Calculating the convolution of piecewise constant functions

Question

Let $f(x) = \frac{1}{2}$ on $[-1,1]$. Find $f*f*f$

$(f*g)(x)=\int\limits_{-\infty}^\infty f(t)g(x-t)\,dt$.

So $(f*f)(x)=\begin{cases} \frac{1}{4}x+\frac{1}{2} & -2\le x \le 0 \\ \frac{-1}{4}x+\frac{1}{2} & 0 < x \le 2 \\ 0 & \text{otherwise} \end{cases}$

This is almost identical to this question

Which this answer in this questions does now clearly explain how to find the limits of integration. I figured it out for $f*f$, but I am now lost in trying to calculate $f*f*f$.

I know that $(f*f*f)(x)$ will be in three pieces on $[-3,-1],[-1,1],$ and $[1,3]$, but I do not know how to set up my integrals.

Answer 1

The convolution is $f*f*f(w)=\int_{x}\int_{y} f(x)f(y)f(w-x-y) dydx$. We seek the integration limits, firstly for $x$ given $w$ (with $-3\lt w\lt 3$), then for $y$ given $w,x$.

Limits for $x$ given $w$:

• $-1\lt x\lt 1$
• $w-2\lt x\lt w+2\qquad \text{(lower bound comes from $x=w-y-z\;$ with $y=z=1$)}$ $\qquad\qquad\qquad\qquad\qquad\text{(upper bound comes from $x=w-y-z\;$ with $y=z=-1$)}$

So the limits for $x$: $\qquad\max\{-1,w-2\}\lt x\lt\min\{1,w+2\}$

Note:

• $-1\lt w-2$ iff $w\gt 1$
• $1\lt w+2$ iff $w\gt -1$

Hence, we split the interval $w\in [-3,3]$ at points $-1,1$.

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Limits for $y$ given $w,x$:

• $-1\lt y\lt 1$
• $w-x-1\lt y\lt w-x+1\qquad \text{(lower bound comes from $y=w-x-z\;$ with $z=1$)}$ $\qquad\qquad\qquad\qquad\qquad\qquad\text{(upper bound comes from $y=w-x-z\;$ with $z=-1$)}$

So the limits for $y$: $\qquad\max\{-1,w-x-1\}\lt y\lt\min\{1,w-x+1\}$

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For interval $w\in(-3,-1)$:

• $-1\lt x\lt w+2$

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• $\max\{-1,w-x-1\} = -1$
• $\min\{1,w-x+1\} = w-x+1$

$\qquad\therefore\quad -1\lt y\lt w-x+1$

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For interval $w\in(-1,1)$:

• $-1\lt x\lt 1$

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• $\max\{-1,w-x-1\} = -1 \text{ only when $x\gt w$}$
• $\min\{1,w-x+1\} = 1 \text{ only when $x\lt w$}$

So if $-1\lt x\lt w$ then $w-x-1\lt y\lt 1$.

And if $w\lt x\lt 1$ then $-1\lt y\lt w-x+1$.

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For interval $w\in(1,3)$:

• $w-2\lt x\lt 1$

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• $\max\{-1,w-x-1\} = w-x-1$
• $\min\{1,w-x+1\} = 1$

$\qquad\therefore\quad w-x-1\lt y\lt 1$

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So our integrals are:

For $-3\lt w\lt -1$:

\begin{align} f*f*f(w) &= \int_{x=-1}^{w+2}\int_{y=-1}^{w-x+1} f(x)f(y)f(w-x-y)\;dydx \\ &= \left(\dfrac{w+3}{4} \right)^2 \end{align}

For $-1\lt w\lt 1$:

\begin{align} f*f*f(w) &= \int_{x=-1}^{w}\int_{y=w-x-1}^{1} f(x)f(y)f(w-x-y)\;dydx \\ &\qquad + \int_{x=w}^{1}\int_{y=-1}^{w-x+1} f(x)f(y)f(w-x-y)\;dydx \\ &= \dfrac{3-w^2}{8} \end{align}

For $1\lt w\lt 3$:

\begin{align} f*f*f(w) &= \int_{x=w-2}^{1}\int_{y=w-x-1}^{1} f(x)f(y)f(w-x-y)\;dydx \\ &= \left(\dfrac{w-3}{4} \right)^2 \end{align}

It might help to see the limits of integration by looking at how the plane $x+y+z=w$ moves through the cube $[-1,1]^3$ as $w$ varies from $-3$ to $3$.