📂Banach Space

What is a Norm Space?

Definition¹

Let’s call $X$ a vector space. If there exists a function $\left\| \cdot \right\| : X \to \mathbb{R}$ that satisfies the following three conditions, then $\left\| \cdot \right\|$ is called the norm of $X$, and $(X,\left\| \cdot \right\| )$ is called a normed space.

(a) $\left\| x \right\| \ge 0,\quad \forall\ x \in X$ and $\left\| x \right\|=0 \iff x = 0$

(b) $\|cx\|=|c|\left\| x \right\|,\quad \forall\ x\in X,\ \forall\ c \in\mathbb{C}$

(c) $\left\| x + y \right\| \le \left\| x \right\| + \left\| y \right\|,\quad \forall\ x,y\in X$

Explanation

• The norm of the normed space $X$ can be represented as below:

  $$\left\| x \right\|_{X},\quad \left\| x, X \right\|, \quad \left\| x ; X \right\|$$

• (a) Without the condition of $\left\| x \right\|=0 \iff x = 0$, it becomes a seminorm.

• (b) means that $\left\| x - y \right\| =\|y -x\|$ holds.

• (c) is called the triangle inequality, and the inequality below is called the reverse triangle inequality. For normed spaces $(X, \left\| \cdot \right\| )$ and $x, y \in X$, the following inequality holds:

  $$\left| \left\| x \right\| - \left\| y \right\|\ \right| \le \left\| x- y \right\|$$

• The norm is a continuous mapping.

Normed Space as a Metric Space and Topological Space

Given a norm, a distance can be naturally defined. Therefore, a normed space becomes a metric space.

$$d(x,y) = d_{X}(x,y) = \left\| x - y \right\|_{X}$$

Given the distance, an open ball can be defined as below:

$$B_{d}(x,r)=B_{r}(x):=\left\{ y\in X\ :\ \left\| x - y \right\|_{X} <r \right\}$$

The collection of all open balls forms a basis in the topological sense on $X$. In other words, the open balls defined by the norm of $X$ can create a topology on $X$. This resulting topology on $X$ is called the norm topology². Additionally, if the topology of topological vector space $X$ is the norm topology, $X$ is called normable.

In summary, saying that $X$ is a normed space implies that $X$ is a vector space, a metric space, and a topological space. Therefore, in functional analysis, a given normed space is naturally treated as a metric space and a topological space as well.

Proof³

By the triangle inequality,

$$\left\| x \right\|= | (x-y) +y| \le |x-y| + \left\| y \right\|$$

holds. Therefore,

$$\begin{equation} \left\| x \right\| - \left\| y \right\| \le \left\| x- y \right\| \end{equation}$$

Similarly,

$$\left\| y \right\| = | (y - x) + x| \le \left\| y- x \right\| + \left\| x \right\|$$

thus,

$$\begin{equation} \left\| y \right\| - \left\| x \right\| \le \left\| y- x \right\|=\left\| x - y \right\| \end{equation}$$

holds. Therefore, by $(1), (2)$,

$$\left| \ \left\| x \right\| -\left\| y \right\|\ \right| \le \left\| x- y \right\|$$

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