I have 2 functions $f(x)=1_{[-1,1]}$ and $g(x)=\frac{1}{2\epsilon}1_{[-\epsilon,\epsilon]}$
My questions is how to calculate the convolution $f\ast g\ast g$. I know that $g \ast g$ is the piecewise function:
$\frac{2\epsilon+x}{4\epsilon^2}$ for $-2\epsilon \leq x < 0$
$\frac{1}{2\epsilon}$ for $x=0$
$\frac{2\epsilon-x}{4\epsilon^2}$ for $0 < x \geq 2\epsilon$
But I don't know how to proceed.
Let $2\epsilon\le1$, we have that \begin{align} (f\ast g\ast g)(x)&=\int_{\Bbb R}f(t)(g\ast g)(x-t)dt\\ &=\int_{\Bbb R}\chi_{[-1,1]}(g\ast g)(x-t)dt\\ &=\int_{-1}^1(g\ast g)(x-t)dt\\ &=\int_{-1}^1(g\ast g)(x-t)\chi_{[x-2\epsilon,x+2\epsilon]}dt\\ &=\int_{\max(x-2\epsilon,-1)}^{\min(x+2\epsilon,1)}(g\ast g)(x-t)dt\\ &=\left\{ \begin{aligned} \frac{(x+1+2\epsilon)^2}{8\epsilon^2}&,\text{ if }-2\epsilon-1\le x\le-1\\ 1-\frac{(x+1-2\epsilon)^2}{8\epsilon^2}&,\text{ if }-1\le x\le-1+2\epsilon\\ 1&,\text{ if }-1+2\epsilon\le x\le1-2\epsilon\\ 1-\frac{(x-1+2\epsilon)^2}{8\epsilon^2}&,\text{ if }1-2\epsilon\le x\le1\\ \frac{(x-1-2\epsilon)^2}{8\epsilon^2}&,\text{ if }1\le x\le1+2\epsilon\\ \end{aligned} \right.\\ \end{align} which gives
I leave the case $2\epsilon>1$ for you.