ℓ1⊗pX is Isometrically Isomorphic to ℓ1(X) for Banach Spaces X

# ℓ1⊗pX is Isometrically Isomorphic to ℓ1(X) for Banach Spaces X

 Theorem 1: Let $X$ be a Banach space. Then $\ell^1 \otimes_p X$ is isometrically isomorphic to $\ell^1(X)$.

Here, $\ell^1$ denotes the space of all complex-valued sequences that are absolutely summable, and $\ell^1(X)$ denotes the space of all sequences in $X$ that are absolutely summable. Recall that both $\ell^1$ and $\ell^1(X)$ are Banach spaces.

• Let $X$ be a Banach space. Tensors in $\ell^1 \otimes X$ can be viewed as sequences in $X$. For each $(a_n) \in \ell^1$ and $x \in X$ we have that:
(1)
\begin{align} \quad (a_n) \otimes x = (a_nx) \end{align}
• Observe that the sequence $(a_nx)$ is indeed a sequence in $X$ since each $a_n \in \mathbf{F}$. Moreover, the sequence $(a_nx)$ is absolutely summable since:
(2)
\begin{align} \quad \sum_{n=1}^{\infty} \| a_nx \| = \sum_{n=1}^{\infty} |a_n| \| x \| = \| (a_n) \|_1 \| x \| \end{align}
• Let $\ell^1(X)$ be the space of all sequences in $X$ that are absolutely summable with norm - also denoted by $\| \cdot \|_1$ defined for all $(x_n) \in \ell^1(X)$ by:
(3)
\begin{align} \quad \| (x_n) \|_1 = \sum_{n=1}^{\infty} \| x_n \| \end{align}
• Let $J : \ell^1 \times X \to \ell^1(X)$ be defined for all $(a_n) \in \ell^1$ and all $x \in X$ by:
(4)
\begin{align} \quad J((a_n), x) = (a_nx) \end{align}
• Then $J$ is certainly a bilinear map since:
(5)
\begin{align} \quad J((a_n) + (b_n), x) &= ([a_n + b_n]x) = (a_nx) + (b_nx) = J((a_n), x) + J((b_n, x) & \forall (a_n), (b_n) \in \ell^1, \: x \in X \\ \quad J(\alpha (a_n), x) &= ([\alpha a_n]x) = \alpha (a_nx) = \alpha J((a_n), x) & \forall (a_n) \in \ell^1, \: x \in X, \: \alpha \in \mathbf{F} \end{align}
(6)
\begin{align} \quad J((a_n), x + y) &= (a_n(x + y)) = (a_nx) + (a_ny) = J((a_n), x) + J((a_n), y) & \forall (a_n) \in \ell^1, \: x, y \in X \\ \quad J((\alpha), \alpha x) &= (a_n \alpha x) = \alpha (a_n x) = \alpha J((a_n), x) & \forall (a_n) \in \ell^1, \: x \in X, \: \alpha \in \mathbf{F} \end{align}
(7)
\begin{align} \quad J((a_n) \otimes x) = (a_nx) \quad \forall (a_n) \in \ell^1, \: \forall x \in X \end{align}
• Let $u \in \ell^1 \otimes X$ with $u = \sum_{i=1}^{m} (a_{in})_{n=1}^{\infty} \otimes x_i$ then observe that:
(8)
\begin{align} \quad \| J(u) \|_1 = \left \| J \left ( \sum_{i=1}^{m} (a_{in})_{n=1}^{\infty} \otimes x_i \right ) \right \| = \left \| \sum_{i=1}^{m} J((a_{in})_{n=1}^{\infty} \otimes x_i) \right \|_1 = \left \| \sum_{i=1}^{m} (a_{in})_{n=1}^{\infty}x_i \right \|_1 = \sum_{n=1}^{\infty} \left \| \sum_{i=1}^{m} a_{in}x_i \right \| \leq \sum_{n=1}^{\infty} \sum_{i=1}^{m} \| a_{in} \| \| x_i \| = \sum_{i=1}^{m} \| (a_{in}) \|_1 \| x_i \| \end{align}
• Since the above inequality holds for any arbitrary representation $\sum_{i=1}^{m} (a_{in})_{n=1}^{\infty} \otimes x_i$ of $u$ we have that:
