Embedding Subspaces A of BL(X) and B of BL(Y) Into BL(X⊗pY)

Proposition: Let $X$ and $Y$ be normed linear spaces and let $A$ be a linear subspace of $\mathrm{BL}(X)$ and let $B$ be a linear subspace of $\mathrm{BL}(Y)$. Then $A \otimes B$ can be embedded in $\mathrm{BL}(X \otimes_p Y)$. Moreover, the norm on $A \otimes B$ is a cross norm.

• Proof: Let $S \in A \subseteq \mathrm{BL}(X)$ and let $T \in B \subseteq \mathrm{BL}(Y)$. Consider the map $\phi: X \times Y \to X \otimes Y$ defined for all $x \in X$, $y \in Y$ by:

(1)

\begin{align} \quad \phi (x, y) = S(x) \otimes T(y) \end{align}

• Note that for each fixed $x \in X$ and all $y_1, y_2 \in Y$, $\alpha \in \mathbf{F}$ we have that:

(2)

\begin{align} \quad \phi(x, y_1 + y_2) = S(x) \otimes T(y_1 + y_2) = S(x) \otimes [T(y_1) + T(y_2)] = S(x) \otimes T(y_1) + S(x) \otimes T(y_2) = \phi(x, y_1) + \phi(x, y_2) \end{align}

(3)

\begin{align} \quad \phi(x, \alpha y_1) = S(x) \otimes T(\alpha y_1) = S(x) \otimes \alpha T(y_1) = \alpha [S(x) \otimes T(y_1)] = \alpha \phi(x, y_1) \end{align}

• So for each fixed $x$ the map $y \to \phi(x, y)$ is linear. Similarly, it can be shown that for each fixed $y$ the map $x \to \phi(x, y)$ is linear. Thus $\phi : X \times Y \to X \otimes Y$ is a bilinear map.

• By the theorem on The Existence of a Linear Map σ on X⊗Y to Z that Matches a Bilinear Map on X×Y to Z page, there exists a unique linear map (dependent of $S$ and $Y$), denote it $S \square T : X \otimes Y \to X \otimes Y$ such that for all $x \in X$ and for all $y \in Y$:

(4)

\begin{align} \quad (S \square T)(x \otimes y) = \phi(x, y) = S(x) \otimes T(y) \end{align}

• Let $u \in X \otimes Y$ be such that $u = \sum_{i} x_i \otimes y_i$. Then observe that:

(5)

\begin{align} \quad (S \square T)(u) = (S \square T) \left ( \sum_{i} x_i \otimes y_{i} \right ) = \sum_{i} (S \square T)(x_i \otimes y_i) = \sum_{i} S(x_i) \otimes T(y_i) \end{align}

• Now observe that:

(6)

\begin{align} \quad p((S \square T)(u)) = p \left ( \sum_{i} S(x_i) \otimes T(y_i) \right ) \leq \sum_{i} p(S(x_i) \otimes T(y_i)) = \sum_{i} \| S(x_i) \| \| T(y_i) \| \leq \sum_{i} \| S \| \| x_i \| \| T \| \| y_i \| = \| S \| \| T \| \sum_{i} \| x_i \| \| y_i \| \end{align}

• Since the above inequality holds true for all representations $\sum_{i} x_i \otimes y_i$ of $u$ we see that $p((S \square T)(u)) \leq \| S \| \| T \| p(u)$. So $(S \square T)$ is bounded with $\| S \square T \| \leq \| S \| \| T \|$.

• Also note that the set of elementary tensors $ \{ x \otimes y : x \in X, y \in Y \} $ is a dense subset of $ X \otimes Y $ and that $S \square T$ is a bounded linear operator defined on $X \otimes Y$. So $S \square T$ can be uniquely extended to a bounded linear operator defined on $X \otimes_p Y$, i.e., an element of $\mathrm{BL}(X \otimes_p Y)$ such that $\| S \square T \| \leq \| S \| \| T \|$.

• Furthermore, we have that:

(7)

\begin{align} \quad \| S \square T \| = \sup_{p(u) = 1} \{ p((S \square T)(u)) \} \geq \sup_{p(x \otimes y) = 1} \{ p((S \square T)(x \otimes y)) \} = \sup_{\| x \|\| y \| = 1} \{ p(S(x) \otimes T(y)) \} &= \sup_{\| x \| \| y \| = 1} \{ \| S(x) \| \|T(y) \| \} \\ & \geq \sup_{\| x \| = \| y \| = 1} \{ \| S(x) \| \| T(y) \| \} \\ &\geq \sup_{\| x \| = 1} \{ \| S(x) \| \} \sup_{\| y \| = 1} \{ \| T(y) \| \} \\ &= \| S \| \| T \| \end{align}

• Since this holds true for all elementary tensors $S \otimes T \in A \otimes B$ we have that the norm on $A \otimes B$ is a cross norm.

• So we have a mapping $\delta : A \times B \to \mathrm{BL}(X \otimes_p Y)$ such that for all $S \in A$ and all $T \in B$, $\delta (S, T) = S \square T$. By the the theorem on The Existence of a Linear Map σ on X⊗Y to Z that Matches a Bilinear Map on X×Y to Z page again, there exists a linear map $\sigma : A \otimes B \to \mathrm{BL}(X \otimes_p Y)$ such that $\sigma(S \otimes T) = S \square T$] for all $S \in A$ and all $T \in B$.

• All that remains to show is that $\sigma$ is injective. Let $u, v \in A \otimes B$ with $u = \sum_{i} S_i \otimes T_i$ and $v = \sum_{j} S_j' \otimes T_j'$. Suppose that $\sigma(u) = \sigma(v)$. Then:

(8)

\begin{align} \quad \sigma (u) &= \sigma(v) \\ \sigma \left ( \sum_{i} S_i \otimes T_i \right ) &= \sigma \left ( \sum_{j} S_j' \otimes T_j' \right ) \\ \sum_{i} \sigma (S_i \otimes T_i) &= \sum_{j} \sigma(S_j' \otimes T_j') \\ \sum_{i}S_i \square T_i &= \sum_{j} S_j' \square T_j' \end{align}

• Let $f \in A^*$ and $g \in B^*$. Then:

(9)

\begin{align} \quad u(f, g) - v(f, g) &= \left ( \sum_{i} S_i \otimes T_i \right )(f, g) - \left ( \sum_{j} S_j' \otimes T_j' \right )(f, g) \\ &= \sum_{i} (S_i \otimes T_j)(f, g) - \sum_{j} (S_j' \otimes T_j')(f, g) \\ &= \sum_{i} (S_i \square T_j)(f, g) - \sum_{j} (S_j' \square T_j')(f, g) \\ &= 0 \end{align}

• So $u = v$. Thus $\sigma$ is injective, and is an embedding of $A \otimes B$ into $\mathrm{BL}(X \otimes_p Y)$. $\blacksquare$