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Let $X$ be a vector space over $\mathbb K \in \{\mathbb R, \mathbb C\}$. Does there exists a pairing $X \times X \rightarrow \mathbb K$ that induces a Hilbert space structure on $X$?

I have been thinking that one may choose an arbitrary basis of $X$, which exists by the axiom of choice, and declare it as the orthonormal basis of the Hilbert space, but then limits of series might become unhandy.

shuhalo
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2 Answers2

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No. If $X$ is a space of countably infinite dimension, then the Baire category theorem implies that $X$ can't be given the structure of a Hilbert space (or even a Banach space). Your idea fails since, in an infinite-dimensional Hilbert space, an orthonormal basis is never an algebraic (Hamel) basis.

Chris Eagle
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You may be interested in the following paper:

Lorenz Halbeisen, and Norbert Hungerbühler. The cardinality of Hamel bases of Banach spaces, East-West Journal of Mathematics, 2, (2000) 153-159.

There, Lorenz and Norbert prove a few results about the size of Hamel bases of arbitrary infinite dimensional Banach spaces. In particular, they show:

Lemma 3.4. If $K\subseteq\mathbb C$ is a field and $E$ is an infinite dimensional Banach space over $K$, then every Hamel basis of $E$ has at least cardinality $\mathfrak c=|\mathbb C|$.

Theorem 3.5. Under the above assumptions, the size of any Hamel basis of $E$ is $|E|$.

Among other things, this implies a negative answer to your question, and addresses Miha Habič's comment to the other answer.

The argument is short, and uses a key set theoretic fact, namely, that there are independent families of size $\mathfrak c$. Recall that an independent family is a collection $I$ of infinite subsets of $\mathbb N$ with the property that for any distinct $A_1,\dots,A_m,B_1,\dots,B_n$ sets in $I$, we have that $$ \bigcap_{i=1}^m A_i\setminus\bigcup_{j=1}^n B_j\quad\mbox{ is infinite.} $$ (Lorenz has a few papers on set theoretic issues in Banach spaces, that can be downloaded from his page.)

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    Theorem 3.5 is mainly due to Löwig, Über die Dimension linearer Räume, Studia Math. 5 (1934), pp. 18–23, Satz 2: If the dimension of a real vector space $E$ is $\geq \mathfrak{c}$ then $|E| = \dim E$. Lemma 3.4 is due to Mackey, On infinite-dimensional linear spaces, Trans. AMS 57 (1945), 155-207, Theorem I-1, page 159. See also Lacey The Hamel Dimension of any Infinite Dimensional Separable Banach Space is $c$ Amer Math. Mon. 80 (1973), 298. – Martin Jun 24 '13 at 02:21
  • @Martin Thanks for the references! In a later paper, Lorenz and his collaborators refer to Mackey's paper, but I hadn't seen Löwig's paper. – Andrés E. Caicedo Jun 24 '13 at 02:39
  • Miha Habič's question doesn't seem to be answered by Theorem 3.5. For example, Halbeisen et al. prove in On Bases in Banach Spaces, Theorem 2.7 that the cardinality of a Banach space must have cofinality $> \omega$, so there are non-trivial restrictions on the possible dimensions of Banach spaces by Theorem 3.5. – Martin Jun 24 '13 at 03:16
  • Yes, quite interesting! There is more to that question than expected. I'll see what I can find beyond the facts we have listed. – Andrés E. Caicedo Jun 24 '13 at 04:02
  • In this answer julien mentions Wolfgang Arendt, Robin Nittka, Equivalent complete norms and positivity, Archiv der Mathematik May 2009, Volume 92, Issue 5, pp 414-427. In section 4.2 they assume a weakening of SCH, namely hypothesis $(H)$: Let $\kappa \gt \mathfrak{c}$ be a cardinal. Then a) if $\operatorname{cf}\kappa > \omega$ then $\kappa^{\omega} = \kappa$ and b) if $\operatorname{cf}\kappa = \omega$ then $\mu^\omega \lt \kappa$ whenever $\mu \lt \kappa$. – Martin Jun 24 '13 at 04:37
  • Using $(H)$ they show that the spaces $\ell^p(\kappa)$ have cardinality $\kappa$ and deduce that $\kappa \geq \mathfrak{c}$ and $\operatorname{cf} \kappa \gt \omega$ are necessary and sufficient for the existence of a Banach space of dimension $\kappa$. Arendt and Nittka say they don't know whether $(H)$ can be proved from $\mathsf{ZFC}$. Apparently this work was done independently of Halbeisen et al. – Martin Jun 24 '13 at 04:37