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The question is simple but I still can't prove it or contradict it. Here it goes:

Suppose $f$ and $g$ are defined on the circle (or, equivalently, $2\pi$ periodic functions) and Lebesgue integrable, is their convolution $(f*g)(x) = \frac{1}{2\pi}\int_{-\pi}^{\pi} f(x-y) g(y) dy $ continuous?

In the case when two functions are bounded, it is proved in Elias Stein's Fourier Analysis (page 47) that their convolution is truly continuous. However, for unbounded functions, I have tried tools in real analysis, say, Lusin's theorem, transition continuity of $L_1$ functions, etc., but can't figure it out.

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    $\begingroup$ Let me remark that it is sufficient that one of the functions is bounded (the convolution of an $L^1$-function and an $L^\infty$-function on a unimodular group is always continuous. $\endgroup$
    – The User
    Commented Jul 14, 2013 at 16:18
  • $\begingroup$ yes, you are right, a similar proof from the book will do. $\endgroup$
    – booksee
    Commented Jul 14, 2013 at 16:50

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Edit: sorry, there was a sign error. It should just be:

$$f(x) = g(x) = \begin{cases} x^{-3/4}&x > 0\cr 0&x \leq 0. \end{cases}$$ Then $\lim_{x \to 0^+}f*g(x) = \infty$ and $\lim_{x \to 0^-} f*g(x) = 0$.

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  • $\begingroup$ Are you sure you have calculated the integrals ? I computed the two integrals in mathematica $f*g(x)$ with $x=10^{-7}$ and $x=-10^{-7}$, both going very large. $\endgroup$
    – booksee
    Commented Jul 14, 2013 at 16:42
  • $\begingroup$ No. The overlap is nearly half the interval. Say, for $x=-0.0001$, the overlap is $[-\pi+0.0001,-0.0001]$ $\endgroup$
    – booksee
    Commented Jul 14, 2013 at 16:56
  • $\begingroup$ @booksee Does it matter? If it goes to infinity the convolution is certainly not in $C(\mathbb{T})$ (it is for the real-line, but that does not really change the game, the large values of $x$ are irrelevant). The example is kinda standard, you could look here, I think you can trust in it. $\endgroup$
    – The User
    Commented Jul 14, 2013 at 17:34
  • $\begingroup$ In this example, $f*g(0)=\infty$, which to some extent really solves my question. I have never imagined that the counterexample will look like this-_-. $\endgroup$
    – booksee
    Commented Jul 14, 2013 at 17:50
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    $\begingroup$ @TheUser, the paper you refer to seems not talk about convolution, since $ab(t)=\int_0^t a(t-s)b(s)ds$, in which the variable 't' is also the upper bound of the integral. $\endgroup$
    – booksee
    Commented Jul 14, 2013 at 17:55
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A less explicit answer: Salem and Zygmund proved that convolution $L^1(\mathbb T) \times L^1(\mathbb T) \to L^1(\mathbb T)$ is onto.

This was shown to hold for all locally compact groups by Paul Cohen in 1959. This result was the starting point of an entire industry establishing "factorization theorems".

A nice survey on this topic is Jan Kisynski, On Cohen's proof of the Factorization Theorem, Annales Polonici Mathematici 75, 2 (2000), 177-192.

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  • $\begingroup$ Cool, I did not know this. $\endgroup$
    – Nik Weaver
    Commented Jul 14, 2013 at 23:10
  • $\begingroup$ Isn't much of that "industry" subsumed by the Cohen Factorization Theorem for appropriate modules over a Banach algebra? $\endgroup$
    – Yemon Choi
    Commented Jul 15, 2013 at 3:14

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