convolution chi function with itself

Question

Find $f*f$ where $f= \chi _{[0,1]}$

Find its convolution

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$$ \chi_{[0,1] } = \begin{Bmatrix} 1 & x \in [0,1] \\ 0 & x \not \in [0,1] \end{Bmatrix} $$ Convolution Def

$$ f*g= \int f(\tau) g(t-\tau) d\tau$$

Sketch $f(-\tau)$

Sketch $f(\tau)$

Sketch $f(t-\tau)$

and let us consider $t\in [0,1]$

I know it makes a rectangle so the area is length times width which is $t*1=1$ but not sure how to interpret the integral

$$ \int^1_0 f(t-\tau)g(\tau) d(\tau)= \int \text{??} d\tau = \int^t_0 1 d \tau =t $$

I am missing stuff for $t[1,2]$. also when its zero for the values at $t>2$ and $t<0$ _______________________-

This is an example 20.1.2 on Gaskett Fourier

the answer is

$$f*g(x) = \begin{Bmatrix} 0 & \text{ if } x\leq 0 \\x & \text{ if } 0\leq x\leq 1 \\2-x & \text{ if } 1\leq x\leq 2 \\ 0 & \text{ if } x\geq 0 \end{Bmatrix} $$

Also do not understand why

$$\int_R f(x-t) g(t) = \int 1_0 \chi _{[0,1]} (x-) dt = meausure ([0,1 ] \cap [x-1,x]) $$

Answer 1

Hint 1: Don't think too hard.

Hint 2: Just write down the integral and compute.

Hint 3: Remember that $\int_\mathbb{R} \chi_S(x) f(x)\ dx = \int_S f(x)\ dx$.

Answer 2

For $t \le 1,$

$$ \int^t_0 f(t-\tau)g(\tau) d(\tau)= \int^t_0 1\times 1 d\tau = \int^t_0 1 d \tau =t$$

For $t>1$,

$$ \int^t_0 f(t-\tau)g(\tau) d(\tau)= \int^1_0 f(t-\tau)g(\tau) d(\tau) + \int^t_1 f(t-\tau)g(\tau) d(\tau)= $$

$$1+\int^t_1 f(t-\tau)g(\tau) d(\tau)=1+\int^t_1 f(t-\tau)\times (0) d(\tau)= 1$$