Tensor Products with Direct Sums of Normed Linear Spaces

# Tensor Products with Direct Sums of Normed Linear Spaces

 Proposition 1: Let $X$ and $Y$ be normed linear spaces and let $Y = Y_1 \oplus Y_2$ where $Y_1$ and $Y_2$ are subspaces of $Y$. Then $X \otimes Y = X \otimes (Y_1 \oplus Y_2) = (X \otimes Y_1) \oplus (X \otimes Y_2)$.
• Proof: To show the above equality we must show that $X \otimes Y = X \otimes Y_1 + X \otimes Y_2$ (that is, every tensor of $X \otimes Y$ can be written as a sum of tensors from $X \otimes Y_1$ and $X \otimes Y_2$), and that $X \otimes Y_1 \cap X \otimes Y_2 = \emptyset$.
• First, let $u = X \otimes Y$ with $u = \sum_{i=1}^{n} x_i \otimes y_i$ and $x_i \in X$, $y_i \in Y$ for all $1 \leq i \leq n$. Since $Y = Y_1 \oplus Y_2$ we have that $Y = Y_1 + Y_2$. So $y_i = y_{i,1} + y_{i,2}$ where $y_{i, 1} \in Y_1$ and $y_{i, 2} \in Y_2$ for each $1 \leq i \leq n$. Thus:
(1)
\begin{align} \quad u = \sum_{i=1}^{n} x_i \otimes y_i = \sum_{i=1}^{n} x_i \otimes (y_{i,1} + y_{i,2}) = \sum_{i=1}^{n} [x_i \otimes y_{i,1} + x_i \otimes y_{i, 2}] = \underbrace{\sum_{i=1}^{n} x_i \otimes y_{i, 1}}_{\in X \otimes Y_1} + \underbrace{\sum_{i=1}^{n} x_i \otimes y_{i, 2}}_{\in X \otimes Y_2} \end{align}
• Thus $X \otimes Y = X \otimes Y_1 + X \otimes Y_2$.
• Now let $u \in (X \otimes Y_1) \cap (X \otimes Y_2)$. Since $u \in X \otimes Y_1$ we have that $u = \sum_{i=1}^{n} x_i \otimes y_i$ where $x_i \in X$ and $y_i \in Y_1$ for all $1 \leq i \leq n$. Similarly, since $u \in X \otimes Y_2$ we have that $u = \sum_{j=1}^{m} x_j' \otimes y_j'$ where $x_j' \in X$ and $y_j' \in Y_2$ for all $1 \leq j \leq m$. Therefore:
(2)
\begin{align} \quad \sum_{i=1}^{n} x_i \otimes y_i - \sum_{j=1}^{m} x_j' \otimes y_j' = 0 \end{align}
(3)
\begin{align} \quad \sum_{i=1}^{n} f(x_i)y_i - \sum_{j=1}^{m} f(x_j')y_j' &= 0 \\ \quad \sum_{i=1}^{n} f(x_i)y_i &= \sum_{j=1}^{m} f(x_j')y_j' \end{align}
• Note that the lefthand side of the equation is in $\mathrm{span} (Y_1)$ while the righthand side of the equation is in $\mathrm{span}(Y_2)$. Since $Y_1 \cap Y_2 = \emptyset$ we see from above that $\sum_{i=1}^{n} f(x_i)y_i = 0$ for all $f \in X^*$. This is equivalent to $u = 0$. Thus $(X \otimes Y_1) \cap (X \otimes Y_2) = \emptyset$.
• We conclude that:
(4)
\begin{align} \quad X \otimes Y = X \otimes (Y_1 \oplus Y_2) = (X \otimes Y_1) \oplus (X \otimes Y_2) \quad \blacksquare \end{align}