I am currently reading through chapter 42 on Tensor products in the text "Complete Normed Algebras" by F.F. Bonsall and J. Duncan. The definitions that are given in the text are as follows:
Definition 1: Let $ X $ and $ Y $ be normed spaces. The Algebraic Tensor Product of $ X $ and $ Y $ denoted $ X \otimes Y $ is the linear span of the set $ \{ x \otimes y : x \in X, y \in Y \} $ where for each $ x \in X $ and for each $ y \in Y $, $ x \otimes y : X^* \times Y^* \to \mathbf{F} $ is the bounded bilinear map defined for all $ (f, g) \in X^* \times Y^* $ by:
$$ (x \otimes y)(f, g) = f(x)g(y) $$
Definition 2: The Weak Tensor Norm on $ X \otimes Y $ is defined for all $ u = \sum_{i=1}^{n} x_i \otimes y_i \in X \otimes Y $ by:
$$ w(u) = \sup \left \{ \left | \sum_{i=1}^{n} f(x_i)g(y_i) \right | : \| f \| \leq 1, \: \| g \| \leq 1 \right \} $$
and the Weak Tensor Product of $ X $ and $ Y $ denoted by $ X \otimes_w Y $ is the completion of $ (X \otimes Y, w) $ in $ \mathrm{BL}(X^*, Y^*; \mathbf{F}) $.
Definition 3: The **Projective Tensor Norm* on $ X \otimes Y $ is defined for all $ u \in X \otimes Y $ by:
$$ p(u) = \inf \left \{ \sum_{i=1}^{n} \| x_i \| \| y_i \| : u = \sum_{i=1}^{n} x_i \otimes y_i \right \} $$
and the Projective Tensor Product of $ X $ and $ Y $ denoted by $ X \otimes_p Y $ is the completion of $ (X \otimes Y, p) $ in $ \mathrm{BL}(X^*, Y^*; \mathbf{F}) $.
Later on in the section, the following proposition is presented:
Proposition: Let $ X $ and $ Y $ be normed algebras over $ \mathbf{F} $. There exists a unique product on $ X \otimes Y $ with respect to which $ X \otimes Y $ is an algebra and:
$$ (x_1 \otimes y_1)(x_2 \otimes y_2) = x_1x_2 \otimes y_1y_2 $$
I have a few (perhaps elementary) questions regarding the definitions and proposition above. I am quite new to this subject so please bear with me.
Question 1: If I am not mistaken, the algebraic tensor product $ X \otimes Y $ is a linear subspace of $ \mathrm{BL}(X^*, Y^*; \mathbf{F}) $. The space $ \mathrm{BL}(X^*, Y^*; \mathbf{F}) $ is a normed space with operator norm defined for all $ T \in \mathrm{BL}(X^*, Y^*; \mathbf{F}) $ by:
$$ \| T \| = \sup \{ T(f, g) : \| f \| \leq 1, \| g \| \leq 1 \} $$
So is the algebraic tensor product $ X \otimes Y $ a normed space with the operator norm?
Question 2: The authors define the projective tensor norm of an element $ u \in X \otimes Y $ to be the infimum of the sums $ \sum_{i=1}^{n} \| x_i \| \| y_i \| $ where $ \sum_{i=1}^{n} x_i \otimes y_i $ is a representation of $ u $. For the weak tensor norm, this is not the case. Is it obvious that $ w(u) $ is well-defined regardless of the choice of representation for $ u $?
Question 3: When $ X $ and $ Y $ are instead algebras, then the above proposition tells us $ X \otimes Y $ can be made into an algebra as well. I am a bit confused as to the wording concerning "uniqueness" though. Is the proposition saying that there is only one way to define a product on $ X \otimes Y $ to make it an algebra? Or is the proposition saying that there is one way to define a product on $ X \otimes Y $ with the exact property that $ (x_1 \otimes y_1)(x_2 \otimes y_2) = x_1x_2 \otimes y_1y_2 $?
Question 4: Suppose $ X $ and $ Y $ are algebras. Then $ X \otimes Y $ is an algebra with the product defined by the above proposition. The authors say that $ X \otimes Y $ can be made into a normed algebra if equip it with the projective tensor norm. It is easy to verify that $ p $ is an algebra norm. The authors then state if further, $ X $ and $ Y $ are Banach algebras then the product on $ X \otimes Y $ can be extended to $ X \otimes_p Y $ so that $ X \otimes_p Y $ becomes a Banach algebra.
My question is, if $ X $ and $ Y $ are Banach algebras, then is $ X \otimes Y $ a Banach algebra to begin with? Furthermore, I am not sure how exactly the product above on $ X \otimes Y $ is extended to $ X \otimes_p Y $. My professor had told me that $ X \otimes Y $ is a dense subspace of $ X \otimes_p Y $ (which I assume follows by definition of $ X \otimes_p Y $ being the completion of $ (X \otimes Y, p) $, but is this extension necessarily unique?
Thanks for any help regarding these questions!
Best Answer
Question 1: Well, you can impose that norm, but normally we don't. The whole point of talking about the "algebraic" tensor product is that we are considering just vector spaces, and forgetting about any norms or topology. In fact, the "operator norm" you propose is exactly the same as the weak tensor norm, since what you call $T(f,g)$ is equal to $\sum_{i=1}^{n} f(x_i)g(y_i)$ if $T=\sum x_i\otimes y_j$.
Question 2: Yes, this is obvious, since as remarked above, the expression $\sum_{i=1}^{n} f(x_i)g(y_i)$ is just the evaluation of the bilinear map $u$ on $(f,g)$.
Question 3: It is saying there is only one product with the property that $(x_1 \otimes y_1)(x_2 \otimes y_2) = x_1x_2 \otimes y_1y_2$. Obviously if you don't impose this condition then there are many many different products, since you just have some abstract vector space and no additional constraints.
Question 4: No, $X\otimes Y$ will usually not be a Banach algebra, because it will not be complete with respect to the projective tensor norm. You extend the product to the completion using density: given $a,b\in X\otimes_p Y$, we can write $a$ and $b$ as the limits of sequences $(a_n)$ and $(b_n)$ in $X\otimes Y$, and then define $ab$ to be the limit of $a_nb_n$. (Of course, you must prove that this limit exists and is independent of the choice of sequences, and that this really does make $X\otimes_p Y$ a Banach algebra.) This extension is the unique extension of the product on $X\otimes Y$ to a product on $X\otimes_p Y$ which is continuous with respect to the norm (indeed, the definition above is forced if you want it to be continuous). If you don't care about continuity, the extension will usually not be unique.