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Let $X$ be a metric space and let $C(X)$ be a family of all bounded and continuous functions from $X$ in $\mathbb{R}$.

We call a positive linear functional $\varphi: C(X) \rightarrow \mathbb{R}$ the functional of Riesz if there is a borel measure $\mu$ on $X$, such that $\varphi(f)=\int_X f \,d\mu$, for $f\in C(X)$.

We call a positive linear functional $\varphi: C(X) \rightarrow \mathbb{R}$ the functional of Banach if for each borel measure $\nu$ on $X$ the condition:$\int_X f d\nu\leq \varphi(f)$, for $f\in C(X)$ - implies that $\nu$ is trivial.

There is a well known theorem :

Let $X$ be a polish space. Then, for each positive linear functional $\varphi: C(X) \rightarrow \mathbb{R}$ there is a unique couple $(\varphi_0,\varphi_*)$ of positive linear functionals defined on $C(X)$, such that $\varphi_0$ is the functional of Riesz, $\varphi_*$ is the functional of Banach and $\varphi=\varphi_*+\varphi_0$. Moreover, the measure $\mu$ related to $\varphi_0$ is defined by: $$\mu(K)=\inf\{\varphi(f): f\in C(X), 1_X\geq f \geq 1_K\},$$ for each compact set $K\subset X$.

More pecisely, for the proof, we define: $$\varphi_{\delta}(f)=\sup\{\varphi(h): \mbox{ supp}\,h\in N(\delta), 0\leq h\leq f\},$$ for $\delta>0$, $$\varphi_{0}(f)=\lim\limits_{\delta \to 0^+}\varphi_{\delta}(f),$$ for $f\in C(X), f\geq 0$, and $$\varphi_{0}(f)=\varphi_{0}(f^+)-\varphi_{0}(f^-),$$ for $f \in C(X)$, where $N(\delta)$ is a family of sets that possess a covering composed of finite number of open balls with a radius equal to $\delta$.

My question concerns the truth of the following sentence: Let $X$ be a $\sigma$-compact and polish space. Assume that $\varphi^x:C(X) \rightarrow \mathbb{R}$ is a positive linear functional, for all $x \in X$ and let $((\varphi^x)_0,(\varphi^x)_*)$ be a couple of Banach-Riesz functionals, for $x \in X$. If the mapping $X \ni x \mapsto \varphi^x(f)$ is continuous for all $f \in C(X)$ and $\varphi^x(1_X)=1$, for $x \in X$, then mapping $X \ni x \mapsto (\varphi^x)_0(f)$ is continuous for all $f \in C(X)$ (or may be for only $f \in C_c(X)$).

I was able to proof only that the mapping $X \ni x \mapsto (\varphi^x)_0(f)$ is upper semi-continuous, for $f\in C_c(X)$.

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    I'm not sure if I understand your definition of $\varphi_{\delta}$. Do you mean you're taking the supremum over all $h$ such that $0 \leq h \leq f$ with the property that $\operatorname{supp}{h}$ can be covered by finitely many balls of radius $\delta$?2011-07-28
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    exactly yes, this is a construction from the proof of the theorem that i gave2011-07-29
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    Now let us get rid of all unnecessary ballast. Fix $f \geq 0$. Then you're taking the supremum over the *continuous* functions $x \mapsto \phi^x (f)$. In my world this gives a *lower* semicontinuous function in general. Now you're taking a monotonically decreasing limit of lower semi-continuous functions to get $\phi_{0}(f)$. So why on earth is that *upper semicontinuous*? Also, where did you get that from? The theorem is definitely not well-known to me.2011-07-29
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    So, at the first, I noticed that the function $x \mapsto \varphi_\delta (f)$ is lower semi-continous (as you said), next, i proved that the same function $x \mapsto \varphi_\delta (f)$ is also upper semi-continous, so it is continous. Finally, $x \mapsto \varphi_0 (f)$ is upper semicontinous as monotonically decreasing limit of upper semi-continuous (even continous) functions.2011-07-29
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    And I assume that you use compact support of $f$ to see that $\phi_{\delta}$ is also upper smicontinuous (after all, you're then working on the compact space given by the support of $f$)?2011-07-29
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    Yes, exactly, I assumed that $f\in C_c(X)$ and used the fact, that for each $\delta>0$ and a compact set $K$ (here I set K=supp f) there is a function $f_\delta\in N(\delta)$, such that $1_K\leq f_\delta$ and the equality $(\varphi^x)_\delta(f_\delta)=\varphi^x(f_\delta)$.2011-07-29
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    I can afford to assume that X is also $\sigma$-compact. Let $\mu_x$ be a measure related to Riesz functional $(\varphi^x)_0$. We have: $(\varphi^x)_0(f)=\mu_x(X)-(\varphi^x)_0(1-f)=\lim\limits_{n\to\infty}\mu_x(K_n)-(\varphi^x)_0(1-f)$, so i think it suffice to prove that $x \mapsto \mu_x(K)$ is lower semi-continous, for compact sets $K$, but i dont know if that is truth.2011-07-29
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    It's getting too late for me to think straight, maybe I'll get back to your problem tomorrow. I'm pretty sure that you're trying to adapt the paper (specifically Appendix A) of Lasota-Szarek *[Lower bound technique in the theory of a stochastic differential equation](http://doi:10.1016/j.jde.2006.04.018)* to your more general situation? It would be extremely helpful if you gave the reference you're working with next time you ask a question.2011-07-29
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    Actually, I am trying to use this theorem to generalize a result in the paper of Meyn and Tweedie "Markov Chains and Stochastic Stability associated with Markov e-chains" on the polish spaces. Sorry I didn't gave this kind of informations. I will be very grateful for your help.2011-07-29

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