Fubini Identities for Convolutions

Convolution Integrals and Fubini

Convolution turns two functions into a third by integrating over all ways to write an input as a sum. Fubini’s theorem then relates the integral of that third function to the integrals of the original functions. The same idea permits an exchange with an integral against a test function, but that exchange requires an integrability condition: three functions in $L^1$ alone do not guarantee that the resulting pairing is finite.

Throughout, the underlying space is $(\mathbb{R},\mathcal{L},m)$, where $m$ is Lebesgue measure. This measure space is sigma-finite and complete, but not finite. Functions may be real- or complex-valued. The space $L^1(m)$ consists of measurable functions $f$ for which $\int_{\mathbb{R}}|f|\,dm<\infty$, with functions equal $m$-a.e. identified. For $f,g\in L^1(m)$, define their convolution, wherever the integral exists, by

(f*g)(x)=\int_{\mathbb{R}} f(y)g(x-y)\,dy.

All integrals below are Lebesgue integrals. When an identity involves $f*g$ as an $L^1$ function, it is an equality of integrals and the convolution is understood up to equality $m$-a.e. We use translation invariance of Lebesgue measure and the Tonelli and Fubini theorems established earlier in the course. For measurable representatives, the relevant functions on $\mathbb{R}^2$ are product-measurable; equivalently, one may first choose Borel representatives of the Lebesgue-measurable $L^1$ functions. Changing representatives does not affect the resulting identities a.e.

The Integral of a Convolution

Theorem: If $f,g\in L^1(m)$, then for $m$-a.e. $x$ the integral defining $(f*g)(x)$ is absolutely convergent, $f*g\in L^1(m)$, and $$ \int_{\mathbb{R}}(f*g)(x)\,dx = \left(\int_{\mathbb{R}}f(y)\,dy\right) \left(\int_{\mathbb{R}}g(z)\,dz\right). $$ The functions may be complex-valued; the displayed integrals are then complex integrals.

Proof. Consider the nonnegative measurable function $(x,y)\mapsto |f(y)||g(x-y)|$ on $\mathbb{R}^2$. Tonelli’s theorem applies without assuming its integral is finite. For each fixed $y$, translation invariance, with $z=x-y$, gives $\int_{\mathbb{R}}|g(x-y)|\,dx=\int_{\mathbb{R}}|g(z)|\,dz$. Therefore

\int_{\mathbb{R}}\int_{\mathbb{R}}|f(y)||g(x-y)|\,dy\,dx = \int_{\mathbb{R}}|f(y)|\,dy\int_{\mathbb{R}}|g(z)|\,dz = \|f\|_1\|g\|_1<\infty.

Tonelli’s theorem also identifies the inner integral $\int |f(y)||g(x-y)|\,dy$ as finite for $m$-a.e. $x$. At every such $x$, the convolution integral is absolutely convergent and $|(f*g)(x)|\le\int |f(y)||g(x-y)|\,dy$. Integrating this inequality and using the displayed calculation proves $\int|f*g|\,dx\le\|f\|_1\|g\|_1$, so $f*g\in L^1(m)$. Finally, the absolute integrability just proved checks the hypothesis of Fubini’s theorem for the complex- or real-valued function $f(y)g(x-y)$. Thus

\begin{aligned} \int_{\mathbb{R}}(f*g)(x)\,dx &=\int_{\mathbb{R}}\int_{\mathbb{R}}f(y)g(x-y)\,dy\,dx\\ &=\int_{\mathbb{R}} f(y)\left(\int_{\mathbb{R}}g(x-y)\,dx\right)dy\\ &=\left(\int_{\mathbb{R}}f(y)\,dy\right) \left(\int_{\mathbb{R}}g(z)\,dz\right). \end{aligned}

The last equality again uses $z=x-y$ and translation invariance. $\square$

Worked Example: Convolving Two Indicator Functions

Let $f=g=\chi_{[0,1]}$ on $\mathbb{R}$. Both functions belong to $L^1(m)$, and we compute their convolution for each $x$.

