How to calculate the third power of convolution of a characteristic function $\chi\ast\chi\ast\chi$

Question

Let $\chi(x)$ be defined as $1$ for $x\in(-1,1)$ and everywhere else it's 0. So I managed to calculate $\chi\ast\chi$ which is equal to $2-\lvert x \rvert$ for $x\in(-2,2)$ and everywhere else it's 0. I did it graphically and also with the integral. But when I try to do $\chi\ast\chi\ast\chi$ I kinda get stuck, because when I plot the function online and try to calculte the integral $\int_{-1}^{1} (\chi\ast\chi)(x-y)dy$ I don't get "smooth" transitions from the x axes. Can someone give me some kind of hint of how to do it? Calculating the integral gives me $3-x^2$ for $x\in[-1,1]$, $2-2\lvert x\rvert$ for $\lvert x\rvert\in[1,3)$ and everywhere else it's 0.

Answer 1

Here is a powerful way for computing this convolution.

Let us first recall three properties, with $H$ the Heaviside step function:

$$\tag{0} \begin{cases}(i) & H'(x-a)=\delta_a(x)\\ (ii) & \delta_a \ast \delta_b=\delta_{a+b}\\ (iii) & (f*g)'=f'*g=f*g'\end{cases}$$

The main idea is to process in two steps

• A) differentiate $\chi\ast\chi\ast\chi$ 3 times, in order to reach the level at which one can compute on $\delta_a$ "functions", and then

• B) integrate 3 times (!) the result of part A).

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Part A):

As a consequence of $(0)(iii)$, the third derivative is

$$\tag{1}(\chi\ast\chi\ast\chi)'''=\chi'\ast\chi'\ast\chi'$$

Knowing that, (using $(0)(i)$):

$$\tag{2}\chi(x):=H(x+1)-H(x-1) \ \ \implies \ \ \chi'=\delta_{-1}-\delta_{1}$$

Plugging the result of $(2)$ into $(1)$, we get:

$$\tag{3}(\chi\ast\chi\ast\chi)'''=(\delta_{-1}-\delta_{1})\ast(\delta_{-1}-\delta_{1})\ast(\delta_{-1}-\delta_{1}) \ = \ \delta_{-1}\ast \delta_{-1}\ast \delta_{-1}+\cdots$$

Using the result $(0)(ii)$, we finally obtain:

$$\tag{4}(\chi\ast\chi\ast\chi)'''=\delta_{-3}-3\delta_{-1}+3\delta_{1}-\delta_{3}.$$

(The sequence of coefficients $1,-3,3,-1$ can be explained by considering binomial expansion.)

Part B) We have to integrate 3 times expression $(3)$:

$$\text{1st int.:} \ (\chi\ast\chi\ast\chi)''=H(x+3)-3H(x+1)+3H(x-1)-H(x)$$

$$(\chi\ast\chi\ast\chi)''=\begin{cases} \ \ \ 1 & \text{if} -3<x<-1\\-2 & \text{if} \ -1<x<1\\ \ \ \ 1 & \text{if} \ 1<x<3\\ \ \ \ 0 \ & \ \text{elsewhere} \end{cases}.$$

$$\text{2nd int.:} \ (\chi\ast\chi\ast\chi)'=\begin{cases} \ \ \ 1x+a & \text{if} \ -3<x<-1\\-2x+b & \text{if} \ -1<x<1\\ \ \ \ 1x+c & \text{if} \ 1<x<3\\ \ \ \ 0 & \text{elsewhere}\end{cases}=\begin{cases} \ \ \ x+3 & \text{if} \ -3<x<-1\\-2x & \text{if} \ -1<x<1\\ \ \ \ x-3 \ & \text{if} \ 1<x<3\\ \ \ \ 0 & \text{elsewhere}\end{cases}$$

(where the integration constants $a,b,c$ have been chosen to grant continuity to the result).

$$\text{3rd integration:} \ \chi\ast\chi\ast\chi=\begin{cases} \ \ \ x^2/2+3x+9/2 & \text{if} \ -3<x<-1\\-x^2+3 & \text{if} \ -1<x<1\\ \ \ \ x^2/2-3x+9/2 & \text{if} \ 1<x<3\\ \ \ \ 0 & \text{elsewhere}\end{cases}.$$

(we have not detailed in this case the computation of integration constants.)

Here is the graphical representation (which belongs to the family of quadratic splines, union of 3 Bezier curves):