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It is known that $L^1(\mathbb{R}) \ast f$ is dense in $L^1(\mathbb{R})$ for some $f\in L^1(\mathbb{R})$. So for such $f$ the closure of $L^1(\mathbb{R}) \ast f$ in the $L^1$ norm is $L^1(\mathbb{R})$. But apparently

(1)$\quad\quad L^1(\mathbb{R}) \ast f \neq L^1(\mathbb{R})$ for every $f\in L^1(\mathbb{R})$.

Is there a simple proof of (1)?

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  • $\begingroup$ Can you give a reference to the fact that "$L^1(\mathbb{R}) \ast f$ is dense in $L^1(\mathbb{R})$ for some $f\in L^1(\mathbb{R})$"? $\endgroup$ – Iosif Pinelis Apr 10 '18 at 23:50
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    $\begingroup$ @Iosif That holds for any $f$ such that the Fourier transform of $f$ is $\neq 0$ everywhere, by Wiener's Tauberian theorem. $\endgroup$ – user95282 Apr 11 '18 at 0:57
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I'm guessing $*$ means convolution (since this is a math forum) and not pointwise multiplication (since this is not a computer forum).

Some steps to try ... suppose $L^1 * f = L^1$

$\widehat{f} \ne 0$ a.e.

There is $g \in L^1$ so that $g * f = f$

$ \widehat{g} \widehat{f} = \widehat{f}$

$\widehat{g} = 1$ identically

$\widehat{g} \notin C_0$

$g \notin L^1$

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    $\begingroup$ Indeed very simple. Note: $\hat f$ cannot have any zeros, since it must be a divisor of the function $t\mapsto e^{-t^2}$ (say). $\endgroup$ – Iosif Pinelis Apr 11 '18 at 0:18
  • $\begingroup$ Yes, very nice. $\endgroup$ – user95282 Apr 11 '18 at 0:58

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