The integrand $f(y)g(x-y)$ equals $1$ exactly when both $0\le y\le1$ and $0\le x-y\le1$. The second pair of inequalities is equivalent to $x-1\le y\le x$. Thus the convolution is the length of $[0,1]\cap[x-1,x]$. For $0\le x\le1$, the intersection is $[0,x]$, of length $x$. For $1\le x\le2$, it is $[x-1,1]$, of length $2-x$. Outside $[0,2]$ the intersection has length zero. Hence

(f*g)(x)= \begin{cases} x,&0\le x\le1,\\ 2-x,&1\le x\le2,\\ 0,&x<0\text{ or }x>2. \end{cases}

At $x=1$, both formulas give $1$, so there is no conflict at the shared endpoint. Integrating the result gives $\int_0^1x\,dx+\int_1^2(2-x)\,dx=\tfrac12+\tfrac12=1$. The theorem gives the same value: $(\int\chi_{[0,1]}\,dx)^2=1\cdot1=1$. This independently checks the computation.

Exchanging Convolution with a Test-Function Integral

A test function in the next result means a bounded measurable function $h:\mathbb{R}\to\mathbb{C}$; it need not be integrable. The boundedness condition provides a direct absolute-integrability estimate. The theorem makes explicit both the double-integral identity and the change of variables that shifts the test function.

Theorem: Let $f,g\in L^1(m)$, and let $h:\mathbb{R}\to\mathbb{C}$ be bounded and measurable. Then $(f*g)h\in L^1(m)$, the double integral below is absolutely convergent, and $$ \int_{\mathbb{R}}(f*g)(x)h(x)\,dx = \int_{\mathbb{R}}\int_{\mathbb{R}} f(y)g(z)h(y+z)\,dz\,dy. $$ The equality is an equality of complex numbers (or real numbers for real-valued functions); no complex conjugate is inserted.

Proof. Write $H=\sup_{x\in\mathbb{R}}|h(x)|<\infty$. Tonelli’s theorem and translation invariance give

\begin{aligned} \int_{\mathbb{R}}\int_{\mathbb{R}} |f(y)||g(x-y)||h(x)|\,dy\,dx &\le H\int_{\mathbb{R}}\int_{\mathbb{R}}|f(y)||g(x-y)|\,dy\,dx\\ &=H\|f\|_1\|g\|_1<\infty. \end{aligned}

This estimate checks absolute integrability, so Fubini’s theorem permits exchanging the $x$- and $y$-integrals. It also implies $\int|(f*g)(x)h(x)|\,dx\le H\|f\|_1\|g\|_1$, using the convolution bound from the first theorem. Consequently

\begin{aligned} \int_{\mathbb{R}}(f*g)(x)h(x)\,dx &=\int_{\mathbb{R}}\int_{\mathbb{R}}f(y)g(x-y)h(x)\,dy\,dx\\ &=\int_{\mathbb{R}} f(y)\left(\int_{\mathbb{R}}g(x-y)h(x)\,dx\right)dy. \end{aligned}

For each fixed $y$, substitute $z=x-y$, so $x=y+z$ and $dx=dz$. The inner integral becomes $\int_{\mathbb{R}}g(z)h(y+z)\,dz$. This proves the stated identity, with the order and substitution justified by the absolute-integrability estimate. $\square$

Worked Example: Pairing an Exponential Convolution

Take $f(x)=g(x)=e^{-x}\chi_{(0,\infty)}(x)$, and $h(x)=e^{-x}\chi_{(0,\infty)}(x)$. Each is bounded and belongs to $L^1(m)$, with integral $\int_0^\infty e^{-x}\,dx=1$. For $x>0$, the convolution integral is over $0<y<x$, so

(f*g)(x)=\int_0^x e^{-y}e^{-(x-y)}\,dy =e^{-x}\int_0^x1\,dy =xe^{-x}.

For $x\le0$, the integrand vanishes a.e., so $(f*g)(x)=0$. The pairing is therefore $\int_0^\infty xe^{-2x}\,dx=\tfrac14$: integration by parts gives $\int_0^\infty xe^{-2x}\,dx=[-\tfrac12xe^{-2x}]_0^\infty+\tfrac12\int_0^\infty e^{-2x}\,dx=0+\tfrac12\cdot\tfrac12=\tfrac14$.

To check by the double-integral identity, the integrand is nonzero only for $y,z>0$. There $f(y)g(z)h(y+z)=e^{-y}e^{-z}e^{-(y+z)}=e^{-2y}e^{-2z}$. Thus the double integral factors into $(\int_0^\infty e^{-2y}\,dy)(\int_0^\infty e^{-2z}\,dz)=\tfrac12\cdot\tfrac12=\tfrac14$, in agreement with the direct computation.

