Non- seperable Hilbert Spaces

Question

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.

Answer 1

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.