Convolution, How to develop this piecewise? This is an exercise from my book on hilbert spaces. I don't understand the last step. Can someone develop the procedure to get that piecewise?.
Convolution $\left(f\ast g\right)$
Indicator function  $\mathcal{X}_{\left[0,1\right]}$
If $f=g=\mathcal{X}_{\left[0,1\right]}$, you have $\left(f\ast g\right)=\int_{-\infty}^{\infty}f(t-u)g(u)du=\int_{0}^{1}\mathcal{X}_{\left[0,1\right]}(t-u)du=length\Big(\left[0,1\right]\cap\left[t-1,t\right]\Big)$
so $t-u\in\left[0,1\right]$ if and only if $u\in\left[t-1,t\right]$. Therefore,
\begin{equation*}
\left(f\ast g\right)= \left\{ 
  \begin{array}{ll} 
  0 \quad &\text{if}\, t\leq 0 \\
  t \quad &\text{if}\, 0\leq t\leq 1 \\
     2-t  \quad &\text{if}\, 1\leq t\leq 2 \\
     0 \quad &\text{if}\, t\geq 2
  \end{array}
 \right.
\end{equation*}
 A: By "last step" I suppose you mean "the last equality". There is also a "(t)" missing on the left hand side of the first equality. Also, be sure not to confuse $\chi_{[0,1]}(t-u)$ with the product of two functions.
To avoid such possible misunderstandings, I rewrite all the equations, using the notation $\chi$ for the function $\chi_{[0,1]}$. So
$$ (f*g)(t)=\int_{-\infty}^{\infty} f(t-u) g(u) du =\int_{-\infty}^{\infty} \chi(t-u) \chi(u) du . $$
Now, the first factor in the integrand is $1$ if $$ \begin{align} 0\leq t-u \leq 1 &\iff 0 \geq u-t \geq -1 \iff t-1 \leq u \leq t \\ &\iff u \in [t-1,t] \end{align}$$ and $0$ otherwise. Likewise, the second factor in the integrand is $1$ if $u\in [0,1]$ and $0$ otherwise. Finally, since the integral is the surface under the integrand and the integrand is a (possibly trivial) rectangle of height $1$, the integral can be replaced by the length of the rectangle which is indeed $$length([0,1] \cap [t-1,t]) = \max(\min(1,t) - \max(0, t-1),0).$$
In the last equality, I gave a solution which avoids piecewise analysis. But if you still want to do the piecewise analysis, with the help of @peek-a-boo's comment, everything should be clear.