When the Test Function Is Integrable

Boundedness of $h$ is sufficient, not necessary. The precise requirement for the same Fubini argument is absolute integrability of the three-variable product after the convolution variables are written as $x=y+z$. In particular, if $f,g,h\in L^1(m)$ and

\int_{\mathbb{R}}\int_{\mathbb{R}} |f(y)||g(z)||h(y+z)|\,dz\,dy<\infty,

then Fubini’s theorem applies and yields $\int(f*g)(x)h(x)\,dx=\int\int f(y)g(z)h(y+z)\,dz\,dy$. The displayed condition must be checked; membership of all three functions in $L^1(m)$ does not imply it. The following example shows that the pairing can even be infinite for nonnegative $L^1$ functions.

Worked Example: Three Integrable Functions with an Infinite Pairing

Set $f(x)=g(x)=h(x)=x^{-3/4}\chi_{(0,1)}(x)$. Each function is nonnegative and measurable, and $\int_{\mathbb{R}}f(x)\,dx=\int_0^1x^{-3/4}\,dx=4$, so all three belong to $L^1(m)$.

For $0<x<1$, the conditions $0<y<1$ and $0<x-y<1$ reduce to $0<y<x$. Therefore, using $y=xt$ and $dy=x\,dt$,

\begin{aligned} (f*g)(x) &=\int_0^x y^{-3/4}(x-y)^{-3/4}\,dy\\ &=x^{-1/2}\int_0^1 t^{-3/4}(1-t)^{-3/4}\,dt\\ &=B(1/4,1/4)x^{-1/2}. \end{aligned}

Here $B(1/4,1/4)$ denotes the integral in the preceding line. It is finite and strictly positive: near $0$ its integrand is bounded by a constant times $t^{-3/4}$, and near $1$ by a constant times $(1-t)^{-3/4}$; both powers are integrable at their respective endpoints. Consequently, on $(0,1)$, $(f*g)(x)h(x)=B(1/4,1/4)x^{-5/4}$, and

\int_0^1(f*g)(x)h(x)\,dx =B(1/4,1/4)\int_0^1x^{-5/4}\,dx =+\infty.

The divergence follows, for example, from $\int_\varepsilon^1x^{-5/4}\,dx=4(\varepsilon^{-1/4}-1)\to+\infty$ as $\varepsilon\downarrow0$. This is a nonnegative example, so Tonelli’s theorem identifies the double integral with the same infinite value; it does not give a finite Fubini interchange. The calculation checks why the extra absolute-integrability or boundedness hypothesis matters.

A Common Mistake: Multiplying Three Separate Integrals

A tempting but incorrect calculation is to assert $\int(f*g)h=(\int f)(\int g)(\int h)$, by treating a convolution pairing as three independent integrals. The variables are not independent in that way: the valid double-integral expression contains $h(y+z)$, which couples the two variables. For the exponential example above, the incorrect product would be $1\cdot1\cdot1=1$, whereas the actual pairing is $\tfrac14$. The corrected calculation is $\int_0^\infty\int_0^\infty e^{-2y}e^{-2z}\,dz\,dy=\tfrac14$. The test-function identity preserves the sum $y+z$; it does not separate $h$ from the convolution variables.

The distinction is useful whenever an integral contains a convolution inside a pairing. First express the convolution using its defining integral. Then verify absolute integrability of the resulting function on the product space before applying Fubini. A bounded test function guarantees that check for $L^1$ inputs; for an unbounded test function, establish the triple-integral condition directly rather than inferring it from separate $L^1$ assumptions.

Check Your Understanding

In each problem, identify the measure space and check the relevant integrability condition before exchanging integrals.

For $f=g=\chi_{[0,1]}$, compute $(f*g)(x)$ and verify its integral using both the explicit formula and the convolution integral identity.
If $f,g\in L^1(m)$ and $h$ is bounded and measurable, write an upper bound for $\int\int|f(y)||g(x-y)||h(x)|\,dy\,dx$.
For the exponential functions in the worked example, compute $\int(f*g)(x)h(x)\,dx$ directly and by the double-integral identity.
Explain why three $L^1(m)$ assumptions do not by themselves justify a finite value for $\int(f*g)h\,dm$.
State the absolute-integrability condition that permits the test-function identity when $h$ is not assumed bounded.