Let $X$ be a closed interval of $\mathbb{R}$, and $C(X)$ be the Banach space of all real-valued continuous functions defined on $X$. Denote by $C(X)_+$ the set of all non-negative functions in $C(X)$.
Let $\pi(x, \mathrm{d} y)$ be a transition probability ($\int \pi(x, \mathrm{d} y) = 1$) that possesses Feller property; that is, the function $Kg(x):= \int_X g(y) \,\pi(x, \mathrm{d} y)$ is continuous on $X$ whenever $g \in C(X)$.
Let an operator $M$ be defined by $$ \left( Mf \right) (x) := \left( \int_X f^{-1} (y) \,\, \pi( x, \mathrm{d} y ) \right)^{-1} \qquad(\forall \, x \in X). $$
I want to verify that for any given function $f$ in $C(X)_+$, is the function $Mf$ also in $C(X)_+$?
It seems to me that the answer for the above question is no. But I could not find a counterexample to prove my conjecture.
In fact, I was thinking to take some continuous function that could reach zero at some point $x$. Because when some function $f$ is close to $0$, the integrand $f^{-1}$ blows up, and hence $\int f^{-1} \pi(x, \mathrm{d} y)$ also blows up and goes to infinity. Under this situation, the dominated convergence theorem may fail and the continuity of $[ Mf(x) ]^{-1}$ may not be obtained. Thus, $Mf$ may fail to be continuous.
That was all my conjecture. But I could not provide a concrete counterexample to show that $Mf$ may not be in $C(X)_+$ for some $f \in C(X)_+$.
Could anyone help me out please? Any idea or suggestions are most welcome!
Thank you very much!
Let $X=[-1,1]$ and let $\pi(x,dy)$ be the integration along $[x,1+x]$ for $x\in[-1,0]$ and integration along $[0,1]$ if $x\in [0,1]$. This measure should have the Feller property.
Now $f(y)=\begin{cases}0& y≤0\\ \sqrt y &y≥0\end{cases}$. Then $\frac1{f(y)}=\begin{cases}\infty & y≤0\\ \frac{1}{\sqrt y}& y≥0\end{cases}$, from which $$\int_X \frac1{f(y)}\pi(x,dy)=\infty \qquad \text{if $x<0$}$$ immediately follows. However for $x≥0$ you have $$\int_0^1\frac1{\sqrt y}dy = \frac12 (\sqrt1-\sqrt0)=\frac12.$$
These calculations give you $$M(f)(x)=\begin{cases} 0& x<0 \\ 2 &x≥0\end{cases},$$ which is not continuous.