Convolution of L1 Functions: Definition and Existence

From Two Integrable Functions to a Convolution

Convolution combines two functions by shifting one of them and integrating their product. For $f,g\in L^1(\mathbb{R})$, the natural formula is $(f*g)(x)=\int_{\mathbb{R}}f(y)g(x-y)\,dy$. The difficulty is that this integral is not automatically finite for every $x$. The key existence statement is that it is absolutely finite for almost every $x$, and that the resulting function can be chosen measurable.

Throughout, $m$ denotes Lebesgue measure on $\mathbb{R}$. The measure space $(\mathbb{R},\mathcal{L},m)$ is complete and sigma-finite, but not finite. Functions $f,g\in L^1(\mathbb{R})$ may be real- or complex-valued and are Lebesgue measurable; here $L^1(\mathbb{R})$ means that $\int_{\mathbb{R}}|f|\,dm<\infty$. Functions in $L^1$ are identified if they agree $m$-almost everywhere. An integral of a complex-valued function is defined when its absolute value is integrable, by integrating its real and imaginary parts.

Definition: For representatives $f,g$ of two $L^1(\mathbb{R})$ functions, define the absolute section integral $$ A(x)=\int_{\mathbb{R}}|f(y)|\,|g(x-y)|\,dy\in[0,\infty]. $$ At any $x$ for which $A(x)<\infty$, define $$ (f*g)(x)=\int_{\mathbb{R}}f(y)g(x-y)\,dy. $$ We will show that $A$ is measurable and finite almost everywhere. Thus convolution is defined almost everywhere; on the exceptional null set we may assign any finite value, such as $0$, to obtain a measurable representative.

Measurability and Borel Representatives

To apply Tonelli’s theorem, the function of two variables $(x,y)\mapsto |f(y)|\,|g(x-y)|$ must be measurable. A Lebesgue measurable function need not itself be Borel measurable, so we first use the standard Borel-version fact: every Lebesgue measurable function on $\mathbb{R}$ that is finite almost everywhere agrees almost everywhere with a Borel measurable function. One way to obtain the fact is to approximate the function pointwise by measurable simple functions, replace each of their level sets by Borel sets differing by null sets, and take the pointwise limit off the countable union of those null sets. Changing the function on that union gives a Borel version.

Choose Borel versions $f_0,g_0$ of $f,g$. The maps $(x,y)\mapsto y$ and $(x,y)\mapsto x-y$ are continuous, so $(x,y)\mapsto |f_0(y)|\,|g_0(x-y)|$ is Borel measurable on $\mathbb{R}^2$. In particular, it is measurable for the product of the Borel sigma-algebras, and Tonelli’s theorem applies to it on the sigma-finite product measure space $(\mathbb{R}^2,\mathcal{B}_{\mathbb{R}}\otimes\mathcal{B}_{\mathbb{R}},m\times m)$.

The choice of representatives does not affect any fixed section integral. Indeed, if $f=f_0$ off a null set $N_f$, the integrands can differ only for $y\in N_f$. If $g=g_0$ off a null set $N_g$, they can differ only when $x-y\in N_g$, or equivalently $y\in x-N_g$. Translation and reflection preserve null sets, so $x-N_g$ is null for each fixed $x$. The original and Borel-version integrands therefore agree for almost every $y$, for every fixed $x$.

The Almost-Everywhere Existence Theorem

Theorem: Let $f,g\in L^1(\mathbb{R})$, with real or complex values. Then $A(x)=\int_{\mathbb{R}}|f(y)|\,|g(x-y)|\,dy$ is measurable and finite for $m$-almost every $x$. Consequently, the convolution integral defining $(f*g)(x)$ is absolutely convergent for almost every $x$, and it has a measurable representative.

Proof. It suffices to use the Borel versions above, since changing representatives does not alter any fixed section integral. The integrand $H(x,y)=|f_0(y)|\,|g_0(x-y)|$ is nonnegative and product measurable. Tonelli’s theorem, which permits exchanging the order of integration for nonnegative measurable functions on this sigma-finite product space, gives

\begin{aligned} \int_{\mathbb{R}} A(x)\,dx &=\int_{\mathbb{R}}\int_{\mathbb{R}}|f_0(y)|\,|g_0(x-y)|\,dy\,dx\\ &=\int_{\mathbb{R}}|f_0(y)|\left(\int_{\mathbb{R}}|g_0(x-y)|\,dx\right)dy\\ &=\int_{\mathbb{R}}|f_0(y)|\,\|g\|_1\,dy\\ &=\|f\|_1\|g\|_1<\infty. \end{aligned}

For the third line, fix $y$ and make the translation $u=x-y$; translation invariance of Lebesgue measure gives $\int_{\mathbb{R}}|g_0(x-y)|\,dx=\int_{\mathbb{R}}|g_0(u)|\,du=\|g\|_1$. The final quantity is finite by the $L^1$ hypotheses. Tonelli’s theorem also gives measurability of the section-integral function $A$. If the measurable set $S=\{x:A(x)=\infty\}$ had positive measure, then for every positive integer $n$, $$ \int_{\mathbb{R}}A(x)\,dx\ge \int_S n\,dx=n\,m(S). $$ If $m(S)>0$, the right side is unbounded as $n$ increases, contradicting the finite integral just obtained. Hence $m(S)=0$, proving absolute finiteness almost everywhere.

