Normed Linear Spaces

# Normed Linear Spaces

Recall from the Linear Spaces page that a linear space over $\mathbb{R}$ (or $\mathbb{C}$) is a set $X$ with a binary operation $+$ defined for elements in $X$ and scalar multiplication defined for numbers in $\mathbb{R}$ (or $\mathbb{C}$) with elements in $X$ that satisfy ten properties (see the aforementioned page).

We said a subset $Y \subseteq X$ is a linear subspace of $X$ if $Y$ with the same binary operation $+$ and scalar multiplication $\cdot$ restricted to $Y$ is itself a linear space.

We are about to define a special type of linear space called a normed linear space. We will first need to define what a norm on a linear space is.

 Definition: Let $X$ be a linear space over $\mathbb{R}$ (or $\mathbb{C}$). A Norm on $X$ is a nonnegative real-valued function on $X$, commonly denoted $\| \cdot \|$ such that: 1) $\| x \| = 0$ if and only if $x = 0$. 2) $\| \alpha x \| = | \alpha | \| x \|$ for all $\alpha \in \mathbb{R}$ (or $\mathbb{C}$) and for all $x \in X$. 3) $\| x + y \| \leq \| x \| + \| y \|$ for all $x, y \in X$. If only properties (2) and (3) from above hold, then $\| \cdot \|$ is instead called a Seminorm.
 Definition: A Normed Linear Space is a pair $(X, \| \cdot \|)$ where $X$ is a linear space and $\| \cdot \| : X \to [0, \infty)$ is a norm on $X$.

The terms "normed linear space", "normed vector space", and "normed space" can be used interchangeably.

When we have a normed linear space $(X, \| \cdot \|)$, we can quite naturally obtain a metric space by defining a metric $d$ on $X$ in terms of the given norm $\| \cdot \|$.

 Theorem 1: If $X$ is a normed linear space over $\mathbb{R}$ (or $\mathbb{C}$) with norm $\| \cdot \|$ and if $d : X \times X \to [0, \infty)$ is defined for all $x, y \in X$ by $d(x, y) = \| x - y \|$ then $(X, d)$ is a metric space.
• Proof: We show that $d$ is a metric on $X$.
• Let $x, y \in X$. Then:
(1)
\begin{align} \quad d(x, y) = \| x - y \| = \| -1(y - x) \| = |-1| \| y - x \| = 1 \| y - x \| = \| y - x \| = d(y, x) \end{align}
• Let $x, y \in X$ and suppose that $d(x, y) = 0$. Then:
(2)
• Lastly, let $x, y, z \in X$. Then:
(3)
\begin{align} \quad d(x, z) = \| x - z \| = \| (x - y) + (y - z) \| \leq \| x - y \| + \| y - z \| = d(x, y) + d(y, z) \end{align}
• Hence $d$ is a metric on $X$ and $(X, d)$ is a metric space. $\blacksquare$

We give a special name to the metric space that can be obtained by a normed linear space.

 Definition: If $X$ is a normed linear space over $\mathbb{R}$ (or $\mathbb{C})$ with norm $\| \cdot \|$ then the Metric Space of $X$ Induced by $\| \cdot \|$ is the metric space $(X, d)$ where $d : X \times X \to [0, \infty)$ is defined for all $x, y \in X$ by $d(x, y) = \| x - y \|$.
 Definition: If $X$ is a normed linear space over $\mathbb{R}$ (or $\mathbb{C})$ with norm $\| \cdot \|$ then $X$ is said to be Complete if the metric space of $X$ induced by $\| \cdot \|$ is a complete metric space.

Recall that a metric space $(X, d)$ is said to be complete if every Cauchy sequence in $X$ converges in $X$.

 Proposition 1 (Reverse Triangle Inequality): Let $(X, \| \cdot \|_X)$ be a normed linear space. Then for all $x, y \in X$ we have that $| \| x \|_X - \| y \|_X | \leq \| x - y \|_X$.
• Proof: We have that:
(4)
\begin{align} \quad \| x \|_X = \| x - y + y \|_X \leq \| x - y \|_X + \| y \|_X \end{align}
• Therefore:
(5)
\begin{align} \quad \| x \|_X - \| y \|_X \leq \| x - y \|_X \quad (*) \end{align}
• Similarly we have that:
(6)
\begin{align} \quad \| y \|_X = \| y - x + x \|_X \leq \| x - y \|_X + \| x \|_X \end{align}
• Therefore:
(7)
\begin{align} \quad \| y \|_X - \| x \|_X \leq \| x - y \|_X \end{align}
• And so:
(8)
\begin{align} \quad \| x \|_X - \| y \|_X \geq -\| x - y \|_X \quad (**) \end{align}
• Hence:
(9)
\begin{align} \quad - \| x - y \|_X \leq \| x \|_X - \| y \|_X \leq \| x - y \|_X \Rightarrow | \| x \|_X - \| y \|_X | \leq \| x - y \|_X \quad \blacksquare \end{align}
 Proposition 2: Let $(X, \| \cdot \|_X)$ be a normed linear space. Then the function $\| \cdot \|_X : X \to \mathbb{R}$ is a continuous function on $X$.
• Proof: Let $x_0 \in X$ and let $\epsilon > 0$ be given. Let $\delta = \epsilon$. Then if $\| x - x_0 \|_X < \delta = \epsilon$ we have that:
(10)
\begin{align} \quad | \| x \|_X - \| x_0 \|_X| \leq \| x - x_0 \|_X < \delta = \epsilon \end{align}
• So $\| \cdot \|_X : X \to \mathbb{R}$ is continuous at $x_0$. Since $x_0$ was arbitrary, $\| \cdot \|_X : X \to \mathbb{R}$ is continuous on $X$. $\blacksquare$
 Proposition 3: Let $(X, \| \cdot \|_X)$ be a normed linear space. If $(x_n)$ converges to $x \in X$ then $\| x_n \|_X$ converges to $\| x \|_X$.
• Proof: Suppose that $(x_n)$ converges to $x$. Then for all $\epsilon > 0$ there exists an $n \in \mathbb{N}$ such that if $n \geq N$ then $\| x_n - x \|_X < \epsilon$. For all $n \in \mathbb{N}$ we have that:
(11)
\begin{align} \quad |\| x_n \|_X - \| x \|_X| \leq \| x_n - x \|_X < \epsilon \end{align}
• So if $n \geq N$ we have that $|\| x_n \|_X - \| x \|_X| < \epsilon$ and so $(\| x_n \|_X)$ converges to $\| x \|_X$. $\blacksquare$