1
$\begingroup$

Let $G$ be the span of $ \left\{ e^{i\lambda t} \right\} _{\lambda\in \mathbb R}$ with inner product $$ \left\langle f,g \right\rangle =\lim _{T\rightarrow \infty}\frac 1{2T}\int_{-T}^Tf\bar g .$$ I need to show the completion of $G$ is not separable. I'm guessing the proof should go by contradiction but I have no clue what to do... Help!

$\endgroup$
  • $\begingroup$ The completion with respect to what norm? $\endgroup$ – David C. Ullrich Jul 10 '16 at 14:58
  • $\begingroup$ @DavidC.Ullrich $L^2$-norm $\endgroup$ – linalg Jul 10 '16 at 15:01
  • 2
    $\begingroup$ ??? What $L^2$ norm? If you're talking about the $L^2$ norm on $[0,2\pi]$ then the completion is separable. If you're talking about $L^2(\Bbb R)$ this makes no sense because those functions are not in $L^2$ in the first place. $\endgroup$ – David C. Ullrich Jul 10 '16 at 15:16
  • $\begingroup$ @DavidC.Ullrich sorry I took a while. I missed crucial details and asked a "wrong question". $\endgroup$ – linalg Jul 10 '16 at 17:34
  • 3
    $\begingroup$ Ok. Actually that's what I assumed the problem was, but I have this thing about how the OP should at least be able to get the question straight. With that inner product the $e_\lambda$ are orthonormal. Hence $||e_\lambda-e_{\lambda'}||=\sqrt 2$ if $\lambda\ne\lambda'$. That makes it clear the space is not separable, right? $\endgroup$ – David C. Ullrich Jul 10 '16 at 17:47
0
$\begingroup$

Define $e_\lambda(t)=e^{i\lambda t}$. One can easily calculate the relevant integrals explicitly, showing that $\{e_\lambda:\lambda\in\Bbb R\}$ is an orthonormal subset of $G$. Hence the completion $H$ of $G$ is a non-separable Hilbert space, since $\{B(e_\lambda,\sqrt 2/2)\}$ is an uncountable collection of pairwise disjoint nonempty open sets.


Comment What's above was worked out in the comments; I'm posting this just because Questions are better with Answers. For the sake of adding something not in the comments:

In fact $H$ is, in a canonical way, $L^2$ of the Bohr compactification of $\Bbb R$. The Bohr compactification of a locally compact abelian group $A$ is defined like so: Let $\hat A$ be the dual group of $A$, and now let $\hat A'$ be the dual group, but with the discrete topology. The Bohr compactification of $A$ is the dual group of $\hat A'$.

(Saying that is a little silly, since it will not be new to most of the readers familiar with the context. Anyone who doesn't follow what I'm going on about in the previous paragraph but finds it interesting might check out, say, Rudin Fourier Analysis on Groups. Or you could start here.)

$\endgroup$

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.