Answer
False.
Work Step by Step
False, $f(x)=x$ on [0,1] has $g(x)=-x$ and $f_{\text {ave}}=0.5$ on [0,1], with $g_{\text {ave}}$ $-0.5,$. However, their product $-x^{2}=(g \cdot f)(x)$ has $(g \cdot f)_{\text {ave }}=\int_{0}^{1}\left(-x^{2}\right) d x=$ $-\frac{1}{3} \neq g_{a v c} \cdot f_{a v c}$
