Prove the completion of the span of $\left\{ e^{i\lambda t} \right\} _{\lambda\in \mathbb R}$ is not separable

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!

• The completion with respect to what norm? – David C. Ullrich Jul 10 '16 at 14:58
• @DavidC.Ullrich $L^2$-norm – linalg Jul 10 '16 at 15:01
• ??? 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. – David C. Ullrich Jul 10 '16 at 15:16
• @DavidC.Ullrich sorry I took a while. I missed crucial details and asked a "wrong question". – linalg Jul 10 '16 at 17:34
• 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? – David C. Ullrich Jul 10 '16 at 17:47

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.
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 $\hat A'$.