Show that convolution of two measurable functions is well-defined

Question

Question:

Recall the definition of the convolution of $f$ and $g$ given by $$(f*g)(x)=\int_{\mathbb{R}^d}f(x-y)g(y)dy.$$ If we only know that $f$ and $g$ are measurable, can we show that $f*g$ is well defined for a.e. $x$, that is, $f(x-y)g(y)$ is integrable?

(Exercise 2.5.21(c) in 'Real Analysis', by Stein and Shakarchi)

Actually, the book writes like this:

Suppose that $f$ and $g$ are measurable functions on $\mathbb{R}^d$.

(a)Prove that $f(x-y)g(y)$ is measurable on $\mathbb{R}^{2d}$.

(b)Show that if $f$ and $g$ are integrable on $\mathbb{R}^d$, then $f(x-y)g(y)$ is integrable on $\mathbb{R}^{2d}$.

(c)Recall the definition of the convolution of $f$ and $g$ given by $$(f*g)(x)=\int_{\mathbb{R}^d}f(x-y)g(y)dy$$ Show that $f*g$ is well defined for a.e. $x$, that is, $f(x-y)g(y)$ is integrable.

Can we use the assumption that $f,g$ are integrable in (c)?

No, we need some "smallness" condition, like that $f$ and $g$ are integrable (other conditions suffice too, $f$ integrable and $g \in L^p$ implies $f\ast g \in L^p$ for $1 \leqslant p \leqslant \infty$). — Daniel Fischer, Oct 08 '13 at 15:07
If $f$ is a distribution and $g$ is a smooth, compactly supported function, then $f*g$ is defined. — Jas Ter, Oct 08 '13 at 15:15

score 5 · Answer 1 · answered May 11 '16 at 03:22

5

The assumption "$f$ and $g$ are integrable" appears in (b). I guess this is why the author (carelessly) omits it in (c). Without such assumption, e.g. $f=g=1$ on ${\bf R}^d$, the convolution might be nonsense.

If $f, g: {\bf R}^d \to {\bf C}$ are absolutely integrable, then by the Fubini-Tonelli theorem, $f(y) g(x-y)$ is absolutely integrable on ${\bf R}^d \times {\bf R}^d$, which by further application of Fubini-Tonelli shows that $f(y) g(x-y)$ is absolutely integrable in $y$ for almost every $x$.

answered May 11 '16 at 03:22

The note in the link has a clear explanation of the Tonelli Theorem, Fubini Theorem and the Fubini-Tonelli Theorem. – May 11 '16 at 03:45
Sorry for being so dense, but I'm having a very tough time grasping how to apply the Fubini-Tonelli theorem. Could you please elaborate why F-T implies that $f(y)g(x -y)$ is absolutely integrable? Apparently I do not follow any application of F-T unless it's completely spelled out for me. – Woett Aug 19 '17 at 16:22
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Nevermind, a detailed explanation is given here (on page 3): https://www.math.ucdavis.edu/~hunter/m127c/hmwk6_solutions.pdf – Woett Aug 19 '17 at 16:46

score 1 · Answer 2 · answered Oct 08 '13 at 16:09

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Hint: do that first when $f, g \ge 0$. Recall that if the integrand function has a sign, you can safely change the order of integration in a double integral. This is sometimes known as Tonelli's theorem.

Tonelli's theorem is easier than the closely related Fubini's theorem, which regards integrand functions which possibly change sign. In the latter, you need to check that the integrand function is absolutely integrable with respect to both variables before you can do anything.

answered Oct 08 '13 at 16:09

Giuseppe Negro

32,319

So you mean that $f$ and $g$ in (c) are integrable? I'm just not sure about this. – user95640 Oct 08 '13 at 16:23
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Yes, of course. If $f$ or $g$ are not absolutely integrable then there is no guarantee that their convolution product makes sense. There are easy counterexamples, see Landscape's and Daniel Fisher's comments above. – Giuseppe Negro Oct 08 '13 at 16:30

Show that convolution of two measurable functions is well-defined

2 Answers2