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EDIT: sorry I realized I made some mistakes asking the question, so I'm fixing them.

Analysis is not really my field, so I hope this question is not too trivial.

Let's consider $X$ a locally compact metric space, i will denote

$C_C(X)$ the space of continuous functions on $X$ with compact support. It has the $L^\infty$ norm.

$H_C(X)$ the space of Holder functions on $X$ with compact support. To be more precise: $H_K^\mu(X)$ is the space of continuous holder functions on $X$ with holder exponent $\mu$ and support in $K$ compact set. Then $H_C(X)$ is the union of $H_K^\mu(X)$ as $\mu$ and $K$ vary. Every space $H_K^\mu(X)$ has the topology given by the norm $||\varphi||_\mu=sup_{x\in X}|\varphi(x)|+sup_{x\neq y}\frac{|\varphi(x)-\varphi(y)|}{d(x,y)^\mu}$

$R(X)$ is the space of Radon measures on $X$: each element of $R(X)$ defines a continuous linear and positive functional on $C_C(X)$ (with values in $\mathbb{R}$). $R(X)$ is endowed with the weak-star topology.

$H(X)$ is the space of linear functionals on $H_C(X)$ (with values in $\mathbb{R}$) whose restriction to each $H_K^\mu(X)$ is continuous. $H(X)$ is endowed with the topology given by the norm $||\quad ||_\mu$.

Every holder-continuous function is also continuous in the usual sense, so we can define a function $\psi: R(X)\rightarrow H(X)$ which to a Radon measure associates its restriction (as a functional) to $H_C(X)$.

What I can't understand is if $\psi$ is open and continuous

User Justthisguy explained why $\psi$ is injective, but it's not totally clear to me

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  • $\begingroup$ The answer to (1) is yes (i.e. $\psi$ is injective) for $X = \mathbb R^n$. To see this, just note that we can determine a Radon measure on a compact set by considering Holder functions which are one on that set, vanish on a slightly larger open set, and are everywhere less than one. If there is a "Holder Urysohn's lemma" for arbitrary compact metric spaces, then this technique will work there as well, but I don't know if that's true. I suspect, but don't know, that (2) is false, and that in particular the complement of the image of $\psi$ is dense. $\endgroup$ – Justthisguy Feb 4 '16 at 14:53
  • $\begingroup$ thank you very much! I'm sorry but I wrote the wrong topology on $H(X)$.. I fixed that. Also, can you explain me why a Radon measure should be determined by such holder functions on compacts? Thank you $\endgroup$ – user3419 Feb 5 '16 at 12:15
  • $\begingroup$ A Radon measure is inner regular: that is, if $\mu$ is a Radon measure, $\mu(E) = \sup\{\mu(K) : K \subset \subset E\}$ where $K \subset \subset E$ means $K$ is a compact subset of $E$. Thus two Radon measures are equal if they are equal on compact sets. Now, given any compact set $K$, there exist open sets $U_i$ with compact closure such that $K \subset \subset U_i$ and $\bigcap\limits_{i=1}^\infty U_i = K$. For each of these, we can find $f_i \in H(\mathbb R^n)$ with $f_i = 1$ on $K$ and $f_i = 0$ outside $U_i$. It follows that $\mu(K) \leq \int f_i d \mu \leq \mu(U_i)$... $\endgroup$ – Justthisguy Feb 5 '16 at 20:43
  • $\begingroup$ Taking limits we find that $\mu(K) \leq \lim\limits_{i \to \infty} \int f_i d \mu \leq \lim\limits_{i \to \infty} \mu (U_i) = \mu(\bigcap\limits_{i=1}^\infty U_i) = \mu(K)$. Thus all of the inequalities were equalities, and we have determined $\mu(K)$ by knowing $\int f_i d \mu$ for the Holder functions $f_i$. I guess you need to know that the $U_i$ are nested, but this obviously can be done. $\endgroup$ – Justthisguy Feb 5 '16 at 20:45

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