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Supposing $f,g, h$ are defined on $\Bbb R$ and are all integrable on $\Bbb R$. Define their convolution as $f * g (z) = \int_{\Bbb R} f(x) g (z-x) dx$. Now I wish to show that this operation is associative, i.e. that $(f*(g*h))(x)=((f*g)*h)(x)$.

I'm not even sure what the left side of the equation should be. Is it that :

$(f*(g*h))(z)=\int_{\Bbb R}f(x) \cdot (\int_{\Bbb R} g(z-y)h(z-(z-y)) dy)dx$ ?

Hints and insights appreciated.

IntegrateThis
  • 3,778
  • The variable inside the second integral is different, but yes, that is how the expression looks. Now, go for a change of variable, and use Fubini's theorem. – Sarvesh Ravichandran Iyer Feb 22 '18 at 05:58
  • @астонвіллаолофмэллбэрг it should be $dz$ ? – IntegrateThis Feb 22 '18 at 06:01
  • See, you are performing integration over two copies of the real numbers. z is some fixed number, while x is one of the variables spanning one copy of the real numbers. A variable like y is used to span the other integral. That is, inside the inner integral, replace x with y, and you are good to go. – Sarvesh Ravichandran Iyer Feb 22 '18 at 06:04
  • Yes, I see the change you have made, and now it correct. Now, write down the expression for the other side of associativity, and see what you need to move around, which variable changes to make etc. and justify switching integrals etc. through Fubini. – Sarvesh Ravichandran Iyer Feb 22 '18 at 06:10

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