Convolution Integral - Shift property

I am trying to prove $f_1(t-T_1)*f_2(t-T_2)=c(t-T_1-T_2)$ given that $f_1(t)*f_2(t)=c(t)$. Here $f_1(t)*f_2(t)=\int_{-\infty}^{\infty}f_1(x)f_2(t-x)dx$.

I can only deduce that $f_1(t-T_1)*f_2(t)=c(t-T_1)$ and $f_1(t)*f_2(t-T_2)=c(t-T_2)$ by directly evaluating the integral and using the commutative property. According to my textbook, the conclusion should follows directly from these two relationships but I am not able to prove it.

If the 'left' function is delayed by $T_1$ seconds, so is $c(t)$. The same thing happens when the 'right' function is delayed by $T_2$ seconds. So it is quite 'logical' for $c(t)$ to be delayed by the sum of the delayed periods of time in $f_1$ and $f_2$. But this verbal explanation does not satisfy me.

I am also having trouble to express $f_1(t-T_1)*f_2(t-T_2)$ in integral form. I cannot only express it because the definition of the convolution integral seems limited. Thanks in advance.

Define a new function via $g(t) = f_2(t - T_2)$. From what you have proven, the following holds:
$f_1(t - T_1) * f_2(t - T_2) = f_1(t - T_1) * g(t) = c_2(t - T_1)$, where $c_2(t) = f_1(t) * g(t) = f_1(t) * f_2(t - T_2) = c(t - T_2)$.
Combining both equations now yields the desired result $$f_1(t - T_1) * f_2(t - T_2) = c_2(t - T_1) = c((t - T_1) - T_2) = c(t - T_1 - T_2).$$
• brilliant! May I ask how do you come up with the idea of defining $c_2$? It might be a trivial question for you but not for me. – Dave Clifford Nov 10 '16 at 13:08
• I just wanted to apply the first proven equation to the functions $f_1(t)$ and $f_2(t - T_2)$, so i have renamed $f_2(t - T_2)$ as $g(t)$. Then $c_2$ just occurs naturally as the result of the equation. – Dominik Nov 10 '16 at 13:11