Given is the function $f(x)= \frac{1}{2} $ for $ -1 \leq x \leq 1$ and $0$ elsewhere. We define the function $g:= f(x)*f(x)$ where $*$ is the convolution operator. So $g$ is the result of the convolution integral of $f(x)$ with itself. Now, I am asked to find $h:=g*f$, i.e. calculate $f(x)$ convolved with its convolution with itself. And then I need to evaluate it at $h(\frac{1}{3})$. My solution says that I should get $h(\frac{1}{3})=\frac{1}{2}$
My attempt:
First, I found $g(t)$ by evaluating $\int_{-\infty}^{\infty}f(\tau) f(t-\tau)d\tau$ and I get
$\frac{1}{2} + \frac{t}{4}, -2 \leq t \leq 0$
$\frac{1}{2} – \frac{t}{4}, 0 \leq t \leq 2$
That seems to be correct. I draw both cases on paper and it seems to make sense. If we draw $g$, we get a triangle of maximum height $\frac{1}{2}$ at $0$, and $0$ for $ t < -2$ and $ t >2$.
But now, finding $h=g*f$ and evaluating it at $h(\frac{1}{3})$ is a lot more troubling. First, I don't think that I need to find all cases given that we are only interested in $h(\frac{1}{3})$, i.e. the region that includes $\frac{1}{3}$. After several trial and errors, I get $\frac{1}{2}+\frac{t}{4}$ for the region $-1 \leq t \leq 1$. And so, if I evaluate it at $t=\frac{1}{3}$ I get $\frac{7}{12}$ which is close to $\frac{1}{2}$ but not equal.
I am 99% sure that my $g(t)$ is correct. So either I got something wrong while finding $h=g*f$ and $h(\frac{1}{3})$, or the solution they gave us is wrong. Could you help me ? Any support would be greatly appreciated. Thank you !
Best Answer
The convolution is an associative and commutative operation. In the following $\chi_A$ is the indicator function for a set $A$ and the integral sign $\int$ can be understood as a shorthand for $\int_{-\infty}^\infty$, then
$$ \begin{align*} f*f*f(x)&=\frac1{8}\iint \chi _{[-1,1]}(s)\chi_{[-1,1]}(t-s)\chi_{[-1,1]}(x-t)\,\mathrm d t\,\mathrm d s\\ &=\frac1{8}\iint\chi_{[-1,1]}(s)\chi_{[s-1,s+1]}(t)\chi_{[x-1,x+1]}(t)\,\mathrm d t\,\mathrm d s\\ &=\frac18\iint\chi_{[-1,1]}(s)\chi_{[s-1,s+1]\cap [x-1,x+1]}(t)\,\mathrm d t\,\mathrm d s \end{align*}\tag1 $$
For $x=1/3$ we get $$ [-2/3,4/3]\cap [s-1,s+1]=[\max\{-2/3,s-1\},\min\{s+1,4/3\}]\\ \qquad =\frac12\big[s-5/3+|s-1/3|,s+7/3-|s-1/3|\big]\tag2$$
where we used the formulas $$ \max\{a,b\}=\frac12(a+b+|a-b|),\quad \min\{a,b\}=\frac12(a+b-|a-b|)\tag3 $$ Thus from $\rm (2)$ and any chosen $s$ the length of the interval for valid values of $t$ in $\rm (1)$ is $2-|s-1/3|$, as far as this value is non-negative. Therefore $$ \int\chi_{[-1,1]}(s)\chi_{[-1,1]}(t-s)\chi_{[-1,1]}(1/3-t)\,\mathrm d t\\ \qquad =\chi_{[-1,1]}(s)(2-|s-1/3|)\chi_{[0,\infty)}(2-|s-1/3|)\tag4 $$ Hence $$ f*f*f(1/3)=\frac1{8}\int\chi_{[-1,1]}(s)(2-|s-1/3|)\chi_{[0,\infty)}(2- |s-1/3|)\,\mathrm d s\\ =\frac1{8}\int_{-1}^{1}(2-|s-1/3|)\,\mathrm d s=\frac12-\frac18\int_{-1}^{1}|s-1/3|\,\mathrm d s=\frac{13}{36}\tag5 $$
Note that
$$ g(t):=f*f(t):=\int f(s)f(t-s)\,\mathrm d s\tag6 $$
Then $$ h(x):=g*f(x)=\int g(t)f(x-t)\,\mathrm d t=\iint f(s)f(t-s)f(x-t)\,\mathrm d s\,\mathrm d t\tag7 $$ and using Fubini's theorem we can change the order of integration as we please, by example to get the expression of $\rm (1)$. Because convolution is commutative then alternatively we can write $$ h(x)=f*g(x)=\int f(t)g(x-t)\,\mathrm d t=\iint f(t)f(s)f(x-t-s)\,\mathrm d t\,\mathrm d s\tag8 $$ for an equivalent expression.