Convolution of characteristic function $\chi_{[0,1]}$ with itself The characteristic function of a set $E$ is defined as follows: $\chi_{E}(x) :=1 \space \text{if} \space  x\in E, \space \text{and} \space \chi_{E}(x) := 0 \space \text{if} \space x \notin E.$ Find a formula for the convolution $\chi_{[0,1]} \star \chi_{[0,1]}(x).$

I have that 
$$\int^{\infty}_{-\infty} \chi_{[0,1]}(x-y)\cdot \chi_{[0,1]}(y)dy=
\begin{cases}
    0 & \text{if }x\notin (0,2) \\
   x & \text{if }x\in (0,1] \\
2-x & \text{if }x\in (1,2).
\end{cases}$$

Is the above computation correct? Thank you for your feedback in advance.
 A: Here is a very different proof based on Dirac $\delta$s.
The first fact we use is the formula for the derivative of a convolution  :
$$(f \star g)'=f' \star g=f \star g'\tag{1}$$
yielding in particular
$$(f \star g)''=f' \star g'\tag{2}$$
Applying (2) to $f=g=\chi_{[0,1]}$ :
$$(\chi_{[0,1]} \star \chi_{[0,1]})''=\chi_{[0,1]}' \star \chi_{[0,1]}'.\tag{3}$$
But (see Fig. 1) : $$\chi_{[0,1]}'=\delta_0-\delta_1 \tag{4}$$

Fig. 1
Plugging (4) into (3) :
$$(\chi_{[0,1]} \star \chi_{[0,1]})''=(\delta_0-\delta_1) \star (\delta_0-\delta_1)\tag{5}$$
$$(\chi_{[0,1]} \star \chi_{[0,1]})''=\underbrace{\delta_0 \star \delta_0}_{\delta_0}-\underbrace{\delta_0 \star \delta_1}_{\delta_1} -\underbrace{\delta_1 \star \delta_0}_{\delta_1} +\underbrace{\delta_1 \star \delta_1}_{\delta_2}=\delta_0-2\delta_1+\delta_2 \tag{6}$$
(we have used the distributivity of convolution and the fundamental formula : $\delta_a \star \delta_b = \delta_{a+b}$).
If we integrate a first time (6), we get a function which is constant by intervals (degree $0$ considered as a polynomial), i.e., whose graphical representation has a staircase pattern.
$$(\chi_{[0,1]} \star \chi_{[0,1]})'=\begin{cases}\ \ 0&if&x<0\\ \ \ 1&if&0<x<1\\-1&if&1<x<2\\ \ \ 0&if&x>2\end{cases}.\tag{7}$$
Integrating once more, one gets a function that is of degree 1 (affine) by intervals :
$$\chi_{[0,1]} \star \chi_{[0,1]}=\begin{cases}\ \ \color{blue}{0}&if&x<0\\ \ \ x\color{red}{+0}&if&0<x<1\\-x\color{red}{+2}&if&1<x<2\\ \ \ \color{red}{0}&if&x>2\end{cases}\tag{8}.$$
How have the integration constants (colored) been chosen ?
The first one (in blue) because we desire that convolution of functions with compact support are themselves with compact support. 
The other ones (in red) in order to ensure the continuity constraint.
Conclusion : we recognize in (8) the looked for result.
What is the interest of all that ? It shows that using Dirac $\delta$s not only is efficient but is even the only way to cope with the complexity of some computations involving convolutions (and also other operations). Imagine for example that you are asked to compute :
$$\chi_{[0,1]} \star \chi_{[0,1]} \star \chi_{[0,1]} \star \chi_{[0,1]}...$$
the only thing you have to do is to reach the level of the fourth derivative, where you are with $\delta$s only and then integrate four times, obtaining t the end a cubic spline function with already a large similitude with a gaussian function.
$Remark :$ Such operations with $\delta$s can be done mechanicaly without a theoretical apparatus till a certain point (which comes very early among demanding students who ask "what are really the objects on which we work ?") ; sooner or later, one has to learn a bit (or more) of a mathematical theory called the "theory of distributions".  
A: Your original answer said the convolution was zero for $x  \geq 0$ or $x \leq 2$, i.e., $x \in [0,2]$. This was not correct, since you have shown that the convolution is equal to $x$ for $0<x \leq 1$ and equal to $2-x$ for $1<x<2$.
The correct equation says that the convolution is zero for $x \notin [0,2]$.
