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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 $\delta$? – t.b. Jul 28 '11 at 23:54
exactly yes, this is a construction from the proof of the theorem that i gave – dawid Jul 29 '11 at 0:02
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. – t.b. Jul 29 '11 at 0:13
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. – dawid Jul 29 '11 at 0:25
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 $f$)? – t.b. Jul 29 '11 at 0:27

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