# Compute convolution $f*g(x)$ such that : $f,g\in L^{p}$

Question :

Compute convolution : $$f*g(x)$$

$$f(x)=\begin{cases}3x^2 & \text{ if } |x|\leq4 \\0 & \text{ otherwise}\end{cases}$$

$$g(x)=\begin{cases}1 &\text{ if }|x|\leq 2 \\0 & \text{ otherwise}\end{cases}$$

My try : \begin{align} f*g(x)&=\int f(x-y)g(y)dy\\ &=\int_{[-4,4]}3(x-y)^{2}1_{[-2,2]}(x-y)dy\\ &=\int_{[-4,4]∩[x-2,x+2]}3(x-y)^{2}dy \end{align}

Now, how I find or discussed with this?

$$[-4,4]\cap[x-2,x+2]=?$$

Please, give me ideas and method to approach it.

• why does this have anything to do with $L^p$ spaces? – Calvin Khor May 24 at 20:56

## 1 Answer

Your bounds aren't quite right. We have $$f(x-y) = 3(x-y)^2\cdot\mathsf 1_{[-4,4]}(x-y)$$ and $$g(y) = \mathsf 1_{[-2,2]}(y)$$, so we have the following inequalities for $$y$$: \begin{align} x-4&\leqslant y\leqslant x+4\\ -2&\leqslant y\leqslant 2. \end{align} For $$-6\leqslant x\leqslant -2$$ we have $$f\star g(x) = \int_{-2}^{x+4} 3(x-y)^2\ \mathsf dy = (x+6)(x^2+12).$$ For $$-2\leqslant x\leqslant 2$$ we have $$f\star g(x) = \int_{-2}^2 3(x-y)^2\ \mathsf dy = 12x^2+16.$$ For $$2\leqslant x\leqslant 6$$ we have $$f\star g(x) = \int_{x-4}^{2} 3(x-y)^2\ \mathsf dy = (-x+10)(x^2-8x+28).$$

• Thank you very much Sir , just the last line $\int_{x-4}^{2}$ – Kînan Jœd May 25 at 2:22
• Indeed, fixed. $\$ – Math1000 May 25 at 17:26