Associativity of L1 Convolution by Fubini

Why Associativity Requires a Product-Space Estimate

For integrable functions, convolution combines two functions by integrating their translated product. Associativity says that three functions can be combined in either grouping. The formal expressions involve two integrations, and changing their order is not automatic: the integrand can have both signs or be complex-valued. The essential step is to prove that the absolute value of the three-factor integrand is integrable on the relevant product space. Fubini’s theorem then applies, and the two groupings agree almost everywhere.

Throughout, the underlying measure space is $(\mathbb{R},\mathcal{L},m)$, with Lebesgue measure; it is complete and sigma-finite. The product spaces use the corresponding product Lebesgue measures. Functions $f,g,h$ are real- or complex-valued measurable representatives of elements of $L^1(m)$, so $\|f\|_1=\int_{\mathbb{R}}|f|\,dm<\infty$, and likewise for $g,h$. Statements about $L^1$ functions are statements about equivalence classes modulo equality $m$-a.e.; a chosen representative can have different values on a null set.

Definition: For $f,g\in L^1(m)$, their convolution is represented, wherever the integral is absolutely convergent, by $$ (f*g)(x)=\int_{\mathbb{R}} f(y)g(x-y)\,dy. $$ The integral is absolutely convergent for $m$-a.e. $x$, and $f*g$ defines an element of $L^1(m)$. We may assign any finite value on the exceptional null set; such a choice does not change the $L^1$ element.

The existence and estimate in this definition follow from Tonelli’s theorem applied to the nonnegative measurable function $(x,y)\mapsto |f(y)||g(x-y)|$: translation invariance of Lebesgue measure gives $\int_{\mathbb{R}^2}|f(y)||g(x-y)|\,dy\,dx=\|f\|_1\|g\|_1$. Consequently, the inner integral is finite for a.e. $x$, and the triangle inequality and Tonelli give $\|f*g\|_1\le\|f\|_1\|g\|_1$.

The Triple-Product Estimate and Associativity

Theorem: If $f,g,h\in L^1(m)$ are real- or complex-valued on $(\mathbb{R},\mathcal{L},m)$, then both $(f*g)*h$ and $f*(g*h)$ belong to $L^1(m)$, and $$ (f*g)*h=f*(g*h)\qquad m\text{-a.e.} $$ In particular, both sides are defined as $L^1$ equivalence classes. The theorem does not assert equality at every point for arbitrary representatives.

Proof. Consider the measurable nonnegative function $|f(y)||g(w)||h(x-y-w)|$ on $\mathbb{R}^3$. Tonelli’s theorem applies without an integrability assumption because the integrand is nonnegative. Integrating first in $x$, and using translation invariance, gives

\begin{aligned} \int_{\mathbb{R}^3}|f(y)||g(w)||h(x-y-w)|\,dx\,dw\,dy &=\int_{\mathbb{R}^2}|f(y)||g(w)| \left(\int_{\mathbb{R}}|h(x-y-w)|\,dx\right)dw\,dy\\ &=\int_{\mathbb{R}^2}|f(y)||g(w)|\,\|h\|_1\,dw\,dy\\ &=\|f\|_1\|g\|_1\|h\|_1<\infty. \end{aligned}

This computation also covers a zero norm: if any of the three norms is zero, the displayed integral is zero. By Tonelli, the section integral $\int_{\mathbb{R}^2}|f(y)||g(w)||h(x-y-w)|\,dw\,dy$ is finite for $m$-a.e. $x$. Fix any such $x$. Fubini’s theorem now applies to the corresponding complex- or real-valued integrand on $\mathbb{R}^2$, since its absolute integral is finite. It permits changing the order of the $y,w$ integrations.

