Why is the convolution of two box kernels a triangle kernel? Can anyone show the mathematical steps proving 
$K_{box}*K_{box} = K_{triangle}$
 A: Consider the simple case where $K_\text{box}(x) = \mathbb{I}(0 \leqslant x \leqslant 1)$.  Then you have:
$$\begin{equation} \begin{aligned}
(K_\text{box}*K_\text{box})(r) 
&= \int \limits_\mathbb{R} K_\text{box}(x) K_\text{box}(r-x) dx \\[6pt]
&= \int \limits_\mathbb{R} \mathbb{I}(0 \leqslant x \leqslant 1) \cdot \mathbb{I}(0 \leqslant r-x \leqslant 1) \ dx \\[6pt]
&= \int \limits_\mathbb{R} \mathbb{I}(0 \leqslant x \leqslant 1) \cdot \mathbb{I}(r-1 \leqslant x \leqslant r) \ dx \\[6pt]
&= \begin{cases}
\int_0^1 \mathbb{I}(0 \leqslant x \leqslant r) \ dx  & & & \text{for } 0 \leqslant r \leqslant 1, \\[10pt]
\int_0^1 \mathbb{I}(r-1 \leqslant x \leqslant 1) \ dx  & & & \text{for } 1 < r \leqslant 2, \\[10pt]
0 & & & \text{otherwise}. \\[10pt]
\end{cases} \\[10pt]
&= \begin{cases}
r  & & & \text{for } 0 \leqslant r \leqslant 1, \\[10pt]
2 - r & & & \text{for } 1 < r \leqslant 2, \\[10pt]
0 & & & \text{otherwise}. \\[10pt]
\end{cases} \\[10pt]
&= K_\text{triangle}(r). \\[6pt]
\end{aligned} \end{equation}$$
The more general case where you have a box kernel with arbitrary limits is a simple extension of this, and the working is analogous.
A: I think a 1D solution would suffice to give you an idea. Let $x(t)=1$ when $0<x<1$. Convolution with itself can be formulated as 
$$y(t)=x(t)*x(t)=\int_{-\infty}^{\infty}x(\tau )x(t-\tau )d\tau =\int_{0}^{1}x(t-\tau )d\tau$$
There are four major cases here. When $t<0$ or $t>2$, $y(t)=0$ since there is no $\tau$ in $[0,1]$ that makes inside of the integral non-zero. For $0<t<1$, we have $y(t)=\int_{0}^{t}d\tau=t$. For $1<t<2$, we have $y(t)=\int_{t-1}^{1}d\tau=2-t$. When you sketch it, you’ll see the triangle.
