convolution
If I have a convolution
$$z(t) = x(t) * y(t)$$
where I know $x(t)$ and $z(t)$, is there a way to determine $y(t)$? Is there a "reverse" convolution theorem for this? I know there are numerical methods used in data processing, but I'm looking for an analytical method.
The integral
\begin{align*} (Box \star Box)(t) = \int_{-\infty}^{\infty}Box(s)Box(t-s)\;ds \end{align*}
is equal to the integral
\begin{align*} (Box \star Box)(t) = \int_{0}^{1}Box(s)Box(t-s)\;ds \end{align*}
since $Box(s)$ is only non-zero on the interval $[0,1)$. On the other hand $Box(t-s)$ will be non-zero when $t-s \in [0,1[$ or $s \in ]t-1,t]$ which depends on the value of $t$. If $t>2$ or $t<0$ there is no overlap between the intervals $[0,1[$ and $]t-1,t]$ and therefore the integral is zero. If $t<1$ the overlap will be $[0,t]$, if $t\geq 1$ the overlap will be $(t-1,1)$. In these cases the integral will simply be the length of the overlap.
So,
$$(Box \star Box)(t) = \begin{cases} \int_{0}^{1}Box(s)Box(t-s)\;ds = \int_{0}^{1} 0 \;ds = 0 & \text{ if } t<0 \; , \\ \int_{0}^{t}Box(s)Box(t-s)\;ds = \int_{0}^{t} 1 \;ds = t & \text{ if } 0\leq t<1 \; ,\\ \int_{t-1}^{1}Box(s)Box(t-s)\;ds = \int_{t-1}^{1} 1 \;ds = 2-t & \text{ if } 1\leq t<2 \; , \\ \int_{0}^{1}Box(s)Box(t-s)\;ds = \int_{0}^{1} 0\;ds = 0 & \text{ if } 2\leq t \; . \end{cases} $$
Perhaps this might soothe some of your discomfort.
Smoothing Action
There are many ways that convolution is useful in mathematics. First of all, as you have noted, $$ \mathrm{D}^\alpha\left(f\ast g\right)=\left(\mathrm{D}^\alpha f\right)\ast g\tag{1} $$ This is simply repeated changes of the order of integration and differentiation: $$ \frac{\mathrm{d}}{\mathrm{d}x}\int f(x-t)\,g(t)\,\mathrm{d}t=\int f'(x-t)\,g(t)\,\mathrm{d}t\tag{2} $$ This step can be justified in different ways depending on the context. For instance, if the limit which defines the derivative of $f$, $$ f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}\tag{3} $$ converges uniformly, then $(2)$ is valid for all $g\in L^1$.
Convolution combines the smoothness of two functions. That is, if both $f$ and $g$, and their first derivatives are in $L^1$, then the second derivative of their convolution is in $L^1$. This is because $f\ast g = g\ast f$, and so we can use $(2)$ twice to get $$ \frac{\mathrm{d}^2}{\mathrm{d}x^2}(f\ast g)=f'\ast g'\tag{4} $$
Fourier Analysis
Convolution plays an important role in Fourier Analysis. The key formulas demonstrate the duality between convolution and multiplication: $$ \mathscr{F}(f\ast g)=\mathscr{F}(f)\mathscr{F}(g)\quad\text{and}\quad\mathscr{F}(fg)=\mathscr{F}(f)\ast\mathscr{F}(g)\tag{5} $$ There also exists a duality between decay at $\infty$ and smoothness. Essentially, one derivative of smoothness of $f$ corresponds to one factor of $1/x$ in the decay of $\mathscr{F}(f)$, and vice versa.
The product of decaying functions decays even faster; e.g. $x^{-n}x^{-m}=x^{-(n+m)}$. The duality demonstrated in $(5)$ then says that the convolution of smooth functions is even smoother.
The Riemann-Lebesgue Lemma says that for $f\in L^1$, $$ \lim_{|x|\to\infty}\mathscr{F}(f)(x)=0\tag{6} $$ However, this is simply decay with no quantification. About all that can be said about $f,g\in L^1$ is that $f\ast g\in L^1$. However, if $f,g\in L^2$, then $f\ast g$ is continuous.
Summing Dice
Perhaps one of the earliest uses of convolution was in probability. If $f_n(k)$ is the number of ways to roll a $k$ on $n$ six-sided dice, then $$ f_n(k)=\sum_jf_{n-1}(k-j)f_1(j)\tag{7} $$ That is, for each way to achieve $k$ on $n$ dice, we must have $k-j$ on $n-1$ dice and $j$ on the remaining die. Equation $(7)$ represents discrete convolution.
The distribution function for the roll of a single six-sided die is evenly distributed among $6$ possibilities. This has discontinuities at $1$ and $6$ ($n=1$). The distribution function for the sum of two six-sided dice is the convolution of two of the one die distributions. This is continuous, but not smooth ($n=2$). The distribution function for the sum of three six-sided dice is the convolution of the one die and two dice distributions. This is smooth ($n=3$). For each die we add, we convolve one more of the one die distributions and the function gets smoother.
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As $n\to\infty$, the distribution approaches a scaled version of the normal distribution: $\frac1{\sqrt{2\pi}}e^{-x^2/2}$.
Best Answer
There is not (generally). What you are looking for is deconvolution.
Consider the simple case where the Fourier transform of $x$ is zero somewhere. Then the other part can be arbitrary, since then:
$$ Y(\omega) = \frac{Z(\omega)}{X(\omega)} $$