Why Associativity Requires a Justification
Convolution combines two functions by integrating over all ways to split an input into two parts. It is natural to expect that combining three functions should not depend on which pair is combined first. But this intuition alone does not permit changing the order of integration: that step requires an integrability condition. For \(L^1(\mathbb{R})\) functions, the needed condition is supplied by a bound on the absolute value of the relevant triple integrand.
Throughout, the measure space is \((\mathbb{R},\mathcal{L},m)\), with Lebesgue measure, which is complete and sigma-finite. Functions may be real-valued or complex-valued and are identified when they agree \(m\)-a.e. The space \(L^1(\mathbb{R})\) consists of measurable functions \(f\) for which \(\|f\|_1=\int_{\mathbb{R}}|f|\,dm<\infty\). All convolution identities below are identities \(m\)-a.e.; no pointwise identity at every \(x\) is asserted.
Here and below, measurable representatives may be chosen finite everywhere by changing their values on null sets. For clarity, the existence and integrability in the definition follow from Tonelli’s theorem for the nonnegative measurable integrand \(|f(y)||g(x-y)|\): translation invariance of Lebesgue measure gives
This quantity is finite. Tonelli’s theorem therefore implies that the inner integral is finite for \(m\)-a.e. \(x\); the triangle inequality for integrals and Tonelli’s theorem also give \(\|f*g\|_1\le\|f\|_1\|g\|_1\). These facts ensure that both iterated convolutions in the associativity statement are defined as \(L^1\) functions.
The Absolute-Integrability Bound
Fix \(f,g,h\in L^1(\mathbb{R})\). For each \(x\), the integrand relevant to associativity is \(f(y)g(z-y)h(x-z)\). Its three arguments are linked, so its factors cannot simply be integrated separately. The following estimate establishes the precise integrability needed.
Proof. Choose Borel measurable representatives of \(f,g,h\); replacing the original representatives by these does not change their \(L^1\) classes or the convolutions \(m\)-a.e. The displayed integrand is then measurable on \(\mathbb{R}^3\). Tonelli’s theorem applies because it is nonnegative. First integrate in \(x\). Translation invariance gives \(\int_{\mathbb{R}}|h(x-z)|\,dx=\|h\|_1\) for every \(z\). Then integrate in \(z\); for every \(y\), \(\int_{\mathbb{R}}|g(z-y)|\,dz=\|g\|_1\). Thus the triple integral is $$ \int_{\mathbb{R}}|f(y)|\|g\|_1\|h\|_1\,dy =\|f\|_1\|g\|_1\|h\|_1, $$ which is finite. Applying Tonelli’s theorem to the outer integral in \(x\) shows that the \((y,z)\) integral is finite for \(m\)-a.e. \(x\). \(\square\)
Associativity of L1 Convolution
Proof. By the lemma, there is a set \(X_0\in\mathcal{L}\) with \(m(\mathbb{R}\setminus X_0)=0\) such that, for each \(x\in X_0\), $$ \int_{\mathbb{R}}\int_{\mathbb{R}}|f(y)g(z-y)h(x-z)|\,dy\,dz<\infty. $$ For such an \(x\), Fubini’s theorem permits either order of integration in \(y,z\).
First consider \((f*g)*h\). The definition of \(f*g\) holds for \(m\)-a.e. \(z\). For fixed \(x\), its exceptional set in \(z\) is null, so it does not affect the integral against \(h(x-z)\). The absolute integrability just established allows Fubini’s theorem to be applied, giving $$ ((f*g)*h)(x) =\int_{\mathbb{R}}\int_{\mathbb{R}} f(y)g(z-y)h(x-z)\,dy\,dz =\int_{\mathbb{R}} f(y)\left(\int_{\mathbb{R}}g(z-y)h(x-z)\,dz\right)dy. $$ In the inner integral set \(t=z-y\), so \(z=t+y\) and \(x-z=(x-y)-t\). Translation invariance gives $$ \int_{\mathbb{R}}g(z-y)h(x-z)\,dz =\int_{\mathbb{R}}g(t)h((x-y)-t)\,dt =(g*h)(x-y) $$ whenever this convolution integral is defined.
To see that this substitution identifies the integrand for \(m\)-a.e. \(y\), note that \(g*h\) is defined for \(m\)-a.e. argument by the \(L^1\) convolution bound. For fixed \(x\), the set of \(y\) for which \(x-y\) belongs to its exceptional null set is also null, by translation and reflection invariance of Lebesgue measure. The outer integral is unaffected by such a set. Hence \(((f*g)*h)(x)=\int_{\mathbb{R}}f(y)(g*h)(x-y)\,dy=(f*(g*h))(x)\) for \(m\)-a.e. \(x\). This proves the asserted \(L^1\) identity. \(\square\)
Worked Examples
Worked Example: Three Interval Indicators
Let \(f=g=h=\chi_{[0,1]}\) on \(\mathbb{R}\). Compute both groupings at \(x=\tfrac32\).
