Compute $1_{[0,n]} * 1_{[0,n]}$

Question

$n$ is a natural number. I want to find the convolution of $f = 1_{[0,n]}$ with itself ($1$ is for indicator). Is my work correct

$$(f *f)(x) = \int_0^n 1_{[0,n]}(x-y)dy = 1_{[0,2n]}(x)$$

thanks

Answer 1

Well, let's find out!

$$(f * f)(t) = \int_{-\infty}^{\infty} f(\tau) f(t-\tau) d \tau \\ = \int_{-\infty}^{\infty} 1_{[0,n]}(\tau) 1_{[0,n]}(t-\tau) d \tau \\ = \int_{0}^{n} 1_{[0,n]}(t-\tau) d \tau$$

Substituting $u = t-\tau$, this is $$\int_{t-n}^{t} 1_{[0,n]}(u) d u$$ which is $$\int_{\max(t-n,0)}^{\min(t, n)} du = \min(t,n) - \max(t-n, 0)$$ if $0 \leq t \leq 2n$. (Otherwise the integral is $0$ because no points simultaneously lie in $[0,n]$ and $[t-n, t]$.)

If $n \geq t>0$, this is $t$. If $n < t < 2n$, this is $2n-t$. Otherwise it is $0$.