Banach–Mazur theorem

In functional analysis, a field of mathematics, the Banach–Mazur theorem is a theorem roughly stating that most well-behaved normed spaces are subspaces of the space of continuous paths. It is named after Stefan Banach and Stanisław Mazur.

Statement
Every real, separable Banach space $(X, ⋅)$ is isometrically isomorphic to a closed subspace of $C^{0}([0, 1], R)$, the space of all continuous functions from the unit interval into the real line.

Comments
On the one hand, the Banach–Mazur theorem seems to tell us that the seemingly vast collection of all separable Banach spaces is not that vast or difficult to work with, since a separable Banach space is "only" a collection of continuous paths. On the other hand, the theorem tells us that $C^{0}([0, 1], R)$ is a "really big" space, big enough to contain every possible separable Banach space.

Non-separable Banach spaces cannot embed isometrically in the separable space $C^{0}([0, 1], R)$, but for every Banach space $X$, one can find a compact Hausdorff space $K$ and an isometric linear embedding $j$ of $X$ into the space $C(K)$ of scalar continuous functions on $K$. The simplest choice is to let $K$ be the unit ball of the continuous dual $X&thinsp;′$, equipped with the w*-topology. This unit ball $K$ is then compact by the Banach–Alaoglu theorem. The embedding $j$ is introduced by saying that for every $x ∈ X$, the continuous function $j(x)$ on $K$ is defined by


 * $$ \forall x' \in K: \qquad j(x)(x') = x'(x).$$

The mapping $j$ is linear, and it is isometric by the Hahn–Banach theorem.

Another generalization was given by Kleiber and Pervin (1969): a metric space of density equal to an infinite cardinal $α$ is isometric to a subspace of $C^{0}([0,1]^{α}, R)$, the space of real continuous functions on the product of $α$ copies of the unit interval.

Stronger versions of the theorem
Let us write $C^{k}[0, 1]$ for $C^{k}([0, 1], R)$. In 1995, Luis Rodríguez-Piazza proved that the isometry $i : X → C^{0}[0, 1]$ can be chosen so that every non-zero function in the image $i(X)$ is nowhere differentiable. Put another way, if $D ⊂ C^{0}[0, 1]$ consists of functions that are differentiable at at least one point of $[0, 1]$, then $i$ can be chosen so that $i(X) ∩ D = {0}.$ This conclusion applies to the space $C^{0}[0, 1]$ itself, hence there exists a linear map $i : C^{0}[0, 1] → C^{0}[0, 1]$ that is an isometry onto its image, such that image under $i$ of $C^{0}[0, 1]$ (the subspace consisting of functions that are everywhere differentiable with continuous derivative) intersects $D$ only at $0$: thus the space of smooth functions (with respect to the uniform distance) is isometrically isomorphic to a space of nowhere-differentiable functions. Note that the (metrically incomplete) space of smooth functions is dense in $C^{0}[0, 1]$.