How to prove that if $F_1(t)$ and $F_2(t)$ are of exponential order when $t \to\infty$, therefore $F_1(t) \cdot F_2(t)$ and $F_1(t) + F_2(t)$ are also of exponential order when $t \to\infty$?
I know that a function F(t) is of exponential order if for constants c > 0 and M >0, there exists T > 0 such that |F(t)| $\leqslant Me^{ct}$ for t > T.
Now, for the first part of the proof I did this:
Let $F_1(t)$ and $F_2(t)$ be of exponential order. For $M_1, M_2, c_1, c_2 >0:$
$|F_1(t)|\leqslant M_1 e^{c_1t}$
$|F_2(t)|\leqslant M_2 e^{c_2t}$
$|F_1(t)| \cdot |F_2(t)| \leqslant M_1 e^{c_1t} \cdot M_2 e^{c_2t}$
$|F_1(t) \cdot F_2(t)| \leqslant M_1 \cdot M_2 \cdot e^{(c_1 + c_2)t}$
Since $M_1, M_2, c_1, c_2 >0$, then $M_1 \cdot M_2 > 0$ and $c_1 + c_2 > 0$ .
This means that if $F_1(t)$ and $F_2(t)$ are of exponential order when $t \to\infty$, therefore $F_1(t) \cdot F_2(t)$ is also of exponential order.
When I try to apply the same procedure to $F_1(t) + F_2(t)$ I end up with this:
$|F_1(t)| + |F_2(t)| \leqslant M_1 \cdot e^{c_1t} + M_2 \cdot e^{c_2t}$
Where do I go from here?
I want to get something of the form $|F_1(t) + F_2(t)| \leqslant Me^{ct}$
Hint 1: Use the triangle inequality to relate $|F_1(t) + F_2(t)|$ to $|F_1(t)| + |F_2(t)|$.
Hint 2: Now you have $M_1 e^{c_1 t} + M_2 e^{c_2 t}$. Since we only care about large $t$, we can simply focus on the one with the bigger exponent. Let's consider the case $c_1 > c_2$. Show that there is some constant $M'_2$ such that $M_2 e^{c_2 t} \le M'_2 e^{c_1 t}$ for all large $t$. Then you have $M_1 e^{c_1 t} + M_2 e^{c_2 t} \le (M_1 + M'_2) e^{c_1 t}$.