convolution problem given $h(x)=1/2$ for $0I have a convolution problem in the form
$$g(x)= \int_{-\infty}^\infty h(y)h(x-y)\,dy$$
where they give me the function $h(x)=1/2$ for $0<x<2$ and $0$ otherwise.
I have never done a convolution problem before so I would like someone to walk me through this. I know I'll need to do a change of variables
 A: $\newcommand{\d}[1][x]{\,\mathbb{d}#1}$
If either $y$ or $x-y$ are outside of the interval $(0,2)$ then the integrand is zero and the value of the integral on such intervals (of $y$) is zero.
So search for ranges of $y$ for which $0 \lt x - y \lt 2$ and $0 \lt y \lt 2$, because these are where the integrand is non-zero, and hence the integral is non-zero.
$x-y \gt 0 \implies y \lt x$ and $x-y \lt 2 \implies y \gt x-2$.
So, our integrand is non-zero for all $y \in (x-2,x) \cap (0,2)$ (and zero everywhere else).
From this, we can construct $g(x)$ for different ranges of $x$.
$$g(x) = \begin{cases}
0 & x \lt 0 \\ \int_0^x \tfrac14 \d[y] & 0 \lt x \lt 2 \\ \int_{x-2}^2 \tfrac14 \d[y] & 2 \lt x \lt 4 \\ 0 & x \gt 4 
\end{cases}$$
A: The product we are integrating is $0$ except when $0\lt y\lt 2$ and $0\lt x-y\lt 2$.
Suppose that $x\le 0$. Then we cannot satisfy the two inequalities. So in this case we our integral is $0$. And we cannot satisfy them when $x\ge 4$. So there again the integral is $0$.
So let $0\lt x\lt 4$. There are two cases, (i) $0\lt x\le 2$ and (ii) $2\lt x\lt 4$.
Case (i): Suppose that $0\lt x\le 2$.  To make $0\lt x-y$, we need $y\lt x$. So $y$ ranges over the interval $(0,x)$. We are integrating $\frac{1}{4}$ over this interval, and get $\frac{x}{4}$.
Case (ii): Suppose now that $2\lt x\lt 4$. To make $x-y\lt 2$, we need $y\gt x-2$. So we are integrating $\frac{1}{4}$ from $y=x-2$ to $y=2$. The interval has length $4-x$, so the integral is $\frac{4-x}{4}$.
