Convolution proof If I have two functions in a convolution like
$$X*Y=1$$
$$X*Z=1$$
then it means (trivially) $Y=Z$.
Is this correct or are there subtleties in the convolution theorem where $Y=Z$ isn't always true?
 A: Here is a counter example. The functions $X,Y,Z$ below will satisfy that $X * Y = X * Z = 1$ but $ Y \not= Z$:
$$ X(t) := 1, \forall \,t \in \mathbb{R},$$
$$
Y(t) := \begin{cases} 
\frac{1}{C_Y} e^{-\frac{1}{1-t^2}} \quad \text{if $-1 \leq t\leq 1$}\\
0 \quad \text{otherwise,}
\end{cases}
$$ 
where $C_Y$ is a (normalization) constant such that the integral of $Y$ over $[-1,1]$ equals 1, and
$$
Z(t) := \begin{cases} 
\frac{1}{C_Z} e^{-\frac{1}{2^2-t^2}} \quad \text{if $-2 \leq t\leq 2$}\\
0 \quad \text{otherwise,}
\end{cases}
$$ 
where, similar to $C_Y$, $C_Z$ is a (normalization) constant such that the integral of $Z$ over $[-2,2]$ equals 1.
Those are concrete functions as an explicit example. Any normalized (integral equals 1) function $f$ that has compact support will satisfy that $X * f = 1$. From there you can choose any $Y$ and $Z$ you want as long as they are different.
A: Good Question :
Here we can use the concept of Laplace Transformation 
$$X(s) Y(s) = X(s) Z(s)$$
$$Y(s) = Z(s)$$
We know that "Laplace Transform of two signals can be same but their ROC will be definitely different"
Here is a example :
$$e^{-at} u(t) \rightleftharpoons \frac{1}{s+a}$$
$$-e^{-at} u(-t) \rightleftharpoons \frac{1}{s+a}$$
HENCE now it is proved that it is not necessary that 
$$Y = Z$$
