Examples of Banach Spaces

• normed linear space
• complete metric space
• the metric induced by a norm
• continuity

A complete normed linear space is called a Banach space.

Some common used functional spaces are listed below.

1. \(Y^X\): the set of all functions from \(X\) to \(Y\).
2. \(\mathcal{B}(X;Y)\): the set of all bounded functions from \(X\) to \(Y\).
3. \(\mathcal{C}(X;Y)\): the set of all continuous functions from \(X\) to \(Y\).

Given a metric space \((Y,d)\), the sup metric on \(\mathcal{B}(X;Y)\) corresponding to \(d\) is defined by \[ \rho(f,g):=\sup_{x\in X}d(f(x),g(x)). \]

The standard bounded metric \(\bar{d}\) derived from \(d\) is defined by \[ \bar{d}(x,y):=\min(d(x,y),1). \]

Given a metric space \((Y,d)\), the uniform metric on \(Y^X\) corresponding to \(d\) is defined by \[ \bar{\rho}(f,g):=\sup_{x\in X}\bar{d}(f(x),g(x)). \] Clearly, the relation between uniform metric and sup metric is \[ \bar{\rho}(f,g) = \min(\rho(f,g),1). \]

Given a normed linear space \((Y,\|\cdot\|)\), the sup norm on \(Y^X\) corresponding to \(\|\cdot\|\) is defined by \[ \|f\|_\infty:=\sup_{x\in X}\|f(x)\|. \]

Lemma 1. If \((Y,d)\) is complete, then \((Y^X,\bar{\rho})\) is also complete.

Lemma 2. Let \(X\) be a topological space and \((Y,d)\) be a metric space (not necessarily complete), then \(\mathcal{B}(X;Y)\) and \(\mathcal{C}(X;Y)\) are both closed sets in \((Y^X,\bar{\rho})\).

Theorem 3. Let Let \(X\) be a topological space and \((Y,d)\) be a complete metric space, then \(\mathcal{B}(X;Y)\) and \(\mathcal{C}(X;Y)\) are both complete in the uniform metric.

Example. \((\mathcal{C}[a,b],\|\cdot\|_\infty)\) is a Banach space.

Example. \((\mathcal{B}[a,b],\|\cdot\|_\infty)\) is a Banach space.

Example. \((\ell_\infty,\|\cdot\|_\infty)\) is a Banach space, where \[ \ell_\infty := \{\mathbf{x}\in\mathbb{R}^\mathbb{N}\mid \|\mathbf{x}\|_\infty < \infty\}. \]

Example. \((\mathcal{C}^1[a,b],\|\cdot\|_\infty)\) is not complete, because \(\mathcal{C}^1[a,b]\) is not closed in \((Y^X,\bar{\rho})\). Consider \(f_n(x):=\sqrt{x^2+\frac{1}{n}}\). It is easy to check that \(f_n\) converges to \(|x|\).