Convolution Integrability from Tonelli

Why Tonelli Controls Convolution

Convolution combines values of two functions across all pairs of points that add to a given location. Even when both functions are integrable, the defining integral need not converge absolutely at every location. Tonelli’s theorem resolves the key issue: it computes the total absolute mass across all locations and shows that the defining integral is finite almost everywhere. The same estimate then gives an L1 norm bound.

In this tutorial, $(\mathbb{R}^n,\mathcal{L},m)$ denotes Euclidean space with Lebesgue measure. This measure space is complete and sigma-finite, but not finite. Let $f,g:\mathbb{R}^n\to\mathbb{C}$ be Lebesgue measurable and integrable, meaning $\int_{\mathbb{R}^n}|f|\,dm<\infty$ and $\int_{\mathbb{R}^n}|g|\,dm<\infty$. Their L1 norms are $\|f\|_1=\int|f|\,dm$ and $\|g\|_1=\int|g|\,dm$.

For measurability in the argument, choose Borel measurable representatives of $f$ and $g$, which is possible because Lebesgue measurable functions have Borel representatives equal to them almost everywhere. The map $(x,y)\mapsto f(y)g(x-y)$ is then Borel measurable. Changing either representative on a null set does not change its defining integral for any fixed $x$: the affected $y$-values lie in a null set or in a translate of a null set. Thus the convolution integral below is independent of these choices.

Definition: For $x\in\mathbb{R}^n$ such that $\int_{\mathbb{R}^n}|f(y)g(x-y)|\,dm(y)<\infty$, define $$ (f*g)(x)=\int_{\mathbb{R}^n}f(y)g(x-y)\,dm(y). $$ At points where this absolute integral is infinite, assign $(f*g)(x)=0$. The theorem below shows that these exceptional points form a Lebesgue null set, so the resulting measurable function represents the convolution almost everywhere.

The L1 Bound from Tonelli

Theorem: If $f,g:\mathbb{R}^n\to\mathbb{C}$ are Lebesgue measurable and integrable on $(\mathbb{R}^n,\mathcal{L},m)$, then $y\mapsto f(y)g(x-y)$ is absolutely integrable for almost every $x\in\mathbb{R}^n$. The convolution defined above is measurable, belongs to $L^1(m)$, and satisfies $$ \|f*g\|_1\le \|f\|_1\|g\|_1. $$

Proof. The function $(x,y)\mapsto |f(y)||g(x-y)|$ is nonnegative and measurable on the product Lebesgue space. Tonelli’s theorem applies to nonnegative measurable functions on product measure spaces; in particular, it permits the order of integration to be exchanged even before finiteness is known. Therefore, using translation invariance of Lebesgue measure for the inner integral,

\begin{aligned} \int_{\mathbb{R}^n}\int_{\mathbb{R}^n}|f(y)||g(x-y)|\,dm(y)\,dm(x) &=\int_{\mathbb{R}^n}|f(y)|\left(\int_{\mathbb{R}^n}|g(x-y)|\,dm(x)\right)dm(y)\\ &=\int_{\mathbb{R}^n}|f(y)|\,\|g\|_1\,dm(y)\\ &=\|f\|_1\|g\|_1<\infty. \end{aligned}

Tonelli also says that the inner integral, as a function of $x$, is measurable. Since its integral is finite, it can be infinite only on a null set: if it were infinite on a set of positive measure, its integral would be infinite. Hence the defining integral is absolutely convergent for almost every $x$. On that full-measure set, the parameter integral is measurable; assigning zero on the measurable exceptional set preserves measurability.

At every point of absolute convergence, the triangle inequality for the integral gives $|(f*g)(x)|\le\int|f(y)||g(x-y)|\,dm(y)$. The same inequality holds at an exceptional point under our zero assignment. Integrating over $x$, then using the computed Tonelli integral, yields $\int|f*g|\,dm\le\|f\|_1\|g\|_1$. Thus $f*g\in L^1(m)$, with the asserted bound. $\square$

The theorem gives an almost-everywhere existence statement, not a guarantee at each individual $x$. Its estimate is global: it controls the integral of the absolute value of the convolution, not just the signed or complex integral. Both features matter in applications.

