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I know that an example of non-separable Hilbert space is the space of almost periodic functions as stated in the book of Israel Gohberg Basic operator theory in p.37, but I need a rigor proof for this, Can anyone help me? Definition of almost periodic functions The rest of the details that I have

thanks.

Emptymind
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    Could you give us the definition of almost periodic function you're using? There are apparently several inequivalent definitions. – mathematician Feb 27 '17 at 03:53
  • I have added the definition in an image at the beginning of my post . – Emptymind Feb 27 '17 at 04:11
  • I do not know why I am voted down on this question can any one explain to me please? – Emptymind Mar 06 '17 at 04:13
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    The image shows how to prove there is an uncountable orthonormal set. I assume the reference to the appendix is a proof that this implies the space is not separable. Which part are you unsure of? Basically, the image you showed is a rigorous (but not completely detailed) proof. – mathematician Mar 07 '17 at 20:41
  • No, the reference to the appendix is not a proof. I mean I want a completely detailed proof. – Emptymind Mar 08 '17 at 00:36
  • So do you follow everything up until uncountable orthonormal set => not separable? – mathematician Mar 08 '17 at 03:35
  • @mathematician ..... No, Frankly speaking the general idea is not clear for me, Could u please explain it for me? – Emptymind Mar 09 '17 at 07:09

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From the image: A function $f\in C(\mathbb{R}:\mathbb{C})$ is almost periodic if it is the uniform limit of functions of the form $\sum_{k=1}^n a_k e^{i\lambda_k x}$ (trigonometric polynomials). Let's call the space of almost periodic functions $E$.

We want to prove that $E$ is a non-separable Hilbert space.

The steps are

  1. $E$ is a vector space.
  2. $E$ is a Hilbert space: There exists an inner product on $E$ that makes it complete.
  3. There is an uncountable orthonormal set using this norm.

The last step proves that $E$ is non-separable using:

Theorem: A Hilbert space is separable iff its dimension is finite or countable.

Proof: Separable Hilbert space have a countable orthonormal basis

Now back to the 3 steps. Step 1: Suppose $f,g\in E$. We want to show $\alpha f+g\in E$ where $\alpha\in\mathbb{C}$. We know there exist sequences of trigonometric polynomials $f_n, g_n$ such that $f_n\rightarrow f$ and $g_n\rightarrow g$ uniformly. Then $f_n+\alpha g_n \rightarrow f+\alpha g$ uniformly (by the properties of limits), where $f_n+\alpha g_n$ are trigonometric polynomials (check this).

Step 2: We are told to use $\langle f, g\rangle=\lim_{T\rightarrow\infty}\frac{1}{T}\int_0^T f(x)\bar{g}(x)\,dx$. Check that this is well-defined (the limit converges for any $f,g\in E$) and is a positive definite sesquilinear form. Is $E$ complete under the norm $\sqrt{\langle f,f\rangle}$? (I feel like there's a clever way to show completeness, but am out of energy at this point. You can try the straightforward way by constructing a limit from a Cauchy sequence.)

Step 3: $\{e^{i\lambda x}:\lambda\in \mathbb{R} \}$ is an orthonormal set: First of all, it is clear that these functions are in $E$. It is easy to check that $\langle e^{i\lambda x}, e^{i\lambda x}\rangle =1$ for any $\lambda\in \mathbb{R}$. For $\lambda\neq\mu$, $\langle e^{i\lambda x},e^{i\mu x}\rangle=$ $\lim_{T\rightarrow\infty}\frac{1}{T}\int_0^T e^{i(\lambda-\mu)x}dx$. That this limit equals zero can be checked with basic calculus.

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