(X⊗pY)* and BL(X, Y; F) are Isometrically Isomorphic

(X⊗pY)* and BL(X, Y; F) are Isometrically Isomorphic

Recall from The Projective Tensor Product of X⊗Y page that if $X$ and $Y$ are normed linear spaces then we defined the projective tensor norm on $X \otimes Y$ for all $u \in X \otimes Y$ by:

(1)
\begin{align} \quad p(u) = \inf \left \{ \sum_{i=1}^{m} \| x_i \| \| y_i \| : u = \sum_{i=1}^{m} x_i \otimes y_i \right \} \end{align}

Where the infimum above varies through all finite representations of $U$ as elements in $X \otimes Y$. We prove that the projective tensor norm is indeed a norm on $X \otimes Y$ and we defined the projective tensor product of $X$ and $Y$ to be the completion $(X \otimes_p Y, p)$. We also proved that if $w$ denotes the weak projective norm on $X \otimes Y$ then $w(u) \leq p(u)$ for all $u \in X \otimes Y$ and furthermore, $p(x \otimes y) = \| x \| \| y \|$.

We will now prove an important result which states that the dual space $(X \otimes_p Y)^*$ is isometrically isomorphic to $\mathrm{BL}(X, Y; \mathbf{F})$.

Theorem 1: Let $X$ and $Y$ be normed spaces. For each $F \in (X \otimes_p Y )^*$ let $\phi_F : X \times Y \to \mathbf{F}$ be defined for all $x \in X$ and all $y \in Y$ by $\phi_F(x, y) = F(x \otimes y)$. Let $T : (X \otimes_p Y)^* \to \mathbf{BL}(X, Y; \mathrm{F})$ be defined for all $F \in (X \otimes_p Y)^*$ by $T(F) = \phi_F$. Then $T$ is an isometric isomorphism from $(X \otimes_p Y)^*$ to $\mathrm{BL}(X, Y; \mathbf{F})$.
  • Proof: There are a few things to verify.
  • 1. Showing that for all $F \in (X \otimes_p Y)^*$ we have that $\phi_F \in \mathrm{BL}(X, Y; \mathbf{F})$: Let $F \in (X \otimes_p Y)^*$ and consider $\phi_F$. We want to show that $\phi_F$ is a bounded bilinear map on $X$ and $Y$ to $\mathbf{F}$.
  • First, fix $x \in X$ and let $y_1, y_2 \in Y$, $\alpha \in \mathbf{F}$. Then by the linearity of $F$ we have that:
(2)
\begin{align} \quad \phi_F(x, y_1 + y_2) = F(x \otimes (y_1 + y_2)) = F(x \otimes y_1 + x \otimes y_2) = F(x \otimes y_1) + F(x \otimes y_2) = \phi_F(x, y_1) + \phi_F(x, y_2) \end{align}
(3)
\begin{align} \quad \phi_F(x, \alpha y_1) = F(x \otimes (\alpha y_1)) = F(\alpha (x \otimes y_1)) = \alpha F(x \otimes y_1) = \alpha \phi_F(x, y_1) \end{align}
  • So for each fixed $x \in X$ the map $y \to \phi_F(x, y)$ is linear. It can similarly be shown that for each fixed $y \in Y$ the map $x \to \phi_F(x, y)$ is linear. Furthermore, the outputs of $\phi_F$ are elements of $\mathbf{F}$. So for all $F \in (X \otimes_p Y)^*$ we have that $\phi_F$ is linear. Furthermore we have that for all $x \in X$, $y \in Y$, since $F$ is bounded:
(4)
\begin{align} \quad \| \phi_F(x, y) \| = \| F(x \otimes y) \| \leq \| F \| p(x \otimes y) = \| F \| \| x \| \| y \| \end{align}