Worked Example: Convolving Two Interval Indicators

On $(\mathbb{R},\mathcal{L},m)$, set $f=\chi_{[0,2]}$ and $g=\chi_{[0,1]}$. These are real-valued integrable functions. Compute their convolution and compare its L1 norm with the bound.

The integrand $f(y)g(x-y)$ equals $1$ precisely when $y\in[0,2]$ and $x-y\in[0,1]$. The second condition is equivalent to $y\in[x-1,x]$. Thus the convolution is the length of the intersection $[0,2]\cap[x-1,x]$. For $x<0$ or $x>3$, that intersection has length zero. For $0\le x\le1$, its length is $x$; for $1\le x\le2$, it is $1$; and for $2\le x\le3$, it is $3-x$. Endpoint choices do not affect the integral.

(f*g)(x)= \begin{cases} x,&0\le x\le1,\\ 1,&1\le x\le2,\\ 3-x,&2\le x\le3,\\ 0,&\text{otherwise}. \end{cases}

The input norms are $\|f\|_1=2$ and $\|g\|_1=1$, since the intervals have lengths $2$ and $1$. The output is nonnegative, so its L1 norm is its integral: $\int_0^1x\,dx+\int_1^2 1\,dx+\int_2^3(3-x)\,dx=\tfrac12+1+\tfrac12=2$. Consequently $\|f*g\|_1=2=\|f\|_1\|g\|_1$. Check: the convolution is supported on an interval of length $3$, and its two sloping pieces have areas $\tfrac12$ each while its constant piece has area $1$, again giving total area $2$.

Worked Example: One-Sided Exponentials Attain the Bound

On $(\mathbb{R},\mathcal{L},m)$, let $f(x)=e^{-x}\chi_{[0,\infty)}(x)$ and $g(x)=e^{-2x}\chi_{[0,\infty)}(x)$. Compute the convolution directly and verify the L1 estimate.

If $x<0$, there is no $y$ for which both $y\ge0$ and $x-y\ge0$, so the convolution is zero. If $x\ge0$, these conditions require $0\le y\le x$, and hence

\begin{aligned} (f*g)(x) &=\int_0^x e^{-y}e^{-2(x-y)}\,dy\\ &=e^{-2x}\int_0^x e^y\,dy\\ &=e^{-2x}(e^x-1)=e^{-x}-e^{-2x}. \end{aligned}

For $x\ge0$, $e^{-x}\ge e^{-2x}$, so the convolution is nonnegative. Its norm is $\int_0^\infty(e^{-x}-e^{-2x})\,dx=1-\tfrac12=\tfrac12$. The input norms are $\|f\|_1=\int_0^\infty e^{-x}\,dx=1$ and $\|g\|_1=\int_0^\infty e^{-2x}\,dx=\tfrac12$. Therefore $\|f*g\|_1=\tfrac12=1\cdot\tfrac12=\|f\|_1\|g\|_1$. Check: integrating the displayed convolution gives $[ -e^{-x}+\tfrac12 e^{-2x}]_0^\infty=0-(-1+\tfrac12)=\tfrac12$, agreeing with the norm computation.

Cancellation Can Make the Bound Strict

The estimate uses absolute values before applying Tonelli. When the functions change sign, the convolution integral can have cancellation, so the resulting L1 norm can be strictly smaller than the product of the input norms.

Worked Example: A Signed Function Gives Strict Inequality

On $(\mathbb{R},\mathcal{L},m)$, let $f=\chi_{[0,1)}-\chi_{[1,2)}$ and $g=\chi_{[0,1)}$. Both are real-valued and integrable. Compute the convolution and its norm.

