In the time domain convolution can be well-defined even if one of the signals $x \notin l^1 \cup l^2$ (not absolutely summable or square summable) and $x$ has no DTFT. This is a category where we cannot talk about a frequency response or fourier transform. Consider where this type of a signal is passed through a system with a finite impulse response:

$$
x(n) = \alpha^n u(n) \to \fbox{FIR} \to y(n)
$$

We know that $y(n)$ is the convolution of the two signals.

\begin{align*}
y(n) &= \sum\limits_{k\in\Z}x(k)h(n-k) \\
y(n) &= \sum\limits_{k\in\Z}\alpha^{k}u(k)h(n-k) \\
\end{align*}

This system has well defined response for this growing exponential. Why is the output bounded even though the input to this system is unbounded and has no transform? We can't talk about frequency response or about multiplying $X(\omega)$ with the filter's frequency response because $X(\omega)$ does not exist. But this system has a well-defined response! Suppose for now that the domain of $h$ is in $\{-1,0,1\}$ (it is a FIR).

\begin{align*}
y(n) &= \sum\limits_{k\in\Z}h(k)x(n-k) \\
y(n) &= h(-1)x(n+1) + h(0)x(n) + h(1)x(n-1) \\
\end{align*}

We have a finite number of terms, so we are only summing a finite number of the values of the input. Another example is when you have two right-sided functions.

\begin{definition}
A function is \emph{right-sided} $\iff x(n) = 0 \mbox{ } \forall n \lt N \lt \infty$
\end{definition}


% \img{images/rightleftsided/def/r.ps}

\psset{unit=1cm}
\begin{center}
\begin{pspicture}(-4,-1)(4,3)
    
\rput(3.3,0.5){$\cdots$}

\psline{-*}(-3,0)(-3,0)
\psline{-*}(-2,0)(-2,0)
\psline{-*}(-1,0)(-1,0)
\psline{-*}(0,0)(0,1)
\rput(0,-0.3){N}
\psline{-*}(1,0)(1,0.1)
\psline{-*}(2,0)(2,0.4)
\psline{-*}(3,0)(3,0.9)

% axes
\psline{->}(-4,0)(4,0)
\rput(4,-0.3){ $n$ }
\rput(4,1){ $x(n)$ }

\end{pspicture}
\end{center}


Note that $x$ is causal if $N=0$. This implies that causal signals are a subset of right-sided signals.

\begin{definition}
A function is \emph{left-sided} $\iff x(n) = 0 \mbox{ } \forall M \lt n \lt \infty$
\end{definition}

% \img{images/rightleftsided/def/l.ps}

\begin{center}
\begin{pspicture}(-4,-1)(4,3)
    
\psline{-*}(-3,0)(-3,0.7)
\psline{-*}(-2,0)(-2,0.4)
\psline{-*}(-1,0)(-1,1)
\psline{-*}(0,0)(0,0.4)
\rput(0,-0.3){M}
\psline{-*}(1,0)(1,0)
\psline{-*}(2,0)(2,0)
\psline{-*}(3,0)(3,0)
% axes
\psline{->}(-4,0)(4,0)
\rput(4,-0.3){ $n$ }
\rput(4,1){ $x(n)$ }
\rput(-3.3,0.5){ $\cdots$ }

\end{pspicture}
\end{center}

Note that $x$ is anti-causal if $M=0$. This implies that anti-causal signals are a subset of left-sided signals. Lets continue with some examples.

\begin{example}

Lets look at a situation where convolution is well-defined but at least one of the functions doesn't have a DTFT.

$$
x(n) = \alpha^n u(n) \to \bbox[5px,border:1px solid black]{\beta^n u(n)} \to y(n)
$$

Note that convolution is defined if $\abs{\alpha} \gt 1$ or $\abs{\beta} \gt 1$ or both.

\begin{align*}
y(n) &= \sum\limits_{k\in\Z} \alpha^k u(k) \beta^{n-k} u(n-k) \\
y(n) &= \sum\limits_{k = 0}^{\infty} \alpha^k \beta^{n-k} u(n-k) \\
y(n) &= \begin{cases} \sum\limits_{k=0}^{n}\alpha^k\beta^{n-k} & n\geq0 \\ 0 & \mbox{otherwise} \\ \end{cases}
\end{align*}

We can also see that the ``overlap'' of the signals is finite because we are convolving two right-sided signals. When you have $x(n)$ and $h(n-m)$, there must be some $N$ that zeros out all values $n \lt N$ due to $x$, and zeros for all $n \gt n-M$ due to $h$. There are two finite endpoints, and they have a finite overlap.

% \img{images/rightleftsided/finite/comb.ps}

\begin{center}
\begin{pspicture}(-3,-4)(6,3)
    
\psframe[linestyle=dashed](0.5,-3.5)(3.5,1.5)

% axes
\psline{->}(-2,-3)(6,-3)
\rput(6,-3.3){ $k$ }
\rput(-1.75,-1.5){ $\cdots$ }
\rput(6,-1.5){ $h(n-k)$ }
% axes
\psline{->}(-2,0)(6,0)
\rput(6,-0.3){ $k$ }
\rput(6,2.5){ $x(k)$ }
\rput(5.75,1.5){ $\cdots$ }

\psline{-*}(-1,-3)(-1,-0.5)
\psline{-*}(0,-3)(0,-1.4)
\psline{-*}(1,-3)(1,-2.1)
\psline{-*}(2,-3)(2,-2.6)
\psline{-*}(3,-3)(3,-2.9)
\rput(3,-3.2){n-M}

\psline{-*}(1,0)(1,0.1)
\rput(1,-0.3){N}
\psline{-*}(2,0)(2,0.4)
\psline{-*}(3,0)(3,0.9)
\psline{-*}(4,0)(4,1.6)
\psline{-*}(5,0)(5,2.5)

\end{pspicture}
\end{center}

\end{example}

Right and Left-sided Functions