# Distributivity of projective tensor product over direct sum

Let $I$ is a non-empty set and $\{A_i\}_{i\in I}$ is a family of Banach algebras and $B$ is a Banach algebra. Define $$\ell^1-\oplus_{i\in I}A_i=\{a=\{a_i\}_{i\in I}: \|a\|_1=\sum_{i\in I}\|a_i\|<\infty,\forall i\in I, a_i\in A_i\}$$ Then $(\ell^1-\oplus_{i\in I}A_i,\|.\|_1)$ is a Banach algebra. Consider $\hat\otimes$ as the projective tensor product. Could we say that the following holds? $$B\hat\otimes(\ell^1-\oplus_{i\in I}A_i)\cong\ell^1-\oplus_{i\in I}B\hat\otimes A_i.$$ $\cong$: isomorphism isometric.

If it is not true, we can suppose that $I$ is finite or $\{A_i\}_{i\in I}$ is a family of unital or commutative or finite-dimentional Banach algebras or all of this conditions together.

• Yes this is true. – Norbert Oct 16 '15 at 11:30
• Could you please give me a proof or a reference? – Hamid Shafie Asl Oct 16 '15 at 14:41

Consider maps $$u:B \times \bigoplus_1\{A_i:i\in I\} \to \bigoplus_1\{B\hat{\otimes}A_i:i\in I\}:\left(x,\bigoplus_1\{a_i:i\in I\}\right)\mapsto\bigoplus_1\{x\hat{\otimes}a_i:i\in I\}$$ This is a bilinear contractive map, so by universal property of projective tensor product we have well defined contractive linear operator $$U:B \hat{\otimes} \bigoplus_1\{A_i:i\in I\} \to \bigoplus_1\{B\hat{\otimes}A_i:i\in I\}$$ For each $j\in I$ consider bilinear operator $$v_j:B\times A_j\to B \hat{\otimes} \bigoplus_1\{A_i:i\in I\}:(x,a_j)\mapsto x\hat{\otimes}\bigoplus_1\{\delta_i^j a_i:i\in I\}$$ Again one can check that this is a contractive bilinear operator. By universal property of projective tensor product we have a contractive linear operator $$V_j:B\hat{\otimes} A_j\to B \hat{\otimes} \bigoplus_1\{A_i:i\in I\}$$ Now you can easily see that the linear operator $$V:\bigoplus_1\{B\hat{\otimes} A_j:j\in I\}\to B \hat{\otimes} \bigoplus_1\{A_i:i\in I\}: \bigoplus_1\{t_j:j\in I\}\mapsto \sum_{j\in I}V_j(t_j)$$ is contractive.
We claim that $U$ and $V$ are inverse to each other. For any tensor $s$ of the form $x\hat{\otimes}\bigoplus_1\{a_i:i\in I\}$ we have $VU(s)=s$. The linear span of such tensors in dense in $B\hat{\otimes}\bigoplus_1\{A_i:i\in I\}$, so $VU=1$. Similarly, $UV(t)=t$ for any tensor $t$ of the form $t=\bigoplus_1\{x\hat{\otimes} a_i:i\in I\}$. Since such tensors are dense in $\bigoplus_1\{B\hat{\otimes}A_i:i\in I\}$, then $UV=1$. By similar density argument shows that $U$ and $V$ are homomorphisms of Banach algebras.
As $U$ and $V$ are contractive and inverse to each other the Banach algebras $\bigoplus_1\{B\hat{\otimes}A_i:i\in I\}$ and $B\hat{\otimes}\bigoplus_1\{A_i:i\in I\}$ are isometrically isomorphic.