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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.

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  • $\begingroup$ why does this have anything to do with $L^p$ spaces? $\endgroup$ – Calvin Khor May 24 at 20:56
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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 $$ f\star g(x) = \int_{x-4}^{2} 3(x-y)^2\ \mathsf dy = (-x+10)(x^2-8x+28). $$

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

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