Convolution of two functions.

Question

$f(x)=2x/3$, $0<x<3$, $f(x)=0$ otherwise

$g(x)=1$, $-1<x<3$, $g(x)=0$ otherwise

I am trying to work out the convolution $h=f*g= \int(f(y)g(x-y))dy$

I am able to show that:

$x - 3 < y < x + 1$

In the mark scheme to this question it then says there are various cases to consider:

$2 < x < 3$

$3 < x < 6$

$6 < x $

$-1 < x < 2$

$x < -1$

I am unsure how to get the final five inequalities listed. I understand how to use them after you have them but I would appreciate it if someone explained how to arrive there.

Answer 1

As you mention, you can easily deduce that $x-3<y<x+1$. Equivalently, you can also arrive to $y-1<x<y+3$.

Now, due to the definition of $f$, you have that the interesting values for $y$ are those in $(0,3)$. For $y=0$, $-1<x<3$ and for $y=3$, $2<x<6$. I guess that the intervales you mention arise from those two $(-1,3)$ and $(2,6)$:

• $2<x<3 \Leftrightarrow x\in(-1,3)\cap(2,6)$
• $3<x<6 \Leftrightarrow x\in(2,6)\setminus(-1,3)$
• $-1<x<3 \Leftrightarrow x\in(-1,3)\setminus(2,6)$
• $6<x \mbox { and } x<-1 \Leftrightarrow x\not\in(-1,3)\cup(2,6)$

In any case, I think it is a bit confusing (and probably, unnecessary) using those intervals.