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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!

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  • $\begingroup$ I imagine $C(X)_+=\{ f\in C(X)\mid f(x)>0\ \forall x\in X\}$. This is useful, because then from $f\in C_+$ you get $\frac1f\in C_+$. From that it follows that $x\mapsto \int_X \frac1{f(y)}\pi(x,dy)$ is continuous by the Feller property. Further since $\pi(x,dy)$ is a probability measure the integral of a positive function is a positive number, thus the above function is also positive (ie in $C_+$). Applying the first statement again gives that $M(f)$ is in $C_+$. $\endgroup$ – s.harp Nov 12 '18 at 19:35
  • $\begingroup$ Thanks @s.harp . Good to see you :-) you’re right. If I confine $C(X)$ to the set $C(X)_{++}$ of all positive functions in $C(X)$, then $M(f)$ is in $C(X)_{++}$ for any $f \in C(X)_{++}$. In fact, I was wondering the case in which $C(X)_+ = \{ f\inC(X) \colon f \geq 0 \}$. The tricky part appears if $f(x)=0$ at some point $x$. Moreover, If $f$ is any constant function, then $Mf=f=c$ for the constant $c$. $\endgroup$ – Paradiesvogel Nov 12 '18 at 19:57
  • $\begingroup$ I see, in this case I think it is false. Lets see for a counterexample. $\endgroup$ – s.harp Nov 12 '18 at 20:36
  • $\begingroup$ Thanks so much @s.harp ! $\endgroup$ – Paradiesvogel Nov 12 '18 at 21:01
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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.

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  • $\begingroup$ Thank you very much, @s.harp ! I might need some time to think about your counterexample, which is very impressive! Really appreciate your huge help! $\endgroup$ – Paradiesvogel Nov 12 '18 at 21:04
  • $\begingroup$ Dear @s.harp, Would you mind explaining a bit more about how the measure you constructed has the Feller property please? Many thanks again :-) $\endgroup$ – Paradiesvogel Nov 12 '18 at 21:24
  • $\begingroup$ Note that $\int_X f(y)\pi(x,dy)=\begin{cases} F(1)-F(0) & x≥0 \\ F(1-x) - F(x) & x≤ 0\end{cases}$ where $F$ is an anti-derivative of $f$. Since anti-derivatives are continuous the resulting function is continuous. $\endgroup$ – s.harp Nov 12 '18 at 22:01
  • $\begingroup$ Thanks a million @s.harp, I learn it now. All the best :-) $\endgroup$ – Paradiesvogel Nov 12 '18 at 22:18

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