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The $X$ is infinite dimension real inner product space, not restricting it as a Hilbert space.

This question has troubled my friend and me a lot of days. It is obviously true when $X$ is infinite dimension Hilbert space. However, I can't answer this in arbitrary space.

I have thought to find all of the style of infinite dimension space. I thought any space may like $l^2(\alpha) \times \mathbb{R}^{⊕\beta}$, but it's wrong, such as $l^2(\mathbb{N})^{⊕ω}$.

Any useful answer will be appreciated.

Add: where isomorphic maintain inner product. It isn't only as a algebraic isomorphic. $X \times \mathbb{R}$ is a inner product space, which inner product is $(x,r) \cdot (y,s) = x \cdot y + rs$.

CTuser_103
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    Do you mean as vector spaces, normed spaces, or something else? – Cameron L. Williams Apr 11 '24 at 14:56
  • Linear homeomorphism? Or just homeomorphism? – freakish Apr 11 '24 at 14:56
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    Pick an orthogonal basis, order it, and use the Hilbert hotel map. – hunter Apr 11 '24 at 14:57
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    @hunter Without being a Hilbert space that doesn’t make much sense because the coefficients for the linear combinations that are in the space could be different. (And without separability an orthogonal basis may not even exist.) – David Gao Apr 11 '24 at 16:20
  • As a vector space this is obviously true. If you need the isomorphism to be more than just a linear isomorphism, you need to specify additional structures on $X \times \mathbb{R}$. Like, is it an inner product space? If so, what is the inner product? Is the isomorphism supposed to be isometric? – David Gao Apr 11 '24 at 16:23
  • @DavidGao and freakish, This question adds more precise restriction, such as isometric and linear condition. – CTuser_103 Apr 12 '24 at 13:59
  • Assuming the Axiom of Choice, infinite dimensional vector spaces still have Hamel bases (every vector has a unique expression as a linear combination of a finite number of basis elements). Gram-Schmidt orthogonalization works on Hamel bases just as well as Schauder bases. Choose an infinite sequence in the basis elements, and consider shifting the indices along the sequence by $1$, while leaving all non-sequence basis elements unchanged. – Paul Sinclair Apr 12 '24 at 15:35
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    @PaulSinclair That wouldn’t preserve the inner product. The remaining basis elements are not necessarily orthogonal to the orthonormal basis obtained through Gram-Schmidt, so their inner product with $e_n$ might be different from their inner product with $e_{n+1}$ (after the shift). – David Gao Apr 12 '24 at 16:11
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    @PaulSinclair In fact, here is a concrete example showing this approach wouldn’t work. Let $X = {(a_n) \in l^2: \sum_n (n!)|a_n|^2 < \infty}$. Suppose the basis is given so that it starts with the standard basis of $l^2$, then extends to a full Hamel basis of $X$. Shifting the indices by $1$ means the map is given by $\mathbb{R} \to \text{span}(e_1)$ and sending $e_i$ to $e_{i+1}$. If the map is to preserve inner products, continuity and density of $c_{00}$ then fully determine this map (so you can’t just say it’s identity on the remaining basis element)… – David Gao Apr 12 '24 at 16:16
  • … So in particular, it must act as the right shift on $l^2$ restricted to $X$. But the right shift does not even preserve $X$. For example, $(\frac{1}{n\sqrt{n!}}) \in X$ but its right shift, $(\frac{1}{(n-1)\sqrt{(n-1)!}}) \notin X$. – David Gao Apr 12 '24 at 16:20
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    Also relevant, this answer here: https://math.stackexchange.com/a/201149/465145 . A non-separable inner product space may have all its orthonormal set being countable. So Gram-Schmidt fails badly if one wants to obtain a full orthonormal basis of an inner product space without completeness. (The reason it fails is because at any infinite ordinal step in the Gram-Schmidt procedure, you need orthogonal projection onto the span of infinitely many previously obtained basis elements, which does not make sense unless in a Hilbert space.) – David Gao Apr 12 '24 at 16:27

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