For measurability of the complex convolution, write the real and imaginary parts of $f_0(y)g_0(x-y)$ as differences of their nonnegative positive and negative parts. Tonelli’s section-measurability conclusion applies to each part. Where $A(x)<\infty$, all four section integrals are finite, so their differences define the real and imaginary parts of the convolution measurably. Assigning $0$ on the measurable null set where $A(x)=\infty$ gives a measurable representative on all of $\mathbb{R}$. $\square$

Worked Examples

Worked Example: Exponential Functions

Take $f(t)=e^{-|t|}$ and $g(t)=e^{-2|t|}$. Both are nonnegative Borel functions. Their $L^1$ norms are $$ \|f\|_1=2\int_0^\infty e^{-t}\,dt=2(1)=2,\qquad \|g\|_1=2\int_0^\infty e^{-2t}\,dt=2\left(\frac12\right)=1. $$ Tonelli’s theorem therefore gives the total integral of the absolute sections as $\int_{\mathbb{R}}A(x)\,dx=2\cdot1=2$. At $x=0$, the section is also directly computable: $$ A(0)=\int_{\mathbb{R}}e^{-|y|}e^{-2|y|}\,dy =2\int_0^\infty e^{-3y}\,dy =2\left(\frac13\right)=\frac23. $$ Since the functions are nonnegative, $(f*g)(0)=A(0)=\frac23$. Check: splitting the section at zero gives $\int_{-\infty}^0 e^{3y}\,dy+\int_0^\infty e^{-3y}\,dy=\frac13+\frac13=\frac23$, as required.

Worked Example: An Infinite Section at One Point

Define $f(t)=g(t)=|t|^{-3/4}$ when $0<|t|<1$, and set both functions equal to $0$ elsewhere, including at $t=0$. These are measurable nonnegative functions. Their norms are finite: $$ \|f\|_1=\|g\|_1 =2\int_0^1 t^{-3/4}\,dt =2\left[4t^{1/4}\right]_0^1 =8. $$ Thus Tonelli gives $\int_{\mathbb{R}}A(x)\,dx=8\cdot8=64$. However, at $x=0$, both factors are nonzero precisely when $0<|y|<1$, and $$ A(0)=\int_{0<|y|<1}|y|^{-3/4}|-y|^{-3/4}\,dy =2\int_0^1 y^{-3/2}\,dy =\infty. $$ For the last equality, $\int_\delta^1 y^{-3/2}\,dy=2(\delta^{-1/2}-1)$, which tends to infinity as $\delta\downarrow0$. Check: each input has finite norm $8$, so the total section integral is $64$, finite, even though the section at $0$ is infinite. This is consistent with almost-everywhere, not everywhere, existence.

Worked Example: A Complex-Valued Convolution at Zero

Let $f(t)=e^{-|t|}$ and $g(t)=e^{-|t|}e^{it}$. These are complex- or real-valued Borel functions as appropriate, and $|g(t)|=e^{-|t|}$, so $\|f\|_1=\|g\|_1=2$. At $x=0$, the absolute section integral is $$ A(0)=\int_{\mathbb{R}}e^{-2|y|}\,dy =2\int_0^\infty e^{-2y}\,dy =2\left(\frac12\right)=1. $$ Thus the complex convolution integral at zero is absolutely convergent. Since $g(-y)=e^{-|y|}e^{-iy}$, splitting at zero and substituting $u=-y$ on the negative half-line gives $$ \begin{aligned} (f*g)(0) &=\int_0^\infty e^{-(2+i)y}\,dy+\int_0^\infty e^{-(2-i)u}\,du\\ &=\frac{1}{2+i}+\frac{1}{2-i} =\frac{2-i}{5}+\frac{2+i}{5} =\frac45. \end{aligned} $$ The improper integrals converge because the real part of each exponent coefficient is $2>0$. Check: the two half-line integrals are complex conjugates, and their sum is twice the real part of $1/(2+i)=(2-i)/5$, namely $2(2/5)=4/5$.

A Common Misreading of Tonelli’s Theorem

In the singular example, one might reason: “Since $A(0)=\infty$, the integral $\int_{\mathbb{R}}A(x)\,dx$ must be infinite.” That conclusion is false. A value of infinity at one point contributes zero to a Lebesgue integral over a singleton of measure zero. The actual calculation gives $\int_{\mathbb{R}}A(x)\,dx=64$. In general, Tonelli’s theorem establishes the integral of the section function and its measurability; finiteness of that integral implies that the section is finite almost everywhere, not at every point.

There is a related representative issue. Because $L^1$ functions are equivalence classes modulo null sets, a convolution should not depend on arbitrary changes on null sets. Here, for each fixed $x$, changing $f$ on a null set changes the section integrand only on that null set of $y$; changing $g$ on a null set changes it only on a translate or reflection of a null set of $y$. Thus the section integral is unchanged for each $x$, including when its value is infinite. The almost-everywhere finiteness conclusion is therefore a property of the $L^1$ classes, not of a specially chosen representative.

Check Your Understanding

Use the definition and the hypotheses of the theorem to decide each question.

For $f,g\in L^1(\mathbb{R})$, what is the total integral of $A(x)=\int_{\mathbb{R}}|f(y)|\,|g(x-y)|\,dy$? Which theorem justifies the order of integration?
Why is $(x,y)\mapsto g(x-y)$ Borel measurable when $g$ is Borel measurable?
In the singular example, why does $A(0)=\infty$ not contradict $\int_{\mathbb{R}}A(x)\,dx=64$?
If $g$ is changed on a null set $N$, describe the set of $y$ where the section integrand at a fixed $x$ could change.
Does the theorem guarantee that the convolution integral is finite for every $x$, or only for almost every $x$?