Basic Theorems Regarding the Alg. Tens. Prod. of Two Norm. Spaces

Basic Theorems Regarding the Algebraic Tensor Product of Two Normed Spaces

Recall from The Algebraic Tensor Product of Two Normed Linear Spaces page that if $X$ and $Y$ are normed linear spaces and if $x \in X$, $y \in Y$ then we defined $x \otimes y : X^* \times Y^* \to \mathbf{F}$ for all $f \in X^*$ and all $g \in Y^*$ by:

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

We noted that $x \otimes y$ is a bilinear map on $X^* \times Y^*$ to $\mathbf{F}$ for each $x \in X$ and each $y \in Y$ and we said that the linear span of $\{ x \otimes y : x \in X, y \in Y \}$ in $\mathrm{BL}(X^*, Y^*, \mathbf{F})$ is called the algebraic tensor product of $X$ and $Y$ and is denoted $X \otimes Y$.

We will now prove some important propositions regarding such objects.

Proposition 1: Let $X$ and $Y$ be normed spaces. If $u \in X \otimes Y$ then there exists linearly independent sets $\{ x_1, x_2, ..., x_n \}$ and $\{ y_1, y_2, ..., y_n \}$ for which $\displaystyle{u = \sum_{i=1}^{n} x_i \otimes y_i}$.
  • Proof: Let $u \in X \otimes Y$. By definition, $X \otimes Y$ is the linear span of the set $\{ x \otimes y : x \in X, y \in Y \}$, and so $u$ can be expressed as a finite linear combination of elements from $X \otimes Y$.
  • Let $\{ x_1, x_2, ..., x_n \} \subset X$ and $\{ y_1, y_2, ..., y_n \} \subset Y$ such that $n$ is chosen as small as possible (which is guaranteed by the well-ordering principle) for which $\displaystyle{u = \sum_{i=1}^{n} x_i \otimes y_i}$. We aim to show that $\{ x_1, x_2, ..., x_n \}$ and $\{ y_1, y_2, ..., y_n \}$ are linearly independent sets.
  • If $n = 1$ then certainly the singleton sets $\{ x_1 \}$ and $\{ y_1 \}$ are linearly independent in $X$ and $Y$ respectively.
  • So let $n \geq 2$ and suppose that $\{ y_1, y_2, ..., y_n \}$ is not linearly independent. Without loss of generality, assume that $y_n$ can be written as a linear combination of $\{ y_1, y_2, ..., y_{n-1} \}$. Then there exists $a_1, a_2, ..., a_{n-1} \in \mathbf{F}$ for which:
(2)
\begin{align} \quad y_n = \sum_{i=1}^{n-1} a_iy_i \end{align}
  • Then observe that for all $f \in X^*$ and $g \in Y^*$ we have that:
(3)
\begin{align} \quad \left [ \sum_{i=1}^{n-1} (x_i + a_ix_n) \otimes y_i \right ](f, g) &= \left [(x_1 + a_1x_n) \otimes y_1 + (x_2 + a_2x_n) \otimes y_2 + ... + (x_{n-1} + a_{n-1}x_n) \otimes y_{n-1} \right ] (f, g) \\ &= [(x_1 + a_1x_n) \otimes y_1](f, g) + [(x_2 + a_2x_n) \otimes y_2](f, g) + ... + [(x_{n-1} + a_{n-1}x_n) \otimes y_{n-1}] (f, g) \\ &= f(x_1 + a_1x_n)g(y_1) + f(x_2 + a_2x_n)g(y_2) + ... + f(x_{n-1} + a_{n-1}x_n)g(y_{n-1}) \\ &= f(x_1)g(y_1) + a_1f(x_n)(y_1) + f(x_2)g(y_2) + a_2f(x_n)g(y_2) + ... + f(x_{n-1})g(y_{n-1}) + a_{n-1}f(x_n)g(y_{n-1}) \\ &= \sum_{i=1}^{n-1} f(x_i)g(y_i) + \sum_{i=1}^{n-1} f(x_n) g(a_i y_i) \\ &= \sum_{i=1}^{n-1} (x_i \otimes y_i)(f, g) +\left (x_n \otimes \sum_{i=1}^{n-1} a_i y_i \right ) (f, g) \\ &= \sum_{i=1}^{n-1} (x_i \otimes y_i)(f, g) + (x_n \otimes y_n)(f, g) \\ &= \sum_{i=1}^{n} (x_i \otimes y_i) (f, g) \\ &= u(f, g) \end{align}
  • Therefore $u = \sum_{i=1}^{n-1} (x_i + a_ix_n) \otimes y_i$. But this contradicts the minimality of $n$. Therefore the assumption that $\{ y_1, y_2, ..., y_n \}$ is linearly dependent is false. Thus $\{ y_1, y_2, ..., y_n \}$ is linearly independent.
  • A similar argument shows that $\{ x_1, x_2, ..., x_n \}$ is linearly independent, for if not, say $x_n = \sum_{i=1}^{n-1} a_ix_i$ then we also have that:
(4)
\begin{align} \quad u = \sum_{i=1}^{n-1} x_i \otimes (y_i + a_iy_n) \end{align}
  • Again, contradicting the minimality of $n$. Thus we conclude that that if $u \in X \otimes Y$ then there exists linearly independent sets $\{ x_1, x_2, ..., x_n \} \subset X$ and $\{ y_1, y_2, ..., y_n \} \subset Y$ for which $\displaystyle{u = \sum_{i=1}^{n} x_i \otimes y_i}$.
Proposition 2: Let $X$ and $Y$ be normed spaces and let $\{ x_1, x_2, ..., x_n \} \subset X$, $\{ y_1, y_2, ..., y_n \} \subset Y$. If $\displaystyle{\sum_{i=1}^{n} x_i \otimes y_i =0}$ and $\{ x_1, x_2, ..., x_n \}$ is linearly independent in $X$ then $y_1, y_2, ..., y_n = 0$.

