Riesz representation theorem for duals of spaces of continuously differentiable functions Let $k$ be a positive integer. I am looking for a possibly exhaustive reference discussing representation of dual spaces of $C^k_b(\mathbb{R}^d)$, $C^k_0(\mathbb{R}^d)$, or at least $C^k(K)$ for compact $K\subset \mathbb{R}^d$, in terms of integrals with respect to measures or additive set functions. I also have some concrete questions, which I formulate below.
By $C^k_b(\mathbb{R}^d)$ I mean the space of $k$-times continuously differentiable real functions with all derivatives up to order $k$ bounded. $C^k_0(\mathbb{R}^d)$ are the functions in $C^k_b(\mathbb{R}^d)$ such that all the derivatives vanish at infinity.
By embedding $C^k_b(\mathbb{R}^d)$ into $(C_b(\mathbb{R}^d))^{1+d+d^2+\ldots +d^k}$ and by using the Hahn--Banach theorem, one can show that for any $\rho\in (C^k_b(\mathbb{R^d}))^\ast$ there exist regular bounded additive ($rba(\mathbb{R}^d)$) set functions $\mu_\alpha$, $|\alpha|\leq k$ ($\alpha$ are multi-indices), such that for all $\phi\in C^k_b(\mathbb{R}^d)$ we have
\begin{align}\rho(\phi) = \sum\limits_{|\alpha|\leq k} \int_{\mathbb{R}^d} \partial^\alpha\phi(x)\, \mu_\alpha (dx).\end{align}
Similar result is valid for $C^k_0(\mathbb{R}^d)$, then $\mu_\alpha$ can be taken to be finite Radon measures.
My main question is: consider the subspace of $C^k_b(\mathbb{R}^d)$ consisting of functionals which have a representation such that all $\mu_\alpha$ are finite Radon measures -- is this subspace closed in the norm topology of $(C^k_b(\mathbb{R}^d))^\ast$?
The problem is that due to the lack of uniqueness (at least that is the case for $C^1([0,1])$) of such representation  we only have $\|\rho\|_{(C^n_b(\mathbb{R}^d))^\ast} \leq \sum\limits_{|\alpha|\leq k}\|\mu_\alpha\|_{TV}$ (the total variation norm), but the reverse inequality need not hold, even with a multiplicative constant. So an intermediate question is: can $\mu_\alpha$ always be chosen in a way that $\|\rho\|_{(C^n_b(\mathbb{R}^d))^\ast} \geq C\sum\limits_{|\alpha|\leq k}\|\mu_\alpha\|_{TV}$ for some $C>0$ independent of $\rho$?
 A: I think I have an answer to my questions, but I will still appreciate suggestions on the existing literature and comments about the following arguments.
Notation: $C^{-k}_b = (C^k_b(\mathbb{R}^d))^\ast$, $C^{-k}_0 = (C^k_0(\mathbb{R}^d))^\ast$, $K = 1 + d + \ldots d^k$.
Claim 1. For every $\rho \in C^{-k}_0$ there exists $\mu \in \mathcal{M}(\mathbb{R}^d)^K$ such that $\|\rho\|_{C^{-k}_0} = \|\mu\|_{\mathcal{M}(\mathbb{R}^d)^K}$ and
\begin{align}\label{repr}\tag{1}
\langle \rho,\phi\rangle = \sum\limits_{|\alpha|\leq k} \int_{\mathbb{R}^d} \partial^{\alpha}\phi\, d\mu_\alpha,\quad \phi \in C^k_0(\mathbb{R}^d).
\end{align}
Proof: we embed $C^k_0(\mathbb{R}^d)$ isometrically into $C_0(\mathbb{R}^d)^K$, we identify $C^k_0(\mathbb{R}^d)$ with that embedding. If $\rho\in C^{-k}_0$, then by the Hahn--Banach theorem it can be extended to $\rho^*$ on $C_0(\mathbb{R}^d)^K$ with $\|\rho^*\|_{(C_0(\mathbb{R}^d)^K)^\ast} = \|\rho\|_{C^{-k}_0}$. By the Riesz representation theorem for $C_0(\mathbb{R}^d)$ there exists $\mu\in \mathcal{M}(\mathbb{R}^d)^K$ such that $\|\rho^\ast\|_{(C_0(\mathbb{R}^d)^K)^*} = \|\mu\|_{TV} := \sum_{|\alpha|\leq k} \|\mu_\alpha\|_{TV}$ and
\begin{align}
\langle \rho^\ast,\phi\rangle = \sum\limits_{|\alpha|\leq k} \int_{\mathbb{R}^d} \partial^{\alpha}\phi\, d\mu_\alpha,\quad \phi \in C_0(\mathbb{R}^d)^K.
