I'm trying to prove the following proposition.
If $x[n] = 0$ for $n < N_1$ and $h[n] = 0$ for $n < N_2$, then $x[n] * h[n] = 0$ for $n < N_1 + N_2$.
This is from “Linear Time-Invariant Systems.” Signals & Systems (International Edition), by A. V. Oppenheim et al., 2nd ed., Prentice-Hall, 1997, p. 140.
The following is my proof. Are there errors or confusing parts? Improvement? Thank you.
Proof
It is to be proven that, the conditions
- $x[n] = 0$ for $n < N_1$;
- $h[n] = 0$ for $n < N_2$;
- $n < N_1 + N_2$
imply that $x[n] * h[n] = 0$.
By the definition of a convolution,
$$x[n] * h[n] = \sum_{k=-\infty}^{\infty} x[k]h[n-k]$$.
The right-hand side will be 0 if, for all $k$, either $x[k]=0$ or $h[n-k]=0$, or, in other words, either $k < N_1$ or $n - k < N_2$.
This is to say that, if $k \ge N_1$, then $n - k < N_2$.
I will prove this directly.
Assume that $k \ge N_1$, in addition to the premise that $n < N_1 + N_2$. Then, since $N_1 + N_2 \le k + N_2$, $n < k + N_2$. In other words, $n - k < N_2$. In summary, if $k \ge N_1$, then $n-k < N_2$.
If $k < N_1$, then $x[k]=0$.
If $k \ge N_1$, then because $n - k < N_2$, $h[n-k] = 0$.
Thus, for all $k$, either $x[k] = 0$ or $h[n - k] = 0$.
Therefore, each term in the convolution is 0 for all $n < N_1 + N_2$, which makes the sum 0, also.