12.2 Isometries[2CH]

We rewrite the definition 297 in the case of normed spaces.

Definition 334

[110]If \(M_ 1\), \(M_ 2\) are vector spaces with norms \(\| \| _{M_ 1}\) and respectively \(\| \| _{M_ 2}\), then \(𝜑\) is an isometry when

\begin{equation} ∀ x,y∈ M_ 1, \| x-y\| _{M_ 1}=\| 𝜑(x)-𝜑(y)\| _{M_ 2} \label{eq:isometria_ su_ normati} \end{equation}
335

(rewriting the definition of distance using norms).

We will compare it with this definition.

Definition 336

[111] Let \(B_ 1, B_ 2\) be two normed vector spaces. A function \(f: B_ 1 → B_ 2\) is a linear isometry if it is linear and if

\begin{equation} \left\| z \right\| _{B_ 1} = \left\| f (z) \right\| _{B_ 2} ∀ \ z ∈ B_ 1~ ~ .\label{eq:isometria_ lineare} \end{equation}
337

If \(𝜑\) is linear then the definition of equation ?? is equivalent to the definition of linear isometry seen in equation ?? (just set \(z=x-y\)). This explains why both are called ”isometries”.

By the Mazur–Ulam theorem [ 60 ] if \(M_ 1, M_ 2\) are vector spaces (on real field) equipped with norm and \(𝜑\) is a surjective isometry, then \(𝜑\) is affine (which means that \(x↦ 𝜑(x)-𝜑(0)\) is linear).

We now wonder if there are isometries that are not linear maps, or more generally affine maps.

E337

[112]Suppose the sphere \(\{ x∈ M_ 2,\| x\| _{M_ 2}=1\} \) contains no non-trivial segments: Then every function that satisfies ?? is necessarily affine.

(See also Exercise 3.)

E337

[114]The condition that \(𝜑\) is surjective cannot be removed from the Mazur–Ulam theorem. Find an example.

Hint. By the previous exercise 1, the sphere \(\{ x∈ M_ 2,\| x\| _{M_ 2}=1\} \) must contain segments.

Hidden solution: [UNACCESSIBLE UUID ’115’]