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It is of practical significance in the design of discrete-time systems that they be "well behaved," meaning that for any "well behaved" input, the system gives a "well behaved" output. Colloquially speaking, we do not want an innocuous input to result in the system "blowing up." The technical term for "well behaved" systems is that they are stable . It is essential for many systems that they be stable, for the sake of safety and proper operation within wider systems. For example, with steering or braking or aircraft control systems, it could be catastrophic if a small input led to a wildly divergent output.
Consider the recursive average system $y[n]=x[n]+\alpha y[n-1]$, with an eminently reasonable and contained input of the step function $u[n]$. For values of $\alpha$ less than $1$, the system is "well behaved," but the output "blows up" for $\alpha\gt 1$:
Consider a discrete-time system $H$ and arbitrary input signal $x[n]$ with $|x[n]|\leq M_1\in R~,~\forall n$. Let $y[n]=H\{x[n]\}$. $H$ is BIBO stable if there exists some $M_2\in R$ such that $|y[n]|\leq M_2 \forall n$.
BIBO stability is a guarantee, a stamp of approval on a discrete-time system, certifying that the output will always be capped by some value, so long as the input also is.
Consider a discrete-time LTI system $H$ with impulse response $h[n]$, and arbitrary input signal $x[n]$ with $|x[n]|\leq M_1~,~\forall M$. $H$ is BIBO stable if and only if there exists some $M_2$ such that $\|h[n]\|_1=M_2$. Furthermore, this $M_2$ (if it exists) bounds the output of the system: $|H\{x[n]\}|\leq M_1 M_2$.
$\begin{align*} |y[n]|&=|H\{x[n]\}|\\&=|\sum_{m=-\infty}^\infty h[n-m]x[m]|\\&=|\sum_{m=-\infty}^\infty x[n-m]h[m]|\\&\leq \sum_{m=-\infty}^\infty|x[n-m]||h[m]|\\&\leq \sum_{m=-\infty}^\infty M_1|h[m]|\\&\leq M_1\sum_{m=-\infty}^\infty |h[m]|\\&\leq M_1M_2 \end{align*}$So, if the impulse response $|\h[n]\|_1$ for an LTI system exists, then the system is BIBO stable.
The other side of the proof is to show that if a system is BIBO stable, the norm $\|h[n]\|_1$ of its impulse response must exist. We will demonstrate this by proving the contrapositive: if $\|h[n]\|_1$ is unbounded, the system is not BIBO stable. It will require the use of a the function $\textrm{sgn}\{x[n]\}$, which outputs 1 when $x[n]$ is positive, -1 when it is negative, and 0 when it is 0.
Consider an arbitrary impulse response $h[n]$ that is not absolutely summable, i.e., $\|h[n]\|_1$ is unbounded. For the system to be BIBO stable, any bounded input must result in a bounded output. So let the input $x[n]=\textrm{sgn}\{h[-n]\}$. Clearly, $x[n]$ is bounded: $\|x[n]\|_\infty$. But $\begin{align*} y[0]&=\sum_{m=-\infty}^\infty h[0-m]x[m]\\&=\sum_{m=-\infty}^\infty h[0-m]\textrm{sgn}\{h[m]\}\\&= \sum_{m=-\infty}^\infty|h[m]|\\&=\|h[n]\|_1 \end{align*}$So in this case $y[0]$ is unbounded, so the system is not BIBO stable.
Because of this impulse response property, it is evident that all FIR systems are BIBO stable (for a finite sums of finite values is finite).
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