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).

  • $\begingroup$ 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. $\endgroup$ Commented Jul 26, 2020 at 18:25
  • 2
    $\begingroup$ 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. $\endgroup$
    – s.harp
    Commented Jul 26, 2020 at 18:28
  • $\begingroup$ So far, the closest I've found is this. Related, but not a duplicate. $\endgroup$ Commented Jul 26, 2020 at 18:37

1 Answer 1


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.


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