(9)
• On the otherhand, let $u \in \ell^1 \otimes X$ with $u = \sum_{i=1}^{m} (a_{in})_{n=1}^{\infty} \otimes x_i$. Then:
(10)
\begin{align} \quad J(u) = J \left ( \sum_{i=1}^{m} (a_{in})_{n=1}^{\infty} \otimes x_i \right ) = \sum_{i=1}^{m} (a_{in}x_i)_{n=1}^{\infty} = \left ( \sum_{i=1}^{m} a_{in}x_i \right)_{n=1}^{\infty} \end{align}
• For each $n \in \mathbb{N}$ let $u_n = \sum_{i=1}^{m} a_{in}x_i$. Then from above we see that $J(u) = (u_n)$.
• Let $\{ e_n \}_{n \in \mathbb{N}}$ be the standard basis for $\ell^1$ where $e_1 = (1, 0, 0, 0, ...)$, $e_2 = (0, 1, 0, 0, ...)$, and so forth.
• We have that:
(11)
\begin{align} \quad p \left ( u - \sum_{n=1}^{k} e_n \otimes u_n \right ) &= p \left ( \sum_{i=1}^{m} (a_{in})_{n=1}^{\infty} \otimes x_i - \sum_{n=1}^{k} e_n \otimes \left ( \sum_{i=1}^{m} a_{in}x_i \right ) \right ) \\ &= p \left ( \sum_{i=1}^{m} (a_{in})_{n=1}^{\infty} \otimes x_i - \sum_{n=1}^{k} \sum_{i=1}^{m} e_n \otimes a_{in}x_i \right ) \\ &= p \left ( \sum_{i=1}^{m} \left ( (a_{in})_{n=1}^{\infty} \otimes x_i - \sum_{n=1}^{k}e_n \otimes a_{in}x_i \right ) \right ) \\ &= p \left ( \sum_{i=1}^{m} \left ( (a_{in})_{n=1}^{\infty} \otimes x_i - \sum_{n=1}^{k} a_{in}e_n \otimes x_i \right ) \right ) \\ &= p \left ( \sum_{i=1}^{m} \left ( \left ((a_{in})_{n=1}^{\infty} - \sum_{n=1}^{k} a_{in}e_n \right ) \otimes x_i \right ) \right ) \\ &\leq \sum_{i=1}^{m} p \left (\left ((a_{in})_{n=1}^{\infty} - \sum_{n=1}^{k} a_{in}e_n \right ) \otimes x_i \right ) \\ &\leq \sum_{i=1}^{m} \left \| (a_{in})_{n=1}^{\infty} - \sum_{n=1}^{k} a_{in}e_n \right \|_1 \| x_i \| \end{align}
• But observe that $\lim_{k \to \infty} \| (a_{in})_{n=1}^{\infty} - \sum_{n=1}^{k} a_{in}e_n \|_1 = 0$. Therefore we see that:
(12)
\begin{align} \quad \lim_{k \to \infty} p \left ( u - \sum_{n=1}^k e_n \otimes u_n \right ) = 0 \end{align}
• This shows us that $\sum_{n=1}^{\infty} e_n \otimes u_n$ converges to $u$ in $\ell^1 \otimes_p X$. Therefore:
(13)
\begin{align} \quad p(u) = p \left ( \sum_{n=1}^{\infty} e_n \otimes u_n \right ) \leq \sum_{n=1}^{\infty} p(e_n \otimes u_n) = \sum_{n=1}^{\infty} \| e_n \|_1 \| u_n \| = \sum_{n=1}^{\infty} \| u_n \| = \| J(u) \|_1 \quad (**) \end{align}
• So we conclude from $(*)$ and $(**)$ that for all $u \in \ell^1 \otimes X$:
(14)
\begin{align} \quad p(u) = \| J(u) \|_1 \end{align}
• So in fact, we actually have that $J$ is an linear isometry of $(\ell^1 \otimes X, p)$ with $\ell^1(X)$. Moreover, $J$ can be extended to a unique linear isometry $\bar{J} : \ell^1 \otimes_p X \to \ell^1(X)$. It can be shown that $\bar{J}$ is surjective by showing that for all $(x_n) \in \ell^1(X)$ we have that $u = \sum_{n=1}^{\infty} e_n \otimes x_n \in \ell^1 \otimes_p X$ is such that $\bar{J}(u) = (x_n)$. So we actually have that $\ell^1 \otimes_p X$ is isometrically isomorphic to $\ell^1(X)$. $\blacksquare$