# Sufficient Condition to be Hilbert Space: Convergence of series of orthonormal vectors with $\ell^2$ coefficients implies completeness?

Let $$X$$ be an inner product space.

Consider the following statements.

(a) $$X$$ is complete

(b) If $$(u_n)$$ is any orthonormal sequence and $$(c_n)$$ is any sequence of scalars with $$\sum |c_n|^2 < \infty$$, then $$\sum c_n u_n$$ converges to some $$x \in X$$.

The implication (a) $$\Rightarrow$$ (b) is a basic result in the theory of Hilbert spaces. Some authors call it the Riesz-Fischer theorem. It is key to the proof that every separable Hilbert space is isomorphic to $$\ell^2$$.

Question: Is the converse (b) $$\Rightarrow$$ (a) true?

Proof Attempt: We try to prove the contrapositive: ~(a) $$\Rightarrow$$ ~(b). Suppose $$X$$ is incomplete. Then $$X$$ is not finite dimensional, so it has an infinite sequence of linearly independent vectors. Use Gram-Schmidt to form an infinite sequence of orthonormal vectors: $$(u_n)$$. Let $$M = \text{span}(u_n)$$. Let $$c_n = 1/n$$. Then $$\sum c_n u_n$$ does not converge to an element of $$M$$. However, this doesn't prove that $$\sum c_n u_n$$ does not converge to an element of $$X$$. $$M$$ may be strictly smaller than $$X$$. (I think $$X=C([0,1])$$ with $$L^2$$ norm is an example. I think other examples can be obtained by looking at a discontinuous linear functional on a Hilbert space and taking $$X$$ to be its kernel).

• The converse is true. I expect that we already have this on the site, so I'm searching rather than writing an answer right now. Commented Jul 26, 2020 at 18:25
• Your are kind of going in the right direction. Let $x_n$ be a Cauchy sequence. Do Gram-Schmidt to get $u_n$ an orthonormal system (suppose $u_n$ is not a finite system). If (b) is true then the closure of the span of the $u_n$ is isometric to $\ell^2(\Bbb N)$ (the map from $\ell^2(\Bbb N)$ into this span, mapping a sequence $c_n$ to $\sum_n c_n \, u_n$ is well defined by (b), isometric and has dense image), hence the closure of the span of the $u_n$ is complete. But $x_n$ is in the span of the $u_n$, hence the Cauchy sequence $x_n$ must admit a limit. Commented Jul 26, 2020 at 18:28
• So far, the closest I've found is this. Related, but not a duplicate. Commented Jul 26, 2020 at 18:37

The implication $$(b) \implies (a)$$ holds. One way to prove it is as follows:
Let $$S$$ be a maximal orthonormal set in $$X$$. Consider the Hilbert space $$\ell^2(S) = \biggl\{ f \colon S \to \mathbb{C} \biggm\vert \sum_{s \in S} \lvert f(s)\rvert^2 < +\infty \biggr\}\,.$$ Condition $$(b)$$ says that $$\Phi \colon f \mapsto \sum_{s \in S} f(s)\cdot s$$ is a well-defined map $$\ell^2(S) \to X$$, since for every $$f \in \ell^2(S)$$ we have $$f(s) \neq 0$$ only for a countable (possibly finite) subset of $$S$$. Since $$S$$ is an orthonormal set we have $$\lVert \Phi(f)\rVert_X = \lVert f\rVert_{\ell^2(S)}$$ for all $$f \in \ell^2(S)$$. Furthermore $$\Phi$$ is linear. Thus $$\operatorname{im} \Phi$$ is a complete subspace of $$X$$. Hence there is an orthogonal projection $$P \colon X \to \operatorname{im} \Phi$$. If there were $$x \in X \setminus \operatorname{im} \Phi$$, then $$\frac{x - Px}{\lVert x - Px\rVert}$$ would be a unit vector orthogonal to $$\operatorname{im} \Phi$$, contradicting the maximality of $$S$$.
Therefore $$\Phi$$ is surjective, whence $$X$$ is complete.