Is there a version of the Titchmarsh Convolution theorem to find singular support? Okay, some terminology, correct me if I'm wrong.
Singular support - the set on which a distribution fails to be smooth. In this case a piecewise function.
Is there a name for $f*f*f$? The convolution of function with itself $f*f$ is called the autocorrelation, right.
I want to know the singular support of $f*f*f$, where $f$ is the indicator function of a regular hexagon in the plane?
I can calculate, in mathematica, $f*f$ - it's a Piecewise function. I can extract the conditions without inequalities (ie line segments) and plot it. (For some reason it seems to include the axes, on which I know $f*f$ is smooth).

However, trying this for $f*f*f$ is overheating my Raspberry Pi.
Also, I know from the Titchmarsh Convolution Thereom, that since supp f is convex.
$$
supp f*f*f \subset supp f +supp f*f  \subset supp f+supp f +supp f,
$$
it seems that $supp f*f*f = 3 supp f$.
Is there a version of the Titchmarsh Convolution theorem for singular support?
 A: The answer is all line segments of the form $p_1 + p_2 + e$ where $p_1$ and $p_2$ are vertices of the hexagon and $e$ is an edge. More carefully, I can prove that the singular locus is contained in this set, and I expect equality. Here is a picture.

Proof: Write $H$ for the filled in hexagon. So $H^3$ is a $6$-dimensional polytope in $\mathbb{R}^6$. There is a linear map $\pi: H^3 \to \mathbb{R^2}$ sending $(x_1, y_1, x_2, y_2, x_3, y_3) \to (x_1+x_2+x_3, y_1+y_2+y_3)$. The (ordinary) support of $f \ast f \ast f$ is the image of $\pi$, and the values of the convolution are the $4$-dimensional volumes of the fibers of $\pi$. 
Now, as $z$ varies through $\mathbb{R}^2$, $\pi^{-1}(z)$ is a four dimensional polytope. Generically, changing $z$ will just slide the defining facets of $\pi^{-1}(z)$ parallel to themselves. The volume of a polytope with fixed facet normals is a polynomial (degree $4$ in this case) in the positions of those facets, hence smooth. 
When do singularities occur? When the combinatorics of $\pi^{-1}(z)$ changes. This happens when $5$ facets of $\pi^{-1}(z)$ pass through a common vertex. (Since these are $5$ hyperplanes in $4$-space, it doesn't happen generically.) Back in the big $6$ dimensional polytope $H^3$, this is the same as saying that $z$ lies below a point of $H^3$ which is on $5$ facets -- in other words, $z$ lies below an edge of $H^3$. That's the condition I have above.
Now, if $z$ only lay below one edge of $H^3$, I could write down the singularity introduced by that edge and show that $f \ast f \ast f$ is indeed singular there. Instead, most points which are below some edge of $H^3$ are below many edges, so I can't guarantee the singularities don't cancel. But I don't expect them to.

To be more general, let $P$ be a full dimensional subpolytope in $\mathbb{R}^d$ and let $\pi: \mathbb{R}^d \to \mathbb{R}^e$ be a linear map. Let $g(z) = \mathrm{Volume}(\pi^{-1}(z))$ for $z \in \mathbb{R}^e$. Then $g$ is singular along the images of the $e-1$ dimensional faces of $P$.
If $f_1$, $f_2$, ..., $f_r$ are the characteristic functions of polytopes $P_1$, $P_2$, ..., $P_r$ in $\mathbb{R}^e$, then $f_1 \ast f_2 \ast \cdots \ast f_r$ is singular along sets of the form $Q_1 + Q_2 + \cdots + Q_r$, where $Q_j$ is a face of $P_j$ and $\sum \dim Q_j = e-1$
A: supp$f*f*f$ is just $H+H+H$ where $H$ is the hexagon. If $C$ is any convex set, then $C+C+\ldots+C$ ($n$ times) is just $nC=\{nx\colon x\in C\}$. 
To see this, notice that if $x_1,\ldots,x_n\in C$, then $x_1+\ldots+x_n=n\cdot \frac1n(x_1+\ldots+x_n)\in nC$. Conversely if $x\in nC$, then $x=ny$ for $y\in C$, so that $x=y+y+\ldots+y$, so that $x\in C+\ldots+C$. 
