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Let $A,B$ be two Banach algebra and $A\hat{\otimes} B$ be their projective tensor product. We know that every element $m$ of $A\hat{\otimes} B$ has the following representation: $$m=\sum_{n=1}^\infty x_n\otimes y_n$$ for some $\{x_n\}\subset A,\{y_n\}\subset B$ (which are not unique essentially). Also norm for elements of $A\hat{\otimes} B$ is defined by $$\|m\|=\inf\left\{\sum_{n=1}^\infty \|a_n\|\|b_n\|: m=\sum_{n=1}^\infty a_n\otimes b_n,\qquad \{a_n\}\subset A,\{b_n\}\subset B\right\}.$$ Define $$am=\sum_{n=1}^\infty (ax_n)\otimes y_n,\qquad a\in A.$$ By definition we know $$\|am\|=\inf\left\{\sum_{n=1}^\infty \|a_n\|\|b_n\|: am=\sum_{n=1}^\infty a_n\otimes b_n,\qquad \{a_n\}\subset A,\{b_n\}\subset B\right\}.$$ Now could we conclude the following statement? $$\|am\|=\inf\left\{\sum_{n=1}^\infty \|aa_n\|\|b_n\|: m=\sum_{n=1}^\infty a_n\otimes b_n,\qquad \{a_n\}\subset A,\{b_n\}\subset B\right\}.$$ If we couldn't, under which condition it is possible?

For definition of projective tensor product see the following from "Complete normed algebras" of Bonsall and Duncan.---Definition 9 to Proposition 12 enter image description here

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  • $\begingroup$ Dear Anton, this definition is in page 233 of "Complete normed algebras" by "Bonsall and Duncan" $\endgroup$ – Hamid Shafie Asl Oct 10 '16 at 9:25
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    $\begingroup$ Perhaps I am not thinking clearly right now, but could someone who voted to close explain why? (Since I don't have an MSE account I cannot answer questions that get migrated there, and my impression is that questions on Banach algebras at MSE don't always get attention from specialists.) $\endgroup$ – Yemon Choi Oct 10 '16 at 14:09
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    $\begingroup$ For the OP: I don't have my copy of B+D here right now, but I don't think their definition is exactly the same as yours, although one can prove they give the same result. The projective tensor product is normally defined as a completion of the algebraic tensor product with respect to a suitable norm, and a separate result is needed to show that every element of the completion has a representation as an absolutely convergent sum of elementary tensors. $\endgroup$ – Yemon Choi Oct 10 '16 at 17:41
  • $\begingroup$ I add the pictures. $\endgroup$ – Hamid Shafie Asl Oct 10 '16 at 18:15
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    $\begingroup$ @HeinrichD That direction is, relatively speaking, fairly trivial, and Hamid's question is clearly much harder. I don't see why one would infer that the inequality you mention is the motivation for the question $\endgroup$ – Yemon Choi Oct 10 '16 at 22:32

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