1
$\begingroup$

This is an exercise from Functional Analysis by Peter Lax:

Let $\{x_j\}$ be an arbitrarily indexed orthonormal set in a Hilbert space(possibly non-separable). Show that the closed linear span of $\{x_j\}$ consists of all vectors of the form:$$x=\sum a_j x_j$$ where the sum converges in the sense of Hilbert space norm and the $a_j$ are complex number so chosen that$$\sum \lvert a_j \rvert^2 < \infty$$ Furthermore:$$\lVert x\rVert^2=\sum \lvert a_j \rvert^2$$ with $a_j=(x,x_j)$. (circular brackets denotes the inner product)

The difficulty here is that the orthonormal set is indexed arbitrarily thus might not be countable, but then you have to show that any element in the the closed linear span is a countable linear combination of the orthonormal set.

Here is my attempt in solving this:

Since closed linear span of $\{x_j\}_{j\in J}$ is the closure of the span of $\{x_j\}_{j\in J}$. for $x$ in the closed linear span we can find a sequence $\{y^{(n)}\}_{n\in\mathbb{N}}\subset cl\left(Span\left(\{x_j\}_{j\in J}\right)\right)$ that converges to $x$ in the Hilbert space norm. For each $n\in\mathbb{N}$ we have: $$y^{(n)}=\sum\limits_{k\in F\left(n\right)}a^{(n)}_k x_k$$ where $F\left(n\right)$ is a finite subset of the index set $J$, and $a^{(n)}_k$ are non-zero complex numbers. I know that the set:$$A := \{j\in J : \exists n\in\mathbb{N}: j \in F\left(n\right)\}$$ is countable since it is contained in the set $$\bigcup_{n\in\mathbb{N}}F\left(n\right)$$ which is a countable union of countable sets and therefore also countable. From here I want to use $A$ as the index set for the sum that converges to $x$ but I don't know how to proceed. Any help will be very much appreciated!

$\endgroup$
1
$\begingroup$

You have $y^{(n)}$ converging to $y$ where $y^{(n)}$ is a finite-linear combination of the $x_i$. So the $y^{(n)}$ lie in the closed subspace generated by the $x_i$ where the $i\in I_0$ where $I_0$ is a countable subset of the index set $I$. So we can represent $y$ as $\sum_{i\in I_0} a_i x_i$.

$\endgroup$
  • $\begingroup$ for each $n$, $y^\left( n\right)$ surely lies in a closed subspace generated by $\{x_i\}_{i\in I_0\left(n\right)}$ where $I_0=I_0\left(n\right)$ depends on $n$, but how can their limit $y$ also being generated from $\{x_i\}_{i\in I_0}$? $\endgroup$ – HSea12345n Jun 28 '17 at 6:04
  • $\begingroup$ Sorry I think I misunderstood your answer, you meant $I_0$ is the countable index set $A$ constructed in my question, then I think the question is solved, thank you very much for the help! $\endgroup$ – HSea12345n Jun 28 '17 at 9:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.