  • So $\phi_F$ is bounded. Thus for all $F \in (X \otimes_p Y)^*$ we have that $\phi_F \in \mathrm{BL}(X, Y; \mathbf{F})$.
  • 2. Showing that $T : (X \otimes_p Y)^* \to \mathrm{BL}(X, Y; \mathbf{F})$ is linear: Let $F, G \in (X \otimes_p Y)^*$ and let $\alpha \in \mathbf{F}$. We first want to show that $T(F + G) = T(F) + T(G)$, that is, $\phi_{F + G} = \phi_F + \phi_G$. We then want to show that $T(\alpha F) = \alpha T(F)$, that is, $\phi_{\alpha F} = \alpha \phi_F$.
  • So let $x \in X$, $y \in Y$. Then:
(5)
\begin{align} \quad \phi_{F + G}(x, y) = (F + G)(x \otimes y) = F(x \otimes y) + G(x \otimes y) = \phi_F(x, y) + \phi_G(x, y) \end{align}
  • So indeed, $\phi_{F + G} = \phi_F + \phi_G$. Now let $x \in X$ and $\alpha \in \mathbf{F}$. Then:
(6)
\begin{align} \quad \phi_{\alpha F}(x, y) = (\alpha F)(x \otimes y) = \alpha F(x \otimes y) = \alpha \phi_F(x, y) \end{align}
  • So indeed $\phi_{\alpha F} = \alpha \phi_F$. Therefore $T$ is linear.
  • 3. Showing that $T$ is bounded with $\| T \| \leq 1$: For all $F \in (X \otimes_p Y)^*$ we have that:
(7)
\begin{align} \quad \| T(F) \| = \| \phi_F \| = \sup_{\| x \| \leq 1, \| y \| \leq 1} \{| \phi_F(x, y) | \} &= \sup_{\| x \| \leq 1, \| y \| \leq 1} \{ |F(x \otimes y)| \} \\ & \leq \sup_{\| x \| \leq 1, \| y \| \leq 1} \{ \| F \| p(x \otimes y) \} \\ & \leq \| F \| \sup_{\| x \| \leq 1, \| y \| \leq 1} \{ \| x \| \| y \| \} \\ & \leq \| F \| \end{align}
  • So $T$ is bounded with $\| T \| \leq 1$.
(8)
\begin{align} \quad \quad \left | F \left ( \sum_{i=1}^{m} x_i \otimes y_i \right ) \right | = \left | \sum_{i=1}^{m} F(x_i \otimes y_i) \right | \leq \sum_{i=1}^{m} |F(x_i \otimes y_i)| = \sum_{i=1}^{m} |\phi(x_i \otimes y_i)| \leq \sum_{i=1}^{m} \| \phi \| p(x_i \otimes y_i) = \| \phi \| \sum_{i=1}^{m} \| x_i \| \| y_i \| \end{align}
  • So $|F(u)| \leq \| \phi \| \sum_{i=1}^{m} \| x_i \| \| y_i \|$ for all representations $\sum_{i=1}^{m} x_i \otimes y_i$ of $u$, and so, $|F(u)| \leq \| \phi \| p(u)$. Hence for all $u \in X \otimes Y$ we have that $|F(u)| \leq \| \phi \| p(u)$, so:
(9)
\begin{align} \quad \| F \| = \sup_{p(u) \leq 1} \{ |F(u)| \} \leq \sup_{p(u) \leq 1} \{ \| \phi \| p(u) \} = \| \phi \| \sup_{p(u) \leq 1} \{ p(u) \} \leq \| \phi \| \end{align}
  • Since this holds for all $\phi \in \mathrm{BL}(X, Y; \mathbf{F})$, it certainly holds for all $T(F) = \phi_F \in \mathrm{BL}(X, Y; \mathbf{F})$, and so $\| F \| \leq \| T(F) \|$ for all $F \in (X \otimes_p Y)^*$.
  • Thus we conclude that $T : (X \otimes_p Y)^* \to \mathrm{BL}(X, Y; \mathbf{F})$ is an isometric isomorphism of $(X \otimes_p Y)^*$ to $\mathrm{BL}(X, Y; \mathbf{F})$. $\blacksquare$
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