The Closed Unit Ball of X⊗pY

# The Closed Unit Ball of X⊗pY

Theorem 1: Let $X$ and $Y$ be normed linear spaces. Let $B_X$ denote the closed unit ball of $X$, $B_Y$ the closed unit ball of $Y$, and $B_{X \otimes Y}$ the closed unit ball of $X \otimes Y$ with respect to the norm $p$. Then $B_{X \otimes Y}$ is the closed convex hull of $B_X \otimes B_Y$. |

*If $X$ and $Y$ are normed linear spaces and $A \subseteq X$, $B \subseteq Y$ then we define $A \otimes B = \{ x \otimes y : x \in A, y \in B \}$. In the statement of the above theorem, $B_X \otimes B_Y = \{ x \otimes y : x \in B_X, y \in B_Y \} = \{ x \otimes y : \| x \| \leq 1, \| y \| \leq 1 \}$.*

**Proof:**Let $u$ be a tensor in the open unit ball of $(X \otimes Y, p)$. Then $p(u) < 1$. So there exists a representation $u = \sum_{i=1}^{n} x_i \otimes y_i$ for which:

\begin{align} \quad p(u) = \sum_{i=1}^{n} \| x_i \| \| y_i \| < 1 \end{align}

- Moreover, we may assume each of the $\| x_i \|$ and $\| y_i \|$ are nonzero by taking the $\{ x_1, x_2, ..., x_n \}$ and $\{ y_1, y_2, ..., y_n \}$ to be linearly independent. For each $1 \leq i \leq n$ let:

\begin{align} \quad w_i = \frac{x_i}{\| x_i \|} \quad \mathrm{and} \quad z_i = \frac{y_i}{\| y_i \|} \end{align}

- Also, for each $1 \leq i \leq n$ let $\lambda_i = \| x_i \| \| y_i \|$. Then we have that:

\begin{align} \quad u = \sum_{i=1}^{n} x_i \otimes y_i = \sum_{i=1}^{n} \| x_i \| \| y_i \| \left ( \frac{x_i}{\| x_i \|} \otimes \frac{y_i}{\| y_i \|} \right ) = \sum_{i=1}^{n} \lambda_i (w_i \otimes z_i) \end{align}

- Where $0 < \lambda_i < 1$, $w_i \in B_X$, and $z_i \in B_Y$ for all $1 \leq i \leq n$. Moreover:

\begin{align} \quad \sum_{i=1}^{n} \lambda_i = \sum_{i=1}^{n} \| x_i \| \| y_i \| = p(u) < 1 \end{align}

- So $u$ is contained in a convex combination of the tensors $w_1 \otimes z_1$, $w_2 \otimes z_2$, …, $w_n \otimes z_n$ in $B_X \otimes B_Y$. Thus $u$ is contained in the convex hull of $B_X \otimes B_Y$.

- Therefore the closure of the open unit ball of $(X \otimes Y, p)$ is contained in the closed convex hull of $B_X \otimes B_Y$. So $B_{X \otimes Y}$ is contained in the closed convex hull of $B_X \otimes B_Y$.

- For the reverse inclusion, observe that $B_X \otimes B_Y \subseteq B_{X \otimes Y}$ sincce given $u \in B_X \otimes B_Y$, say $u = x \otimes y$ with $\| x \| \leq 1$, $\| y \| \leq 1$ we have that $p(u) = p(x \otimes y) = \| x \| \| y \| \leq 1$ and so $u \in B_{X \otimes Y}$. So indeed, $B_X \otimes B_Y \subseteq B_{X \otimes Y}$. So the closed convex hull of $B_X \otimes B_Y$ is also contained in $B_{X \otimes Y}$.

- Thus $B_{X \otimes Y}$ is the closed convex hull of $B_X \otimes B_Y$. $\blacksquare$

Corollary 2: Let $X$ and $Y$ be normed linear spaces. Then $B_{X \otimes_p Y}$ is the closed convex hull of $B_X \otimes B_Y$. |

**Proof:**Again, let $B_{X \otimes Y}$ denote the closed unit ball in $X \otimes Y$ with respect to the projective tensor norm $p$. By definition, $X \otimes_p Y$ is the completion of $X \otimes Y$ with the projective tensor norm. So the closed unit ball of $X \otimes_p Y$ - call it $B_{X \otimes_p Y}$ is also the closed convex hull of $B_X \otimes B_Y$. $\blacksquare$