Recall from the Corollaries to the Hahn-Banach Theorem page that if $X$ is a normed space and $x, y \in X$ are such that $f(x) = f(y)$ for all $f \in X^*$ then $x = y$. In the case when $y = 0$ we have that if $f(x) = 0$ for all $f \in X^*$ then $x = 0$. We use this important corollary in the proof below.

  • Proof; Let $f \in X^*$ and let $g \in Y^*$. Since $\displaystyle{\sum_{i=1}^{n} x_i \otimes y_i = 0}$ we have that:
(5)
\begin{align} \quad 0(f, g) &= \left ( \sum_{i=1}^{n} x_i \otimes y_i \right )(f, g) \\ &= ( x_1 \otimes y_1 + x_2 \otimes y_2 + ... + x_n \otimes y_n)(f, g) \\ &= f(x_1)g(y_1) + f(x_2)g(y_2) + ... + f(x_n)g(y_n) \\ &= f \left ( g(y_1)x_1 + g(y_2)x_2 + ... + g(y_n)x_n \right ) \\ &= f \left ( \sum_{i=1}^{n} g(y_i)x_i \right ) \end{align}
  • But observe that $0(f, g) = f(0)g(0) = 0$ for all $f \in X^*$ and for all $g \in Y^*$. Thus for all $f \in X^*$ we have that:
(6)
\begin{align} \quad f \left ( \sum_{i=1}^{n} g(y_i)x_i \right ) = 0 \end{align}
  • This implies that:
(7)
\begin{align} \quad \sum_{i=1}^{n} g(y_i)x_i = 0 \end{align}
  • But since $\{ x_1, x_2, ..., x_n \}$ is a linearly independent set, the above equation implies that $g(y_i) = 0$ for each $1 \leq i \leq n$. Since this holds for all $g \in Y^*$ we must have that $y_i = 0$ for each $1 \leq i \leq n$. $\blacksquare$
Proposition 3: Let $X$ and $Y$ be normed spaces. If $\{ x_1, x_2, ..., x_m \} \subset X$ and $\{ y_1, y_2, ..., y_n \} \subset Y$ are linearly independent in $X$ and $Y$ respectively, then $\{ x_i \otimes y_j : 1 \leq i \leq m, 1 \leq j \leq n \}$ is linearly independent in the algebraic tensor product $X \otimes Y$.
  • Proof: Let $a_{ij} \in \mathbf{F}$, $1 \leq i \leq m$, $1 \leq j \leq n$ and suppose that:
(8)
\begin{align} \quad 0 &= \sum_{i=1}^{m} \sum_{j=1}^{n} a_{ij} x_i \otimes y_j \\ &= \sum_{i=1}^{m} x_i \otimes \sum_{j=1}^{n} a_{ij} y_j \end{align}
  • Since $\{ x_1, x_2, ..., x_m \}$ is linearly independent in $X$, with the above equation and proposition 2 we must have that for each $1 \leq i \leq m$ that:
(9)
\begin{align} \quad \sum_{j=1}^{n} a_{ij} y_j = 0 \end{align}
  • But since $\{ y_1, y_2, ..., y_n \}$ is linearly independent, we have that the above equation implies that $a_{ij} = 0$ for each $1 \leq j \leq n$ as well. Thus we conclude that $\{ x_i \otimes y_j : 1 \leq i \leq m, 1 \leq j \leq n \}$ is linearly independent in $X \otimes Y$. $\blacksquare$
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