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Let $\mathbb T$ be a circle group.

In 1939, Salem, has shown that, every member of $L^{1}(\mathbb T)$ can written as a product(convolution) some other two members of $L^{1}(\mathbb T),$ that is, $L^{1}(\mathbb T)\subset L^{1}(\mathbb T)\ast L^{1}(\mathbb T);$ other way one can get by Fubini's theorem, so we have, $L^{1}(\mathbb T)=L^{1}(\mathbb T)\ast L^{1}(\mathbb T).$

I have learn from the book(R. E. Edwards, Fourier Series A Modern Introduction Volume 1, p.124), $L^{p}(\mathbb T)=L^{1}(\mathbb T)\ast L^{p}(\mathbb T), (1\leq p < \infty).$

In 1959, Walter Rudin has shown for the real line that, $L^{1}(\mathbb R)=L^{1}(\mathbb R)\ast L^{1}(\mathbb R).$

By Young's inequality, we have, $L^{1}(\mathbb R)\ast L^{p}(\mathbb R) \subset L^{p}(\mathbb R), (1\leq p < \infty).$

My Question is: Is it well-known in a literature that, $L^{p}(\mathbb R)\subset L^{1}(\mathbb R) \ast L^{p}(\mathbb R), (1< p< \infty)$ ? If answer is yes, please let me know some references; If not, what one can expect ?

Thanks,

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    $\begingroup$ Since the Banach algebra L^1(R) has a bounded approximate identity, surely this follows from Cohen's factorization theorem $\endgroup$
    – Yemon Choi
    Jul 10, 2014 at 14:32
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    $\begingroup$ Also, you do not need Young's inequality to prove that $L^1(R)$ convolves $L^p(R)$ into itself, this follows from the triangle inequality. $\endgroup$
    – Yemon Choi
    Jul 10, 2014 at 14:33
  • $\begingroup$ @YemonChoi; thanks; but I could not follow you;(see,If I understand correctly, Cohen's factorization theorem, tells that, Suppose $B$ is Banach algebra(suppose with respect to convolution), if $B$ has left approximate identity, then $B= B\ast B.$ So we get, $L^{1}=L^{1}\ast L^{1}$; but $L^{p}(\mathbb R),p>1$ is not an algebra with respect to convolution; so how to conclude that, $L^{p}\subset L^{1}\ast L^{p} , (p>1)$; please correct me If I am wrong some where here, thanks ) $\endgroup$ Jul 11, 2014 at 4:48
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    $\begingroup$ Cohen's factorization theorem has a more general form, which applies to any "essential" Banach module over a Banach algebra with a bounded approximate identity. This can be found in the book of Bonsall and Duncan, or the book of Dales, for instance $\endgroup$
    – Yemon Choi
    Jul 11, 2014 at 8:50
  • $\begingroup$ @YemonChoi; Ah....; Indeed, thanks a lot; $\endgroup$ Jul 11, 2014 at 10:27

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