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Why are the limits of the convolution of two positive functions $0$ and $t$?

$$(f \star g)(t) = \int_0^t f(x)g(t-x)\,dx \text{ for } f,g : [0,\infty) \rightarrow \mathbb{R}$$

I understand the lower limit of 0 because both are non-negative functions, but why $t$?

Thank you!

WiPU
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    You have your convolution slightly ajar. It should read $g(t-x)$, hopefully that answers your question. – Ninad Munshi Dec 15 '20 at 17:05
  • See my edits for proper MathJax usage. In particular in expressions like $f,dy,dx$ the expressions like $dy$ and $dx$ are separated from their surroundings by a small space (likewise with $\Delta x$ in $f(x_i),\Delta x_i$ and in the denominator of the expression $\dfrac{\partial^2f}{\partial x, \partial y},$ etc.) and \text{} is used with the word "for" between the integral and $f,g. \qquad$ – Michael Hardy Dec 15 '20 at 17:09
  • @NinadMunshi: Yes, you are right. I copied the expression from Wikipedia, and it seems as I have made an error while switching tabs. No, the questions would be, then why not $\infty$ instead of $t$? I really can't wrap my head around convolution limits. – WiPU Dec 15 '20 at 17:49
  • @MichaelHardy: Thank you for your corrections and especially for the comment. It really helps me to get better with my expressions! Until now I wasn't aware of , – WiPU Dec 15 '20 at 17:49
  • This isn't about positive functions; it's about functions whose domain is the set of positive numbers. The values of the function don't have to be positive. – Michael Hardy Dec 15 '20 at 20:42

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Consider the integrand term $f(x)g(t-x)$. If $x<0$, then $f(x)=0$. If $x>t$, $g(t-x)=0$. Thus for a fixed $t$, $f(x)g(t-x)$ is 0 outside the interval $x\in [0,t]$.

Alex R.
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