The first convolution is \(q=f*g\). For \(0\le t\le1\), the set of \(y\in[0,1]\) with \(t-y\in[0,1]\) has length \(t\). For \(1\le t\le2\), its length is \(2-t\), and outside \([0,2]\) its length is zero. Thus, including the endpoints where the value is immaterial to the integrals, $$ q(t)= \begin{cases} t,&0\le t\le1,\\ 2-t,&1\le t\le2,\\ 0,&t<0\text{ or }t>2. \end{cases} $$ Since \(h(\tfrac32-y)=1\) exactly when \(y\in[\tfrac12,\tfrac32]\), up to endpoints, we obtain $$ ((f*g)*h)\left(\tfrac32\right) =\int_{1/2}^{3/2}q(y)\,dy =\int_{1/2}^{1}y\,dy+\int_{1}^{3/2}(2-y)\,dy =\frac{3}{8}+\frac{3}{8} =\frac{3}{4}. $$ The other grouping is the same integral because \(g=h\): \(f*(g*h)=q*h\), so its value at \(\tfrac32\) is also \(\tfrac34\).
Check: In the triple integral, \(u,v\in[0,1]\) and the third variable \(\tfrac32-u-v\) lies in \([0,1]\) exactly when \(\tfrac12\le u+v\le\tfrac32\). The excluded regions in the unit square are two right triangles, each of area \(\tfrac12(\tfrac12)^2=\tfrac18\). The remaining area is \(1-\tfrac18-\tfrac18=\tfrac34\), agreeing with both groupings.
Worked Example: Three One-Sided Exponentials
Let \(f(x)=g(x)=h(x)=e^{-x}\chi_{[0,\infty)}(x)\). Each function is nonnegative and has \(L^1\) norm \(\int_0^\infty e^{-x}\,dx=1\). Compute the two groupings at \(x=2\).
For \(t\ge0\), the first convolution is $$ (f*g)(t)=\int_0^t e^{-y}e^{-(t-y)}\,dy =e^{-t}\int_0^t1\,dy =te^{-t}; $$ for \(t<0\) it is zero. Therefore $$ ((f*g)*h)(2) =\int_0^2 ye^{-y}e^{-(2-y)}\,dy =e^{-2}\int_0^2y\,dy =e^{-2}\left[\frac{y^2}{2}\right]_0^2 =2e^{-2}. $$ Since \(g=h\), the same calculation gives \((f*(g*h))(2)=2e^{-2}\).
Check: In the triple integral at \(x=2\), nonzero contributions require \(0\le y\le z\le2\). On this triangle the product is \(e^{-y}e^{-(z-y)}e^{-(2-z)}=e^{-2}\). The triangle has area \(\tfrac12(2)(2)=2\), so its integral is \(2e^{-2}\), as computed.
Worked Example: A Shifted Interval
Let \(f=g=\chi_{[0,1]}\) and \(h=\chi_{[2,3]}\). Compute the two groupings at \(x=3\), keeping track of the shift.
As above, \(q=f*g\) satisfies \(q(t)=t\) for \(0\le t\le1\). For the first grouping, the condition \(3-z\in[2,3]\) is equivalent to \(z\in[0,1]\). Hence $$ ((f*g)*h)(3)=\int_0^1q(z)\,dz =\int_0^1z\,dz=\frac12. $$ For the second grouping, the substitution \(s=t-2\) in the convolution integral shows \((g*h)(t)=q(t-2)\). Therefore $$ (f*(g*h))(3)=\int_0^1q(3-y-2)\,dy =\int_0^1q(1-y)\,dy =\int_0^1(1-y)\,dy =\frac12, $$ because \(1-y\in[0,1]\).
Check: In the triple integral, the first two interval variables \(u,v\) lie in \([0,1]\), and the third variable \(3-u-v\) lies in \([2,3]\) exactly when \(u+v\le1\). This region is a right triangle of area \(\tfrac12(1)(1)=\tfrac12\), confirming the result.
A Common Pitfall: Separating Coupled Factors
Even when the triple integral is absolutely integrable, its factors are not independent: \(f(y)\), \(g(z-y)\), and \(h(x-z)\) depend on linked variables. For the exponential example at \(x=2\), an incorrect calculation might replace the triple integral by the product of the three \(L^1\) norms and claim its value is \(1\). That calculation drops the constraints on the arguments and is false. The corrected computation restricts the nonzero integrand to \(0\le y\le z\le2\); there it equals \(e^{-2}\), and integration over the triangle gives \(2e^{-2}\), not \(1\).
The essential distinction is that Fubini’s theorem allows a change in the order of integration when absolute integrability has been established; it does not allow the variables inside the factors to be separated. Here the lemma supplies that absolute-integrability condition for every triple of \(L^1\) functions. It also explains why the theorem concludes equality almost everywhere: the defining convolution integrals may fail on null sets, and choosing different values there does not change the \(L^1\) functions.
Check Your Understanding
In each question, identify the integrability condition or the relevant variables before computing.
- For \(f=g=h=\chi_{[0,1]}\), compute \((f*g)*h(1)\) using the piecewise formula for \(f*g\).
- Explain why \(\int_{\mathbb{R}}|f(y)||g(z-y)||h(x-z)|\,dy\,dz\,dx\) is finite when all three functions belong to \(L^1(\mathbb{R})\).
- For the one-sided exponential example, compute the common value of the two groupings at \(x=1\).
- Why does a null exceptional set for the definition of \(g*h\) remain harmless when its argument is \(x-y\) for a fixed \(x\)?
- State precisely whether associativity asserts equality everywhere or almost everywhere, and explain why arbitrary values on exceptional null sets do not alter the \(L^1\) identity.