Banach Space
Definition1
A complete normed space is called a Banach space.
Explanation
A complete space is a space in which every Cauchy sequence converges.
A Banach space is a space satisfying all of the following conditions; since a distance function is defined on it and it is equipped with completeness, it is an extremely useful space.
- It is a vector space.
- It is a normed space. $\implies$ It is a metric space.
- It is a complete space.
Meanwhile, as an example of a Banach space, one can think of the set of continuous functions whose domain is a closed interval. This is a very simple example, and at the same time it supports various important theorems, making it a very important fact. Examples of Banach spaces include the following.
- $C[a,b]$
- $\R^{n}$
- $\mathbb{C}^{n}$
We introduce the proof for $C[a,b]$.
Proof1
Part 1. Vector Space
A continuous function defined on a closed interval has the constant function $f(x) = 0$ as its identity element and $f(x) = - f(x)$ as its inverse element. Beyond this, $C[a,b]$ satisfies the conditions of a vector space over the scalar field $\mathbb{R}$ without difficulty.
Part 2. Normed Space
For $f \in C [a,b]$, defining $\| \cdot \|$ as $\displaystyle \| f \| := \sup_{ a \le t \le b } | f (t) |$ satisfies the conditions of a norm without difficulty.
Part 3. Completeness
Let $\left\{ f_{n} \right\}_{n \in \mathbb{N}}$ be a Cauchy sequence in $C [a,b]$. In other words, for every $\varepsilon / 3 > 0$, there exists $N_{1} \in \mathbb{N}$ such that $\| f_{n} (t) - f_{m} (t) \| < \varepsilon / 3$ whenever $n,m > N_{1}$.
Since $\mathbb{R}$ is a complete space, whenever a fixed $t_{0} \in [a,b]$ is given, $\displaystyle \lim_{n \to \infty} f_{n} (t_{0})$ can be expressed, for some $f : [a,b] \to \mathbb{R}$, as
$$ \lim_{n \to \infty} f_{n} ( t_{0} ) = f ( t_{0} ) $$
Then, since $f_{n}$ is a Cauchy sequence, for arbitrary $t \in [a,b]$ there exists $N_{2}$ such that, when $m \ge N_{2}$,
$$ \begin{align*} | f(t) - f_{m} (t) | =& \left| \lim_{n \to \infty} f_{n} (t) - f_{m} (t) \right| \\ =& \lim_{n \to \infty} | f_{n} (t) - f_{m} (t) | \\ \le & \lim_{n \to \infty} \sup_{t \in [a,b] } | f_{n} (t) - f_{m} (t) | \\ =& \lim_{n \to \infty} \| f_{n} - f_{m} \| \\ <& \varepsilon / 3 \end{align*} $$
holds. Of course, there is no guarantee yet that the function $f$ is a continuous function; it has merely been defined so that, for every $t \in [a,b]$, it takes as its function value the value to which $f_{n} (t)$ eventually converges. However, from this definition we can guarantee that $f_{n}$ converges uniformly to $f$, that is, for all $x,y \in [a,b]$ and $\varepsilon / 3 > 0$, there exists $N_{3} \in \mathbb{N}$ satisfying the following simultaneously when $n \ge N_{3}$.
$$ \left| f_{n} (x) - f(x) \right| < \varepsilon / 3 \\ \left| f_{n} (y) - f(y) \right| < \varepsilon / 3 $$
Now it suffices to show that $f$ is a continuous function.
For a non-empty $E \subset \mathbb{R}$, let $f : E \to \mathbb{R}$.
If $f$ is continuous and $E$ is a bounded closed interval, then $f$ is uniformly continuous.
Since $f_{n} : [a,b] \to \mathbb{R}$ is continuous and $[a,b] \subset \mathbb{R}$ is compact, $f_{n}$ is uniformly continuous on $[a,b]$. That is, for all $x,y \in [a,b]$ and $\varepsilon / 3 > 0$, we can guarantee that there exists $\delta > 0$ such that the following holds when $|x-y| < \delta$.
$$ \left| f_{n}(x) - f_{n}(y) \right| < \varepsilon / 3 $$
Putting the above results together, when $|x-y| < \delta$ and $n \ge N_{3}$, there exist $\delta > 0$ and $N_{3} \in \mathbb{N}$ satisfying
$$ \begin{align*} |f(x) - f(y)| \le & \left| f (x) - f_{n} (x) \right| + \left| f_{n}(x) - f_{n}(y) \right| + \left| f_{n} (y) - f(y) \right| \\ =& \varepsilon / 3 + \varepsilon / 3 + \varepsilon / 3 \\ =& \varepsilon \end{align*} $$
so $f$ is uniformly continuous on $[a,b]$ and $f \in C[a,b]$. Since an arbitrary Cauchy sequence of continuous functions $\left\{ f_{n} \right\}_{n \in \mathbb{N}}$ converges uniformly to some $f \in C[a,b]$, $C[a,b]$ possesses completeness.
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