The Weak Tensor Product of X⊗Y

# The Weak Tensor Product of X⊗Y

 Definition: Let $X$ and $Y$ be normed linear spaces. The Weak Tensor Norm on $X \otimes Y$ is the function $w : X \otimes Y \to [0, \infty)$ defined for all $u = \sum_{i=1}^{m} x_i \otimes y_i \in X \otimes Y$ by $\displaystyle{w(u) = \sup_{\| f \| \leq 1, \| g \| \leq 1} \left \{ \left | \sum_{i=1}^{m} f(x_i)g(y_i) \right | \right \}}$. The Weak Tensor Product of $X$ and $Y$ denoted $X \otimes_w Y$ is the completion of $(X \otimes Y, w)$ in $\mathrm{BL}(X^*, Y^*, \mathbf{F})$.

The following proposition tells us what the weak tensor product of $x \otimes y$ ($x \in X$, $y \in Y$) is.

 Proposition 1: Let $X$ and $Y$ be normed linear spaces. Then $w(x \otimes y) = \| x \| \| y \|$ for all $x \in X$ and for all $y \in Y$.
• Proof: Let $x \in X$, $y \in Y$. Then:
(1)
\begin{align} \quad w(x \otimes y) = \sup_{\| f \| \leq 1, \| g \| \leq 1} \{ |f(x)g(y)| \} = \sup_{\| f \| \leq 1} \{ |f(x)| \} \sup_{\| g \| \leq 1} |g(y)| = \sup_{\| f \| \leq 1} \{ |\hat{x}(f)| \} \sup_{\| g \| \leq 1} \{ |\hat{y}(g)| \} = \| \hat{x} \| \| \hat{y} \| = \| x \| \| y \| \quad \blacksquare \end{align}

Before we move on, we should indeed verify that the weak tensor norm is a norm on $X \otimes Y$.

 Proposition 1: Let $X$ and $Y$ be normed linear spaces. Then the weak tensor norm on $X \otimes Y$ is a norm.
• Proof: There are three things to prove.
• 1. Showing that $w(u) = 0$ if and only if $u = 0$: Suppose that $w(u) = 0$. Then
(2)
\begin{align} \quad 0 = \sup_{\| f \| \leq 1, \| g \| \leq 1} \left \{ \left | \sum_{i=1} f(x_i)g(y_i) \right | \right \} = \sup_{\| f \| \leq 1, \| g \| \leq 1} \left \{ \left | \sum_{i=1} x_i \otimes y_i (f, g) \right | \right \} \end{align}
• So for all $f \in X^*$ and $g \in Y^*$ with $\| f \| \leq 1$, $\| g \| \leq 1$ we have that $\left | \sum_{i=1} (x_i \otimes y_i)(f, g) \right | \left | u(f, g) \right | = 0$. So $u(f, g) = 0$ for all $f \in X^*$, $g \in Y^*$ with $\| f \| \leq 1$, $\| g \| \leq 1$.
• Note that if $f \in X^*$ with $\| f \| > 1$ and $g \in Y^*$ with $\| g \| > 1$ then from above we have that:
(3)
\begin{align} \quad 0 = u \left ( \frac{f}{\| f \|}, \frac{g}{\| g \|} \right ) = \| f \| \| g \| u(f, g) \end{align}
• Since $\| f \|, \| g \| > 1$, the above equation shows us that $u(f, g) = 0$ for all $f \in X^*$ and all $g \in Y^*$. Therefore $u = 0$.
• On the otherhand, suppose that $u = 0$. Then $u = 0_X \otimes 0_Y$ and is such that $w(u) = w(0_X \otimes 0_Y) = \sup_{\| f \| \leq 1, \| g \| \leq 1} \left \{ 0_X \otimes 0_Y (f, g) \right \} = 0$.
• 2. Showing that $w(\alpha u) = |\alpha| w(u)$: Let $u \in X \otimes Y$ with $u = \sum_{i=1}^{m} x_i \otimes y_i$ and let $\alpha \in \mathbf{F}$ so that $\alpha u = \alpha \sum_{i=1}^{m} x_i \otimes y_i$. Then:
(4)
\begin{align} \quad w (\alpha u) = \sup_{\| f \| \leq 1, \| g \| \leq 1} \left \{ \left | \alpha \sum_{i=1}^{m} f(x_i)g(y_i) \right | \right \} = |\alpha| \sup_{\| f \| \leq 1, \| g \| \leq 1} \left \{ \left | \sum_{i=1}^{m} f(x_i)g(y_i) \right | \right \} = |\alpha| w(u) \end{align}
• 3. Showing that $w(u + v) \leq w(u) + w(v)$: Let $u, v \in X \otimes Y$ with $u = \sum_{i=1}^{m} x_i \otimes y_i$ and $v = \sum_{j=1}^{n} x_j' \otimes y_j'$. Then $u + v = \sum_{i=1}^{m} x_i \otimes y_i + \sum_{j=1}^{n} x_j' \otimes y_j'$ and:
(5)
\begin{align} \quad w(u + v) = \sup_{\| f \| \leq 1, \| g \| \leq 1} \left \{ \left | \sum_{i=1}^{m} f(x_i)g(y_i) + \sum_{j=1}^{n} f(x_j')g(y_j') \right | \right \} &\leq \sup_{\| f \| \leq 1, \| g \| \leq 1} \left \{ \left | \sum_{i=1}^{m} f(x_i)g(y_i) \right | + \left | \sum_{j=1}^{n} f(x_j')g(y_j') \right | \right \} \\ & \leq \sup_{\| f \| \leq 1, \| g \| \leq 1} \left \{ \left | \sum_{i=1}^{m} f(x_i)g(y_i) \right | \right \} + \sup_{\| f \| \leq 1, \| g \| \leq 1} \left \{ \left | \sum_{j=1}^{n} f(x_j') g(y_j') \right | \right \} = w(u) + w(v) \end{align}
• So we conclude that $w$ is a norm on $X \otimes Y$. $\blacksquare$