Show that there exists no continuous 1-periodic function $f_0$ so that $f\star f_0=f$ holds for all continuous 1-periodic functions $f$. No Lebesgue here. We're dealing with $L^2$, the space of continuous Riemann square integrable functions.

Having trouble getting any intuition on convolution integrals. Any one have a hint to point me in the right direction? I'm given as a suggestion to use Riemann-Lebesgue lemma, but I cannot see how it applies.


1 Answer 1


Suppose $f_0$ exists. Then the convolution identity means that for $x\in [0,1]$ $$\int_0^1f(x-t)f_0(t)dt=f(x)$$ Let's apply this to $f(t)=e^{2i\pi nt}$. We obtain $$\int_0^1 e^{2i\pi n(x-t)}f_0(t)dt=e^{2i\pi nx}$$ That is, for all $n\in\mathbb Z$, $$ e^{2i\pi nx}\int_0^1 e^{-2i\pi nt}f_0(t)dt=e^{2i\pi nx}$$ In other words, $$\int_0^1 e^{-2i\pi nt}f_0(t)dt=1$$ Therefore all Fourier coefficients of $f_0$ are equal to 1. This is a contradiction since $f_0$ is assumed to be a continuous function (so its Fourier coefficients must converge to 0 as $|n|\rightarrow +\infty$ by Riemann-Lebesgue).

  • 1
    $\begingroup$ +1 really nice answer Stefan $\endgroup$ Jan 27, 2020 at 16:56
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    $\begingroup$ Thank you! Doing my best ;-) $\endgroup$ Jan 27, 2020 at 17:07

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