Question about a paper on approximate identities I am currently reading this paper on approximate identities of ternary Banach algebras. Assume that $(A, [.,.,.])$ is a ternary Banach algebra. A bounded net $(e_{\alpha}, f_{\alpha})$ is said to be left-bounded approximate identity for $A$ if $\lim_{\alpha}[e_{\alpha}, f_{\alpha},a]=a$ for all $a \in A$. Can someone clarify my following questions on the same paper:

What is the definition of bounded net? Also at many places in the same paper for instance in Theorem $2.2$, the product $e_{\alpha}f_{\alpha}$ is used. What is the meaning of product in ternary algebra?

All these terms are not explained in paper. Thank you very much in advance!
P.S: The same question was posted on here too but unfortunately dint get any answer.
 A: A set $\Lambda$ together with a partial ordering $≤$ is called directed if for any two $\alpha, \beta\in \Lambda$ there is a $\gamma\in\Lambda$ with $\alpha≤\gamma$ and $\beta≤\gamma$. A net is map $\Lambda\to X$, $\alpha\mapsto x_\alpha$ from a directed set into another set (usually a topological space). A net $x_\alpha$ is said to converge to a point $x$ if for any neighbourhood $U$ of $x$ there a $\gamma\in \Lambda$ so that $x_\alpha\in U$ for all $\alpha≥\gamma$.
A net $x_\alpha$ into normed vector space is called bounded if the image $\{ x_\alpha \mid \alpha\in\Lambda\}$ is a bounded set.
Next the mysterious products that appear in the paper. There are two products that appear, once a "contraction" written as $xy$, another time a quasiproduct written as $x\circ y = x+y-xy$. I will have to guess what the product is. I suspect it is a very liberal use of "abuse of notation".
For this call the elements of $A$ "singles" and pure tensors of $A\otimes A$ "pairs". Now define a "product" on $A$ landing in $A\otimes A$ by
$$x\cdot y := x\otimes y$$
Further extend the meaning of "$\cdot$" for a product of an element of $A\otimes A$ with an element of $A$ in an associative manner:
$$(x\cdot y)\cdot z = [x,y,z], \qquad z\cdot (x\cdot y) = [z,x,y]$$
additionally define the product of two pairs to be another pair:
$$(x\cdot y)\cdot (a\cdot b) = [x,y,a]\cdot b$$
you can check that this operator "$\cdot$", which has different meanings, is associative with regard to any intelligible composition.
Now look at equation 2.3 in the paper. It is:
$$\|[uv\circ uv,a]-a\|=\|[u,v,[u,v,a]-a]-([u,v,a]-a)\|$$
Lets attempt to interpret it with our definiton:
$$uv\circ vu = uv+uv-uvuv=2uv-uvuv$$
which is a pair minus a pair, hence a pair. Now the contraction with $a$ on the left hand side makes sense, it will become:
$$[uv\circ uv , a]-a=2[u,v,a]-[u,v,[u,v,a]] -a = -([u,v,[u,v,a]-a]-([u,v,a]-a))$$
and we see that this understanding of product will be able to prove the equation 2.3. (Take the norm and the initial minus drops out).
This convinces me that this "abuse of notation" is what the product is supposed to mean. You should calculate a few more of the expressions that are there to gather more evidence that this is the correct understanding.

This stuff could be made a bit more explicit: Look at the tensor algebra $T(A)=\bigoplus_{n=0}^\infty A^{\otimes n}$ and consider the closed two-sided ideal $J$ generated by elements of the form $a\otimes b \otimes c - [a,b,c]$. Then $T(A)/J$ contains $A$ and has a product which is compatible with the tri-product on $A$. This is probably a universal bi-algebra so that every tri-algebra morphism into a bi-algebra factors over this thing.
In this bigger algebra products make sense and it contains $A$. So you can occasionally go "up" to the bigger algebra in calculations and everything will work. So you could be careful to make sure you are only making statements about $A$, but in your calculations you may assume that $A$ has a bi-product by going up to this algebra. Just make sure that in the end you get an element of $A$, rather than an element of $T(A)/J$.