\end{align}
Then \eqref{repr} holds and $\|\mu\|_{TV} = \|\rho^*\|_{(C_0(\mathbb{R}^d)^K)^\ast} = \|\rho\|_{C^{-k}_0}$, which proves the claim.
Claim 2. Let $(\rho_n)\subset C^{-n}_b$ and $\rho\in C^{-n}_b$. Assume that $\|\rho_n - \rho\|_{C^{-n}_b} \to 0$ as $n\to\infty$ and for every $n\in \mathbb{N}$ there exists $\mu^n\in \mathcal{M}(\mathbb{R}^d)^K$ such that
\begin{align}\label{reprn}\tag{2}
\langle \rho_n,\phi\rangle = \sum\limits_{|\alpha|\leq k} \int_{\mathbb{R}^d} \partial^{\alpha}\phi\, d\mu^n_\alpha,\quad \phi \in C^k_b(\mathbb{R}^d).
\end{align}
Then there exists $\mu \in \mathcal{M}(\mathbb{R}^d)^K$ such that
\begin{align}
\langle \rho,\phi\rangle = \sum\limits_{|\alpha|\leq k} \int_{\mathbb{R}^d} \partial^{\alpha}\phi\, d\mu_\alpha,\quad \phi \in C^k_b(\mathbb{R}^d).
\end{align}
Proof: By assumptions and Claim 1, for every $n\in \mathbb{N}$ there exists $\tilde{\mu}^n$ such that $\|\tilde{\mu}^n\|_{TV} = \|\rho_n|_{C^k_0(\mathbb{R}^d)}\|_{C^{-k}_0}$ and \begin{align}\label{reprn2}\tag{3}
\langle \rho_n,\phi\rangle = \sum\limits_{|\alpha|\leq k} \int_{\mathbb{R}^d} \partial^{\alpha}\phi\, d\tilde{\mu}^n_\alpha = \sum\limits_{|\alpha|\leq k} \int_{\mathbb{R}^d} \partial^{\alpha}\phi\, d\mu^n_\alpha,\quad \phi \in C^k_0(\mathbb{R}^d).
\end{align}
We will show that $\tilde{\mu}^n$ can be used instead of $\mu^n$ in \eqref{reprn}.  For $R>1$, let $\tau_R$ be a smooth cut-off function, i.e., $0\leq\tau_R\leq 1$, $\tau_R = 1$ on $B_R$, $\tau_R = 0$ on $B_{R+1}^c$ and $\| \tau_R\|_{C^{k+1}_b(\mathbb{R}^d)} \leq M$, with $M$ independent of $R$. Then for every $\phi\in C^k_b(\mathbb{R}^d)$ and $\alpha$ we have $\partial^{\alpha}(\tau_R\phi) \to \partial^{\alpha}\phi$ pointwise as $R\to \infty$, and by the dominated convergence theorem,
\begin{align}
\lim\limits_{R\to\infty} \int_{\mathbb{R}^d} \partial^{\alpha}(\tau_R\phi)\, d\mu^n_\alpha = \int_{\mathbb{R}^d} \partial^{\alpha}\phi\, d\mu^n_\alpha.
\end{align}
Since $\tau_R\phi\in C^k_0(\mathbb{R}^d)$, this together with \eqref{reprn2} gives
\begin{align}\label{reprn3}\tag{4}
\langle \rho_n,\phi\rangle = \sum\limits_{|\alpha|\leq k} \int_{\mathbb{R}^d} \partial^{\alpha}\phi\, d\tilde{\mu}^n_\alpha,\quad \phi \in C^k_b(\mathbb{R}^d).
\end{align}
Furthermore, we have
\begin{align}
\|\tilde{\mu}^n\|_{TV} = \|\rho_n|_{C^k_0(\mathbb{R}^d)}\|_{C^{-k}_0} \leq \|\rho_n\|_{C^{-k}_b} \leq \|\tilde{\mu}^n\|_{TV},
\end{align}
where the last inequality can be checked by hand using \eqref{reprn3}. Thus, $\|\rho_n\|_{C^{-k}_b} = \|\tilde{\mu}^n\|_{TV}$. Therefore, since $\rho_n \to \rho$ in $C^{-k}_b$ we find that $\tilde{\mu}^n$ converge to some $\mu\in \mathcal{M}(\mathbb{R}^d)^K$. Since norm convergence implies weak$^*$ convergence, \eqref{reprn3} yields
\begin{align}
\langle \rho,\phi\rangle = \sum\limits_{|\alpha|\leq k} \int_{\mathbb{R}^d} \partial^{\alpha}\phi\, d\mu_\alpha,\quad \phi \in C^k_b(\mathbb{R}^d),
\end{align}
which ends the proof.
