# Convolution of bounded functions

I have two functions $f(x)$ and $g(x)$ that are nonzero everywhere except in the interval $[a,b]$ (i.e. there are a $h(x)$ and $j(x)$ such that $f(x) = \mathbb{1}_{a\leq x\leq b} \cdot h(x)$ and $g(x) = \mathbb{1}_{a\leq x\leq b} \cdot j(x)$).

The general convolution definition is $$\left(f*g\right)(x)=\int_{-\infty}^{+\infty}f(y) g(x-y)\text dy$$

When $a=0$ and $b=+\infty$ $$\left(f*g\right)(x)=\int_{0}^{x}f(y) g(x-y)\text dy=\int_{0}^{x}f(x-y) g(y)\text dy$$

What are the integration limits for arbitrary $a$ and $b$ (with $a < b$)?

You just have to check where the functions could be both nonzero. In particular you need $$y\in [a,b]\quad\mathrm{and}\quad x-y\in[a,b]$$ that is $$a\leq y\leq b\quad \mathrm{and}\quad x-b\leq y\leq x-a$$ that is equivalent to $$y\in[a,b]\cap[x-b,x-a].$$ Depending on $x$ this could be even the emptyset.
• So the limits would be $\max(a,x-b)$ and $\min(x-a,b)$ for $x\in[2a,2b]$, seems like. – Pedro Carvalho Sep 23 '16 at 18:38
• @Pedrocarvalho it should be that for $x\in [2b,2a]$ – Del Sep 24 '16 at 7:03
• I don't think so? First because, since $a<b$, then $[2b,2a]=\varnothing$. Second, if $f$ is the pdf of a random variable $X$ and $g$ that of a random variable $Y$ then the convolution is the pdf of the random variable $X + Y$ which can take values between $2a$ and $2b$ in that order, so that's also the support of the convolution. – Pedro Carvalho Sep 24 '16 at 13:53