For this $x$, the change of variable $z=y+w$ in the inner integral is a translation in $w$, so it preserves Lebesgue measure. Thus the absolutely convergent double integral can be written in either form:

\begin{aligned} \int_{\mathbb{R}}\int_{\mathbb{R}} f(y)g(z-y)h(x-z)\,dy\,dz &=\int_{\mathbb{R}}\int_{\mathbb{R}} f(y)g(w)h(x-y-w)\,dw\,dy. \end{aligned}

To identify the first expression with $(f*g)*h$, Fubini’s theorem for the pair $f,g$ gives $f*g(z)=\int f(y)g(z-y)\,dy$ for a.e. $z$. For the fixed $x$ under consideration, the absolute double integral in $(y,z)$ is finite: it equals the absolute integral just considered after the translation $w=z-y$. Therefore the inner integral exists for a.e. $z$, and its product with $h(x-z)$ is integrable in $z$. Replacing the inner integral by the representative of $f*g$ changes it only on a null set of $z$, which does not change the outer integral. Hence the first expression is $\int_{\mathbb{R}}(f*g)(z)h(x-z)\,dz$ for a.e. $x$.

Likewise, the second expression is $\int_{\mathbb{R}} f(y)(g*h)(x-y)\,dy$ for a.e. $x$. Indeed, for a.e. $x-y$, the inner integral defining $g*h$ equals $\int g(w)h(x-y-w)\,dw$; Fubini and the finite absolute double integral justify this identification in the outer integral. The exceptional sets in these identifications are null sets in the corresponding integration variables and do not affect the integrals. The two grouped convolutions therefore agree for a.e. $x$. Each is in $L^1(m)$ by the pairwise convolution estimate, applied twice. This proves the theorem. $\square$

Worked Calculations

Worked Example: Three Interval Indicators

Let $f=g=h=\chi_{[0,1]}$ on $\mathbb{R}$. First, $f*g$ is the length of the overlap of $[0,1]$ with $[x-1,x]$. That overlap has length $x$ when $0\le x\le1$, length $2-x$ when $1\le x\le2$, and length zero outside $[0,2]$. Thus, apart from endpoint values that do not affect later integrals, $(f*g)(u)=u$ for $0<u<1$, $(f*g)(u)=2-u$ for $1\le u<2$, and it is zero outside $[0,2]$.

Convolving once more with $\chi_{[0,1]}$ gives $\int_0^1(f*g)(x-t)\,dt$. For $0\le x\le1$, this is $\int_0^x u\,du=x^2/2$. For $1\le x\le2$, split the integral at $u=1$: \[ \int_{x-1}^1u\,du+\int_1^x(2-u)\,du =\frac{1-(x-1)^2}{2}+2x-\frac{x^2}{2}-\frac32 =3x-x^2-\frac32. \] For $2\le x\le3$, only the second part of $f*g$ contributes: \[ \int_{x-1}^2(2-u)\,du =2x-\frac{x^2}{2}-\left(2(x-1)-\frac{(x-1)^2}{2}\right) =\frac{x^2}{2}-3x+\frac92. \] The result is zero outside $[0,3]$.

Since $g=h$, the other grouping has the same two integrations, and the theorem also guarantees the grouping is valid. As a check, the three pieces have values $1/2$ at $x=1$, $1/2$ at $x=2$, and zero at $x=0,3$, matching at the breakpoints. The total integral is $\int_0^1x^2/2\,dx+\int_1^2(3x-x^2-3/2)\,dx+\int_2^3(x^2/2-3x+9/2)\,dx =1/6+2/3+1/6=1$. This equals $\|f\|_1\|g\|_1\|h\|_1=1$, as also follows by Tonelli.

Worked Example: Three One-Sided Exponentials

Take $f(x)=e^{-x}\chi_{[0,\infty)}(x)$, $g(x)=e^{-2x}\chi_{[0,\infty)}(x)$, and $h(x)=e^{-3x}\chi_{[0,\infty)}(x)$. Their norms are $1$, $1/2$, and $1/3$, respectively, so all three are in $L^1(m)$. For $x\ge0$, \[ (f*g)(x)=\int_0^x e^{-y}e^{-2(x-y)}\,dy =e^{-2x}\int_0^xe^y\,dy=e^{-x}-e^{-2x}; \] for $x<0$, it is zero.

