does anybody know...
can (f(t) x g(t))*h(t) be expressed as f(t)*h(t) x g(t)*h(t)? where * is convolution?
No.
Counter example let $\displaystyle h(t)$ be a Gaussian kernel with unit spread parameter (normalise it as you wish). Let $\displaystyle f(t) $ and $\displaystyle g(t)$ both be $\displaystyle 1$ on disjoint intervals and zero elsewhere.
Then $\displaystyle f(t) \times g(t) \equiv 0$, and so:
$\displaystyle [(f(.) \times g(.))*h](t) \equiv 0,$
but $\displaystyle (f*h)(t) (g*h)(t) >0$
RonL
no.. in LHS: f and g is inside the integral sign. in RHS, f will be outside of the integral sign.
these are the some properties (as far as i know) that the convolution function have:
symmetric: f*g=g*f
associative (in convolution): (f*g)*h=f*(g*h)
distributive over addition: f*(g+h) = f*g + f*h
associative and symmetric in real multiplication: a(f*g)=(af)*g=f*(ag)
also: (f*g)=(Df)*g = f* (Dg)