Criteria for a Tensor u in X⊗Y to be 0

# Criteria for a Tensor u in X⊗Y to be 0

 Proposition 1: Let $X$ and $Y$ be normed linear spaces. Let $u = \sum_{i=1}^{n} x_i \otimes y_i \in X \otimes Y$. Then the following statements are equivalent: a) $u = 0$. b) $\sum_{i=1}^{n} f(x_i)g(y_i) = 0$ for all $f \in X^*$ and for all $g \in Y^*$. c) $\sum_{i=1}^{n} f(x_i)y_i = 0$ for all $f \in X^*$. d) $\sum_{i=1}^{n} x_ig(y_i) = 0$ for all $g \in Y^*$.
• Proof: $(a) \Rightarrow (b)$ Suppose that $u = 0$. Then for all $f \in X^*$ and $g \in Y^*$ we have that $u(f, g) = 0$. In particular, for all $f \in X^*$ and $g \in Y^*$ we have that:
(1)
\begin{align} \quad \sum_{i=1}^{n} f(x_i)g(y_i) = \sum_{i=1}^{n} (x_i \otimes y_i)(f, g) = u(f, g) = 0 \end{align}
• $(b) \Rightarrow (c)$ Suppose that $\sum_{i=1}^{n} f(x_i)g(y_i) = 0$ for all $f \in X^*$ and all $g \in Y^*$. Then we have that:
(2)
\begin{align} \quad 0 = \sum_{i=1}^{n} f(x_i)g(y_i) = g \left ( \sum_{i=1}^{n} f(x_i)y_i \right ) \end{align}
• Since the above equality holds for all $g \in Y^*$ we have that $\sum_{i=1}^{n} f(x_i)y_i = 0$ for all $f \in X^*$.
• $(b) \Rightarrow (d)$ Suppose that $\sum_{i=1}^{n} f(x_i)y_i = 0$ for all $f \in X^*$. Then for all $g \in Y^*$ we have that:
(3)
\begin{align} \quad 0 = g \left ( \underbrace{\sum_{i=1}^{n} f(x_i)y_i}_{=0} \right ) = \sum_{i=1}^{n} f(x_i)g(y_i) = f \left ( \sum_{i=1}^{n} x_ig(y_i) \right ) \end{align}
• Since the above equality holds for all $f \in X^*$ we have that $\sum_{i=1}^{n} x_ig(y_i) = 0$ for all $g \in Y^*$.
• $(d) \Rightarrow (a)$ Suppose that $\sum_{i=1}^{n} x_ig(y_i) = 0$ for all $g \in Y^*$. Then for all $f \in X^*$ we have that:
(4)
\begin{align} \quad 0 = f \left ( \underbrace{\sum_{i=1}^{n} x_ig(y_i)}_{=0} \right ) = \sum_{i=1}^{n} f(x_i)g(y_i) = \sum_{i=1}^{n} (x_i \otimes y_i)(f, g) = u(f, g) \end{align}
• So $u(f, g) = 0$ for all $f \in X^*$ and for all $g \in Y^*$, so $u = 0$. $\blacksquare$