The convolution equals $\int_{x-1}^{x} f(y)\,dy$, because $g(x-y)=1$ exactly when $x-1\le y<x$. For $0\le x\le1$, this interval overlaps the positive part of $f$ in length $x$, giving $x$. For $1\le x\le2$, the positive overlap has length $2-x$ and the negative overlap has length $x-1$, giving $(2-x)-(x-1)=3-2x$. For $2\le x\le3$, only the negative part contributes, with length $3-x$, giving $-(3-x)$. Outside $[0,3]$, the overlap is empty.

(f*g)(x)= \begin{cases} x,&0\le x\le1,\\ 3-2x,&1\le x\le2,\\ -(3-x),&2\le x\le3,\\ 0,&\text{otherwise}. \end{cases}

The positive area is $\int_0^1x\,dx+\int_1^{3/2}(3-2x)\,dx=\tfrac12+\tfrac14=\tfrac34$. The negative area in absolute value is $\int_{3/2}^2(2x-3)\,dx+\int_2^3(3-x)\,dx=\tfrac14+\tfrac12=\tfrac34$. Thus $\|f*g\|_1=\tfrac34+\tfrac34=\tfrac32$. Meanwhile $\|f\|_1=1+1=2$ and $\|g\|_1=1$, so the product is $2$, strictly greater than $\tfrac32$. Check: the signed integral of the convolution is $\tfrac34-\tfrac34=0$, consistent with the cancellation between the positive and negative parts.

A Pointwise Finiteness Pitfall

A common mistake is to infer that the convolution integral is finite for every $x$ because both inputs are in $L^1(m)$. Tonelli proves finiteness almost everywhere, and that qualification cannot in general be removed.

Worked Example: An Infinite Value at One Point

On $(\mathbb{R},\mathcal{L},m)$, define the real-valued nonnegative measurable function $f(t)=|t|^{-3/4}$ when $0<|t|<1$, and $f(t)=0$ when $|t|\ge1$; its value at $t=0$ may be set to zero. Let $g=f$. Each function is integrable, since $\|f\|_1=2\int_0^1t^{-3/4}\,dt=2\cdot4=8$.

At $x=0$, the defining integral is $\int_{\mathbb{R}}f(y)f(-y)\,dy=2\int_0^1y^{-3/2}\,dy$, which is infinite. Indeed, for $0<\varepsilon<1$, $\int_\varepsilon^1y^{-3/2}\,dy=2(\varepsilon^{-1/2}-1)$, and this tends to infinity as $\varepsilon$ decreases to zero. Thus the convolution integral diverges at $x=0$, even though each input has norm $8$. This does not contradict the theorem: a single point has Lebesgue measure zero, and Tonelli guarantees absolute convergence for almost every $x$, not at every $x$.

In practice, the useful sequence of steps is to establish product measurability, apply Tonelli to the nonnegative absolute-value integrand, use translation invariance to identify its total integral, and only then conclude that almost every section is integrable. The triangle inequality then transfers that total bound to the L1 norm of the convolution. Attempting to estimate a signed or complex double integral before establishing absolute integrability would miss the central justification.

Check Your Understanding

Use the definitions and Tonelli estimate above to answer the following questions.

For integrable complex-valued $f,g$ on $(\mathbb{R}^n,\mathcal{L},m)$, which nonnegative function on the product space should be integrated first to prove the L1 convolution bound?
Why does finiteness of the integral of $x\mapsto\int|f(y)g(x-y)|\,dm(y)$ imply that the inner integral is finite almost everywhere?
For $f=\chi_{[0,1]}$ and $g=\chi_{[0,1]}$ on $\mathbb{R}$, compute the convolution and its L1 norm.
In the signed interval example, where does the convolution change sign, and why does this make the norm bound strict?
Does the theorem guarantee that $\int|f(y)g(x-y)|\,dm(y)$ is finite for every $x$? Explain what its conclusion guarantees instead.