Products of Linear Spaces

# Products of Linear Spaces

Definition: Let $X$ and $Y$ be linear spaces. The Product $X \times Y$ is defined to be the space with the operators of addition and scalar multiplication defined for all $(x_1, y_1), (x_2, y_2) \in X \times Y$ and $\alpha \in \mathbb{R}$ by $(x_1, y_1) + (x_2, y_2) = (x_1 + x_2, y_1 + y_2)$ and $\alpha (x_1, y_1) = (\alpha x_1, \alpha y_1)$. |

It can be shown without too much trouble that if $X$ and $Y$ are linear spaces then their product $X \times Y$ is also a linear space.

Proposition 1: Let $(X, \| \cdot \|_X)$ and $(Y, \| \cdot \|_Y)$ be normed linear spaces. Define a norm $\| \cdot \| : X \times Y \to [0, \infty)$ for all $(x, y) \in X \times Y$ by $\| (x, y) \| = \| x \|_X + \| y \|_Y$. Then $(X \times Y, \| \cdot \|)$ is a normed linear space. |

*There are a few other ways in which we can define a norm on $X \times Y$. For example, we could define $\| (x, y) \| = \max \{ \| x \|_X, \| y \|_Y \}$.*

**Proof:**We only need to show that $\| \cdot \|$ is a norm on $X$.

- First, suppose that $\| (x, y) \| = 0$. Then $\| x \|_X + \| y \|_Y = 0$. A sum of two nonnegative values equaling zero implies that both values equal $0$. That is, $\| x \|_X = 0$ and $\| y \|_Y = 0$. But this implies that $x = 0$ and $y = 0$. So $(x, y) = (0, 0)$. Conversely, suppose that $(x, y) = (0, 0)$. Then $\| (x, y) \| = \| (0, 0) \| = \| 0 \|_X + \| 0 \|_Y = 0$.

- Now let $\alpha \in \mathbb{R}$. Then:

\begin{align} \quad \| \alpha (x, y) \| = \| (\alpha x, \alpha y) \| = \| \alpha x \|_X + \| \alpha y \|_Y = |\alpha| \| x \|_X + |\alpha| \| y \|_Y = |\alpha|(\| x \|_X + \| y \|_Y) = |\alpha| \| (x, y) \| \end{align}

- Lastly, let $(x, y), (z, w) \in X \times Y$. Then:

\begin{align} \quad \| (x, y) + (z, w) \| = \| (x + z, y + w) \| = \| x + z \|_X + \| y + w \|_Y & \leq \| x \|_X + \| z \|_X + \| y \|_Y + \| w \|_Y \\ & \leq [ \| x \|_X + \| y \|_Y] + [\| z \|_X + \| w \|_Y] \\ & \leq \| (x, y) \| + \| (z, w) \| \end{align}

- Therefore, $\| \cdot \|$ is a norm on $X \times Y$. $\blacksquare$

Proposition 1: Let $(X, \| \cdot \|_X)$ be a normed linear space. Then:a) The function of addition $+ : X \times X \to X$ is continuous on $X \times X$.b) The function of scalar multiplication $\cdot : \mathbb{R} \times X \to X$ is continuous on $\mathbb{R} \times X$. |

*Recall the sequential criterion for continuity: A function $f : X \to Y$ is continuous at $x$ if and only if for all sequences $(x_n)$ in $X$ that converges to $x$ we have that $(f(x_n))$ converges to $f(x)$.*

**Proof of a)**Let $(x_n, x_n')$ be a sequence in $X \times X$ that converges to $(y, y')$. Then $(x_n)$ and $(x_n')$ are sequences in $X$ that converge to $y$ and $y'$ respectively. So $(x_n + x_n')$ converges to $y + y'$. So $+$ is continuous on $X \times X$.

**Proof of b)**Let $(\alpha_n, x_n)$ be a sequence in $\mathbb{R} \times X$ that converges to $(\alpha, x)$. Then $(\alpha_n)$ is a sequence in $\mathbb{R}$ that converges to $\alpha$ and $(x_n)$ is a sequence in $X$ that converges to $X$. So $(\alpha_nx_n)$ converges to $\alpha x$, which shows that $\cdot$ is continuous on $\mathbb{R} \times X$. $\blacksquare$