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$n$ 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

sisiinfinity
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  • Not quite correct. Remember, the convolution of two integrable functions is continuous. You're answer isn't continuous. – User8128 Mar 25 '16 at 19:50

1 Answers1

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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 $0$.

Patrick Stevens
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  • thanks a lot. but I don't understand how you went from $\int_{t-n}^t 1_{[0,n]}(u)du$ to the next step. can you please explain – sisiinfinity Mar 25 '16 at 19:52
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    @sisiinfinity This is the kind of thing you might want to draw pictures for, in the cases that $n < t$, $n \geq t$, and so on. The $\max(t-n, 0)$ term is there because there are two ways a point could fail to contribute to the integral by "being too small": if $u < t-n$ (because then it falls outside the range of integration) or if $u < 0$ (because then the indicator function sends it to $0$). Likewise, the $\min(t, n)$ term is because there are two ways a point can fail to contribute by "being too big": if $u > t$ (falls outside the range of integration) or $u > n$ (sent to $0$ by indicator). – Patrick Stevens Mar 25 '16 at 19:56
  • just one more question: shouldn't it be $2n -t$ for any $t > n$? i don't understand why you insisted that $t < 2n$ – sisiinfinity Mar 25 '16 at 20:13
  • The point is that if $t > 2n$ then my last formulation $\int_{\max(t-n,0)}^{\min(t, n)} du$ no longer holds, because if $t > 2n$ then there's no points of the integrand which come to $1$ at all. I'll clarify that. – Patrick Stevens Mar 25 '16 at 20:53