I know that any non-separable Hilbert space $\mathcal{H}$ does not have a Schauder basis.

However, does a non-separable Hilbert space $\mathcal{H}$ have a "basis" $B$ in the following sense?

  • $B$ is finite, countable or uncountable (i.e., no restrictions on the size of $B$, unlike for Schauder bases)
  • $B$ is orthonormal: For all $b,b' \in B$, $\langle b,b' \rangle = 0$ for $b \neq b'$ and $1$ for $b=b'$
  • $B$ spans $\mathcal{H}$ in the following sense: Any $x \in \mathcal{H}$ can be written as $\sum_{k \in \mathbb{N}} b_k \langle b_k ,x \rangle$ for a countable subset of $B$: $\{ b_k \}_{k \in \mathbb{N}} \subseteq B$ (I assume $b_k \neq b_l$ for $k \neq l$)

If yes, what is the correct name to refer to such a basis? If no, are there additional assumptions on a non-separable $\mathcal{H}$ that allow such a basis?


This is known as an "orthonormal basis". See any book or https://en.wikipedia.org/wiki/Orthonormal_basis. Each Hilbert space has such an orthonormal basis $B$. Each $x$ can be written as $x = \Sigma_{b \in B} \langle x, b \rangle b$. You can verify that only countably many $\langle x, b \rangle$ are non-zero (otherwise the sum would not be well-defined).

  • $\begingroup$ The Wikipedia article you cite claims that not every Hilbert space has an orthonormal basis, contradicting your own claim. Do you know which is correct? $\endgroup$ – Anonymous Feb 10 at 7:30
  • $\begingroup$ @Anonymous It is true. See math.stackexchange.com/q/218092. However, it seems that I did not properly read the Wikipedia article stating "that every Hilbert space admits a basis, but not orthonormal base". This is a mistake. What is true is that not every pre-Hilbert space has an orthonormal basis. $\endgroup$ – Paul Frost Feb 10 at 9:19
  • $\begingroup$ Interesting. What definition of Hilbert space are you using? Do you assume separable? $\endgroup$ – Anonymous Feb 10 at 12:39
  • $\begingroup$ A Hilbert space need not be separable. If it separable if and only if it has a countable orthonormal basis. $\endgroup$ – Paul Frost Feb 10 at 17:50
  • $\begingroup$ Okay, that’s the definition I assumed OP was working with (some books seem to assume separable in their definition). That’s fascinating that every Hilbert space has an orthonormal basis. $\endgroup$ – Anonymous Feb 11 at 3:56

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