Yes, there exists a finite subset ( F ) of ( \ell_2 ) and an infinite-dimensional Banach space ( X ) such that ( F ) does not admit an isometric embedding into ( X ).

Example: Let ( F ) be the set of four vertices of a square in ( \mathbb{R}^2 ), specifically ( F = {(0,0), (1,0), (1,1), (0,1)} ). This is a finite subset of ( \ell_2 ) (since ( \mathbb{R}^2 ) is a subspace...

Short answer: This is open.

More precisely, it is not hard to see that many natural infinite-dimensional Banach spaces cannot serve as counterexamples. In particular, no Lp-space (1 ≤ p < ∞) can be a witness X, because the whole Hilbert space embeds isometrically (up to a fixed multiplicative constant, which can be absorbed by scaling the image) into Lp via the Gaussian (or Rademacher) constructi...

At the moment nobody knows the answer – no counter-example is known and no general positive theorem has been proved.

What is known so far

1. Almost–isometric embeddings are always possible.
   • By Dvoretzky’s theorem (or by the Johnson–Lindenstrauss lemma) every infinite–dimensional Banach space X contains, for every k∈ℕ and every ε>0, a k–dimensional subspace which is (1+ε)–isomorphic to ℓ...

Isometric Embeddings of Finite Subsets of ℓ₂ into Infinite-Dimensional Banach Spaces

This question asks whether there exists a finite subset of the Hilbert space and an infinite-dimensional Banach space such that cannot be isometrically embedded into .

Current State of the Problem

Based on the information provided, this appears to be an open problem in metric geomet...

No genuinely “new’’ counter‐example is known. In fact the question

“Does there exist a finite Euclidean metric space (F\subset \ell_2) and an infinite–dimensional Banach space (X) so that no isometric embedding (F\to X) exists?”

is still open. What one knows is roughly the following:

1. (Dvoretzky–Milman) For every infinite–dimensional (X) and every finite (F\subset\ell_2), and...

Excellent question. The answer is yes, such a finite subset and a Banach space exist.

This question touches upon a fundamental difference between the metric geometry of Hilbert spaces and that of general Banach spaces. While Hilbert spaces are uniquely characterized by the parallelogram law, other Banach spaces can have very different "local" geometries, which can prevent them from accommodating ...

Short answer
At present nobody knows the answer; no concrete pair
  F ⊂ ℓ₂ (finite) and X (infinite–dimensional Banach)
with “F does not embed isometrically into X’’ has been produced.
Equally, nobody has been able to prove that such a counter-example does not exist. The problem is therefore still open.

What is known

1. Positive results (existence of many embeddings).
   • Any finite...

<think> We are given a question: Does there exist a finite subset \( F \) of \( \ell_2 \) and an infinite-dimensional Banach space \( X \) such that \( F \) does not admit an isometric embedding into \( X \)?

First, note that ( \ell_2 ) is the space of square-summable sequences, which is a Hilbert space. The subset ( F ) is finite, so it is a finite metric space with the metric induced by the...

This is a fascinating open problem in the geometry of Banach spaces. Let me analyze what we know about this question.

The Problem

The question asks whether every finite subset of can be isometrically embedded into any infinite-dimensional Banach space. This is a natural extension of known positive results for special classes of metric spaces.

What We Know Works

1. **Ultrametric ...