The Algebraic Tensor Product of Two Normed Linear Spaces

# The Algebraic Tensor Product of Two Normed Linear Spaces

 Definition: Let $X$ and $Y$ be normed linear spaces over $\mathbf{F}$. For each $x \in X$ and for each $y \in Y$ define $(x \otimes y) : X^* \times Y^* \to \mathbf{F}$ for all $f \in X^*$, $g \in Y^*$ by $(x \otimes y)(f, g) = f(x)g(y)$. Elements of the form $x \otimes y$ are called Elementary Tensors. The Algebraic Tensor Product of $X$ and $Y$ denoted $X \otimes Y$ is defined to be the linear span of the set of elementary tensors, that is, $\{ x \otimes y : x \in X, y \in Y \}$ in $\mathrm{BL}(X^*, Y^*, \mathbf{F})$.

Recall that if $X$ is a linear space and $Y \subseteq X$ then the linear span of $Y$ defined to be the set of finite linear combinations of elements of $Y$. Thus, when we consider the linear space $\mathrm{BL}(X^*, Y^*, \mathbf{F})$ and the subset $\{ x \otimes y : x \in X, y \in Y \}$ of $\mathrm{BL}(X^*, Y^*; \mathbf{F})$ then the linear span of $\{ x \otimes y : x \in X, y \in Y \}$ is the set $X \otimes Y$ where $u \in X \otimes Y$ if $u = \sum_{i=1}^{m} x_i \otimes y_i$ where $x_i \in X$, $y_i \in Y$ for all $1 \leq i \leq m$. Note that the representation of $u$ is not necessarily unique.

Note that for each $x \in X$ and each $y \in Y$ we have that:

(1)
\begin{align} \quad \| x \otimes y \| = \sup_{\| f \| \leq 1, \| g \| \leq 1} \{ |(x \otimes y)(f, g)| \} = \sup_{\| f \| \leq 1, \| g \| \leq 1} \{|f(x)g(y)| \} = \sup_{\| f \| \leq 1} \{ |\hat{x}(f)| \} \sup_{\| g \| \leq 1} \{ |\hat{y}(g)| \} = \| \hat{x} \| \| \hat{y} \| = \| x \| \| y \| \end{align}

Before we go any further it will first be important to verify that if $x \in X$, $y \in Y$, then $x \otimes y \in \mathrm{BL}(X^*, Y^*, \mathbf{F})$ so that $X \otimes Y$ is indeed a subspace of $\mathrm{BL}(X^*, Y^*; \mathbf{F})$.

 Proposition 1: Let $X$ and $Y$ be normed linear spaces over $\mathbf{F}$. Then for each $x \in X$ and for each $y \in Y$ we have that $x \otimes y \in \mathrm{BL} (X^*, Y^*; \mathbf{F})$.
• Proof: Fix $f_0 \in X^*$ and consider the map $g \to (x \otimes y)(f_0, g)$. For all $g_1, g_2 \in Y^*$ and all $\alpha \in \mathbf{F}$ we have that:
(2)
\begin{align} \quad (x \otimes y)(f_0, g_1 + g_2) = f_0(x)[g_1(y) + g_2(y)] = f_0(x)g_1(y) + f_0(x)g_2(y) = (x \otimes y)(f_0, g_1) + (x \otimes y)(f_0, g_2) \end{align}
(3)
\begin{align} \quad (x \otimes y) (f_0, \alpha g_1) = f_0(x) \alpha g_1(y) = \alpha f_0(x)g_1(y) = \alpha (x \oplus y) (f_0, g_1) \end{align}
• Thus for each fixed $f_0 \in X^*$ we have that $g \to (x \otimes y)(f_0, g)$ is linear. Similarly, it is easy to show that for each fixed $g_0 \in Y^*$ we have that $f \to (x \otimes y)(f, g_0)$ is linear.
• Lastly we show that $x \otimes y$ is bounded. For all $f \in X^*$, $g \in Y^*$ we have that:
(4)
\begin{align} \quad |(x \otimes y)(f, g)| = |f(x)g(y)| =|f(x)||g(y)| \leq \| f \| \| x \| \| g \| \| y \| = \underbrace{[\| x \| \| y \|]}_{\mathrm{fixed}} \| f \| \| g \| \end{align}
• So for all $x \in X$ and all $y \in Y$ we see that $x \otimes y$ is bounded and that $\| x \otimes y \| \leq \| x \| \| y \|$. Thus for each $x \in X$ and each $y \in Y$, $x \otimes y \in \mathrm{BL}(X^*, Y^*; \mathbf{F})$. $\blacksquare$