Convolution of characteristic function

Question

I am trying to figure out following problem.

Let A ⊂ R. Then we can define the characteristic function:

Let a be bigger than 0. I am trying to find a following convolution: \begin{align} \chi_{[-a,a]} * \chi_{[-a,a]} *\chi_{[-a,a]} \end{align}

Answer 1

Unfortunately, your result for the convolution of the first two terms is not correct. We have

$$\begin{align*} \chi_{[-a,a]}(x-y) \chi_{[-a,a]}(y) &= \chi_{[-a,a]}(y-x) \chi_{[-a,a]}(y) \\ &= \chi_{[-a+x,a+x]}(y) \chi_{[-a,a]}(y). \end{align*}$$

This product equals $1$ if, and only if,

$$-a + x \leq y \leq a+x \qquad \text{and} \qquad -a \leq y \leq a. \tag{1} $$

We consider two cases separately:

1. If $x \geq 0$, then $-a+x \geq -a$ and $a+x \geq a$, so $(1)$ is equivalent to $$-a+x \leq y \leq a.$$ If $x >2a$ there exists no such $y$; if $x \in [0,2a]$, then $(1)$ holds for any $y \in [-a+x,a]$.
2. Similarly, if $x < 0$, then $-a+x < -a$ and $a+x <a$ and so $(1)$ is equivalent to $$a \leq y \leq a+x.$$ If $x <-2a$ there is no such $y$; if $x \in [-2a,0]$, then $(1)$ holds for any $y \in [a,a+x]$.

Consequently,

$$ \chi_{[-a,a]}(x-y) \chi_{[-a,a]}(y) = \begin{cases} 0, & x \notin [-2a,2a], \\ 1_{[a,a+x]}(y), & x \in [-2a,0], \\ 1_{[-a+x,a]}(y), & x \in [0,2a] \end{cases}$$

Integrating with respect to the Lebesgue measure gives

$$\chi_{[-a,a]} \ast \chi_{[-a,a]}(x) = (x+2a) \chi_{[-2a,0)}(x) + (2a-x) \chi_{[0,2a]}(x). \tag{2}$$

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In order to calculate the convolution of this function with the indicator function, we have to determine $\chi_{[-2a,0]}(x-y) \chi_{[-a,a]}(y)$ and $\chi_{[0,2a]}(x-y) \chi_{[-a,a]}(y)$. Note that

$$\begin{align*} \chi_{[-2a,0]}(x-y) \chi_{[-a,a]}(y) &= \chi_{[0,2a]}(y-x) \chi_{[-a,a]}(y) \\ &= \chi_{[x,x+2a]}(y) \chi_{[-a,a]}(y). \end{align*}$$

This expression equals $1$ if, and only if,

$$x \leq y \leq x+2a \qquad \text{and} \qquad - a \leq y \leq a. \tag{3}$$

1. If $x \leq -a$, then $x \leq -a$ and $x+2a \leq -a$, so $(3)$ is equivalent to $$-a \leq y \leq x+2a.$$ If $x < -2a$ there is no such $y$; if $x \in [-2a,-a]$, then $(3)$ holds for $y \in [-a,x+2a]$.
2. If $x>-a$, then $x > -a$ and $x+2a > -a$, so $(3)$ is equivalent to $$x \leq y \leq a.$$ If $x >a$ there is no such $y$; if $x \in [-a,a]$, then $(3)$ holds for $y \in [-a,2a+x]$.

Consequently,

$$\chi_{[-2a,0]}(x-y) \chi_{[-a,a]}(y) = \begin{cases} 1_{[x,a]}(y), & -a \leq x \leq a, \\ 1_{[-a,2a+x]}(y) & -2a \leq x \leq -a. \end{cases}$$

A very similar argumentation shows

$$\chi_{[0,2a]}(x-y) \chi_{[-a,a]}(y) = \begin{cases} 1_{[-a,x]}(y), & -a \leq x \leq a, \\ 1_{[-2a+x,a]}(y) & a \leq x \leq 2a. \end{cases}$$

Adding all up, we get

$$\chi \ast \chi \ast \chi(x) = \begin{cases} 0, & x \notin [-2a,2a], \\ \int 1_{[-a,2a+x]}(y) \cdot (x+2a) \, dy, & x \in [-2a,-a], \\ \int 1_{[x,a]}(y) \cdot (x+2a) + 1_{[-a,x]}(y) \cdot (2a-x) \, dy, & x \in [-a,a] \\ \int 1_{[-2a+x,a]}(y) \cdot (2a-x) \,dy, & x \in [a,2a]. \end{cases}$$

I leave the calculation of this simple integrals to you.