I want to know the following:

If $x_1, x_2, \cdots, x_n, y_1,y_2, \cdots, y_n \in \ell_p$ satisfies $\|x_i-x_j\|_p=\|y_i-y_j\|_p$ for any $i,j$, then does there exist isometry $F$ of $\ell_p$ which send each $x_i$ to $y_i$ ?

Also do you know the precise description of the isometry group of $\ell_p$ ?


2 Answers 2


To answer your last question, you must combine two classical results:

1) First, the Mazur-Ulam theorem tells you that every (surjective) isometry of a real Banach space, is affine. This reduces the description of the isometry group, to the group of linear isometries.

2) By a corollary of the Banach-Lamperti theorem, every linear isometry $T$ of $\ell^p=\ell^p(\mathbb{N})$ (with $1\leq p<\infty,p\neq 2$) is of the form $T:(a_n)\mapsto (\epsilon(n)a_{\sigma(n)})$, where $\sigma$ is a permutation of $\mathbb{N}$, and $\epsilon(n)=\pm 1$ for every $n$.

  • $\begingroup$ Thus your answer also settle the first question (with the same $p$): the condition for the existence of an isometry $F:x^i\mapsto y^i$ becomes a purely combinatorial compatibility condition on suitable subsets of $\mathbb{N}$ for the existence of $\epsilon$ and $\sigma$. $\endgroup$ Commented Nov 24, 2011 at 10:17
  • 1
    $\begingroup$ You can find the Banach-Lamperti theorem in text books, such as Royden's Real Analysis. $\endgroup$ Commented Nov 24, 2011 at 16:33
  • $\begingroup$ >Alain Valette Thank you very much! Thanks to you, I understand precisely. $\endgroup$
    – Ema
    Commented Dec 14, 2011 at 18:22

The answer to the first question is NO. Even among norms on $\mathbb R^2$, the only ones that have this amazing property (any isometry defined on a finite set extends to an isometry defined on the whole space) are those norms that make $\mathbb R^2$ isometrically into Euclidean space.


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