# Question about why Fubini's theorem can be applied to show convolution is associative

I am trying to understand the proof that convolution is associative and the proof requires Fubini's theorem. Lets assume that f,g, and h are in $$L^1(R)$$. The proof goes as follows:

Now the version of Fubini's theorem I am using is:

So my problem with the application of Fubini's theorem is that it doesn't seem like the conditions are satisfied. Fubini's theorem states that if we have a multivariable function and the multivariable integral exist, then is is equal to the iterated integral. But here, for the proof that convolution is associated, we don't have the assumption that the multivariable function is interable (actually do we even have a multivariable function? The convolution (f*g)*h is just a single variable function of $$u$$ correct?) , but we are assuming in the third line of the proof that the iterated integral with the order $$dydx$$ exist and is equal to the integral with order $$dxdy$$. So is the Fubini's theorem in this proof some other version than the one I am using?

actually do we even have a multivariable function? The convolution $$(f*g)*h$$ is just a single variable function of u correct?

What matters is not how many variables the convolution is a function of, but rather how many variables are involved in the function you are integrating over. Indeed, for fixed $$u\in \mathbb{R}$$ you can define a function

$$\phi_u : \mathbb{R}\times\mathbb{R}\to\mathbb{R}, \quad (x,y)\mapsto f(y)g(x-y)h(u-x).$$

The integral you are dealing with and want to apply Fubini to is then

$$\int_{\mathbb{R}^2}\phi_u\,\mathrm{d}\lambda.$$

Now as $$f,g,h\in L^1(\mathbb{R})$$ you also have that $$\phi_u$$ is integrable over $$\mathbb{R}^2$$ and hence satisfies the assumption of Fubini's theorem. This means that we can apply Fubini and get that

\begin{align*} ((f*g)*h)(u)&=\int_\mathbb{R}\left(\int_\mathbb{R}f(y)g(x-y)\,\mathrm{d}\lambda(y)\right)h(u-x)\,\mathrm{d}\lambda(x) \\ &=\int_\mathbb{R}\left(\int_\mathbb{R}f(y)g(x-y)h(u-x)\,\mathrm{d}\lambda(y)\right)\,\mathrm{d}\lambda(x) \\ &=\int_\mathbb{R}\left(\int_\mathbb{R}\phi_u(x,y)\,\mathrm{d}\lambda(y)\right)\,\mathrm{d}\lambda(x) \\ &=\int_\mathbb{R}\left(\int_\mathbb{R}\phi_u(x,y)\,\mathrm{d}\lambda(x)\right)\,\mathrm{d}\lambda(y) \\ &=\int_\mathbb{R}\left(\int_\mathbb{R}f(y)g(x-y)h(u-x)\,\mathrm{d}\lambda(x)\right)\,\mathrm{d}\lambda(y) \\ &=\int_\mathbb{R}f(y)\left(\int_\mathbb{R}g(x-y)h(u-x)\,\mathrm{d}\lambda(x)\right)\,\mathrm{d}\lambda(y) \end{align*}

which is the application of Fubini you were wondering about.

• thank you for the reply. It makes a lot of sense now. Why is it true that: "Now as f,g,h∈L1(R) you also have that $ϕ_u$ is integrable over $R^2$"? I remember there is a similar theorem that says if f(x) is integrable over $R$, then f(x-y) is integrable over $R^2$.
– Bill
Mar 10, 2022 at 13:50

If I understand correctly, the fact that $$(f*g)*h$$ exists means that the third integral is finite and hence you can apply Fubini's. So we did not even use $$f,g,h\in L^1$$ if I am not mistaken.

As for variables, we need to look at variables of integration. So we can think of $$f(y)g(x-y)h(u-x)$$ as some function $$k(x,y)$$ which is integrable.