangle]~\text{dV} \\ & = \cfrac{1}{V}\int_{\Omega}\boldsymbol{A}\cdot\boldsymbol{B}~\text{dV} - \cfrac{1}{V}\int_{\Omega}\langle \boldsymbol{A} \rangle\cdot\boldsymbol{B}~\text{dV} - \cfrac{1}{V}\int_{\Omega}\boldsymbol{A}\cdot\langle \boldsymbol{B} \rangle~\text{dV} + \cfrac{1}{V}\int_{\Omega}\langle \boldsymbol{A} \rangle\cdot\langle \boldsymbol{B} \rangle~\text{dV} ~. \end{align} $$ Now $$\langle \boldsymbol{A} \rangle$$ and $$\langle \boldsymbol{B} \rangle$$ are constants with respect to the integration. Hence,

\begin{align} \langle [\boldsymbol{A} - \langle\boldsymbol{A}\rangle]\cdot[\boldsymbol{B} - \langle \boldsymbol{B} \rangle] \rangle & = \cfrac{1}{V}\int_{\Omega}\boldsymbol{A}\cdot\boldsymbol{B}~\text{dV} - \langle \boldsymbol{A} \rangle\cdot\left(\cfrac{1}{V}\int_{\Omega}\boldsymbol{B}~\text{dV}\right) - \left(\cfrac{1}{V}\int_{\Omega}\boldsymbol{A}~\text{dV}\right)\cdot\langle \boldsymbol{B} \rangle + \langle \boldsymbol{A} \rangle\cdot\langle \boldsymbol{B} \rangle\left(\cfrac{1}{V}\int_{\Omega}~\text{dV}\right)\\ & = \langle \boldsymbol{A}\cdot\boldsymbol{B} \rangle - \langle \boldsymbol{A} \rangle\cdot\langle \boldsymbol{B} \rangle - \langle \boldsymbol{A} \rangle\cdot\langle \boldsymbol{B} \rangle + \langle \boldsymbol{A} \rangle\cdot\langle \boldsymbol{B} \rangle \\ & = \langle \boldsymbol{A}\cdot\boldsymbol{B} \rangle - \langle \boldsymbol{A} \rangle\cdot\langle \boldsymbol{B} \rangle ~. \end{align} $$ Therefore,

{ \langle \boldsymbol{A}\cdot\boldsymbol{B} \rangle - \langle \boldsymbol{A} \rangle\cdot\langle \boldsymbol{B} \rangle = \langle [\boldsymbol{A} - \langle\boldsymbol{A}\rangle]\cdot[\boldsymbol{B} - \langle \boldsymbol{B} \rangle]\rangle \qquad \square } $$