For $x\ge0$, the first grouping is \[ \begin{aligned} ((f*g)*h)(x) &=\int_0^x(e^{-(x-t)}-e^{-2(x-t)})e^{-3t}\,dt\\ &=e^{-x}\int_0^xe^{-2t}\,dt-e^{-2x}\int_0^xe^{-t}\,dt\\ &=\frac12e^{-x}(1-e^{-2x})-e^{-2x}(1-e^{-x})\\ &=\frac12e^{-x}-e^{-2x}+\frac12e^{-3x}. \end{aligned} \] Also, for $x\ge0$, \[ (g*h)(x)=e^{-3x}\int_0^xe^t\,dt=e^{-2x}-e^{-3x}. \] Consequently, \[ \begin{aligned} (f*(g*h))(x) &=\int_0^xe^{-y}\big(e^{-2(x-y)}-e^{-3(x-y)}\big)\,dy\\ &=e^{-2x}(e^x-1)-e^{-3x}\frac{e^{2x}-1}{2}\\ &=\frac12e^{-x}-e^{-2x}+\frac12e^{-3x}. \end{aligned} \] Both convolutions vanish for $x<0$. The expressions agree, and their common integral is $\tfrac12-\tfrac12+\tfrac16=\tfrac16$, which matches the product of the three $L^1$ norms.

Worked Example: A Signed Function and a Shift

Let $q=\chi_{[0,1]}*\chi_{[0,1]}*\chi_{[0,1]}$, computed in the first example, and set $f=\chi_{[0,1]}-\chi_{[1,2]}$, $g=h=\chi_{[0,1]}$. By linearity of integration and translation in the convolution variable, $(f*g)*h=q(x)-q(x-1)$ a.e. For $0\le x\le1$, the value is $x^2/2$. For $1\le x\le2$, substitute the two relevant formulas for $q$: \[ 3x-x^2-\frac32-\frac{(x-1)^2}{2} =-\frac32x^2+4x-2. \] For $2\le x\le3$, it is \[ \left(\frac{x^2}{2}-3x+\frac92\right) -\left(3(x-1)-(x-1)^2-\frac32\right) =\frac32x^2-8x+10. \] For $3\le x\le4$, $q(x)=0$, so the value is \[ -\left(\frac{(x-1)^2}{2}-3(x-1)+\frac92\right) =-\frac{x^2}{2}+4x-8. \] It is zero outside $[0,4]$. This signed example has positive and negative values, so Tonelli cannot be applied directly to its signed integrand. The theorem applies because its absolute value is bounded by the corresponding triple product formed from $|f|,|g|,|h|$, whose integral is $\|f\|_1\|g\|_1\|h\|_1=2$. As a check, the integral of the output is zero: it is $\int q(x)\,dx-\int q(x-1)\,dx=1-1=0$, agreeing with $\int f\,dm=0$ and the product-of-integrals identity justified by Fubini.

Almost-Everywhere Equality and a Common Pitfall

The triple-product estimate gives finite absolute section integrals for a.e. $x$, not necessarily for every $x$. It is therefore not justified to claim that the Fubini calculation works pointwise everywhere. There is a second reason to formulate associativity a.e.: an $L^1$ function is an equivalence class, and changing its representative on a null set can change selected point values without changing the $L^1$ element.

Common mistake. After proving two grouped convolutions represent the same $L^1$ function, one might conclude that any selected representatives agree at every $x$. For example, with all three functions equal to $\chi_{[0,1]}$, the first example gives a representative $q$ with $q(0)=0$. The function $\widetilde q$ defined by $\widetilde q(0)=7$ and $\widetilde q(x)=q(x)$ for $x\ne0$ is the same $L^1$ element, since $\{0\}$ is null. Yet $\widetilde q(0)\ne q(0)$. The correction is to assert equality of the $L^1$ elements, or equality $m$-a.e. for representatives. The absolute-integrability argument proves precisely that, without requiring the section integrals to be finite at every $x$.

Check Your Understanding

In each question, keep track of which statements concern representatives and which concern $L^1$ equivalence classes.

For $f,g,h\in L^1(m)$, write the triple-product absolute-integrability identity used to prove associativity.
Why does finiteness of the triple integral imply that the relevant two-dimensional section integral is finite for a.e. $x$, rather than necessarily for every $x$?
For the one-sided exponential functions in the second example, compute $(g*h)(x)$ for $x\ge0$ and verify its $L^1$ norm.
Explain why changing an $L^1$ representative at one point cannot affect its equivalence class, even though it changes a pointwise value.
In the signed example, why is the absolute-integrability estimate needed before changing the order of integration?

Associativity of L1 Convolution by Fubini

What You'll Learn