Separability of $C^2_b(\Omega)$

Question

Let $\Omega \subseteq \mathbb R$ be open and bounded. Denote by $C_b^2(\Omega)$ the space of functions $f:\Omega\rightarrow \mathbb R$ such that $f, f', f''\in C_b(\Omega)$. Endow this space with the norm $$\|f\|_{C^2_b}:=\|f\|_\infty + \|f'\|_\infty + \|f''\|_\infty.$$ Is the space $(C_b^2(\Omega), \|\cdot\|_{C^2_b})$ separable?

Answer 1

No. For example take the family $\{ \sin(\alpha x) \}_{\alpha\in \mathrm{R}}$. Because $C_b(\Omega)$ is not separable, it admits a family $\{f_\alpha\}_{\alpha\in\mathrm{R}}$ such that $\|f_\alpha-f_\beta\|\geq 1$ for all $\alpha\neq \beta$ (upon renormalizing it). Let $A= \inf \Omega$ and $B=\sup \Omega$ and define $$ g_\alpha(x) = \int_A^x \int_A^y f_\alpha(z)dz dy, $$ where $f_\alpha$ are extended to $0$ on $(A,B)\setminus \Omega$. It is easy to see that $g_\alpha \in C_b^2(\Omega)$ and that $$ \|g_\alpha-g_\beta\|_{C^2} \geq \|f_\alpha-f_\beta\|_C \geq 1. $$

If you want a general result on the non-separability of $C_b(\Omega)$, you can check this post.

On the other side, if you ask that the family has derivatives continuous up to $\partial \Omega$, then you have separability. Indeed you can consider the injection $$ T:C^2(\overline{\Omega})\to C(\overline{\Omega})^3, f\mapsto [f,f',f'']. $$ Then the range of $T$ is a closed subset of the separable metric space $C(\overline{\Omega})^3$, and so separable.

Notice that $C(\overline{\Omega})$ is the Hausdorff compactification of the family $$ \{f\in C(\Omega): f\text{ is uniformly continuous}\} \subset C_b(\Omega), $$ so you can similarly construct space $\tilde{\Omega}$ by considering certain separable sub-families of $C_b(\Omega)$ (or $C_b^2(\Omega)$) and obtain $$ C(\tilde{\Omega}) \text{ separable subspace of } C_b(\Omega). $$