# How to show the continuity of a specific function?

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!

• 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 $C_+$. – s.harp Nov 12 '18 at 19:35
• 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 $c$. – Paradiesvogel Nov 12 '18 at 19:57
• I see, in this case I think it is false. Lets see for a counterexample. – s.harp Nov 12 '18 at 20:36
• Thanks so much @s.harp ! – Paradiesvogel Nov 12 '18 at 21:01

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.
• 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. – s.harp Nov 12 '18 at 22:01