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On September 24, 2022, I asked the question below on Mathematics Stack Exchange, linked here:

Link to question on Mathematics Stack Exchange.

I received two up-votes, but no comments or answer. I offered a bounty which has now expired.

I am not a mathematician. This question is related to my research as a macroeconomist. If it is inappropriate for this site, I am happy to remove it.

My Question:

Let $I=[-1,1]$ and let $f:I\to\mathbb{R}$ be a density function on $I$, i.e. $f\ge 0$ and $\int_I f(x)dx = 1$. If it's helpful, assume $f$ is continuous. Let $\mu_f = \int_I xf(x)dx$ be the associated mean. For $\beta>0$ define $T_\beta(f):I\to \mathbb{R}$ by $$ T(f)(x) = \left\{ \begin{array}{ll} 2\left( 1+F(x)-F(2\beta \mu_f -x) \right)f(x) & \text{ if } x\le \beta \mu_f\\ 2\left( 1+F(2\beta \mu_f -x)-F(x) \right)f(x) & \text{ if } x> \beta \mu_f \end{array}\right. $$ Here $F(x)=\int_{-1}^xf(t)dt$ is the distribution function associated with $f$. Unless I have made an error, I can show that $T(f)$ is a density on $I$. Also, if it is of interest, I can explain why I want to understand this density.

My general question is this: what happens to the density upon repeated application of $T$? To be more explicit, given $f$, let $m_f$ be the induced measure on $I$; and, for $x\in I$, let the measure $\delta_x$ be the atom at $x$, i.e. for any Borel set $A$, $\delta_x(A)=1$ if $x\in A$ and $\delta_x(A)=0$ otherwise. Finally, let $T^n(f)\equiv T\circ T^{n-1}(f)$. I would like to know if (any of) the following are true:

  • Weak result #1. If $0<\beta<1$ then $\mu_{T(f)}<\mu_f$.
  • Weak result #2. If $\beta>1$ then $\mu_{T(f)}>\mu_f$.
  • Strong result #1. If $0<\beta<1$ then $m_{T^n(f)}$ converges weakly to $\delta_{0}$.
  • Strong result #2. If $\beta>1$ then $m_{T^n(f)}$ converges weakly to $\delta_{1}$.

Any help, from references to study to explicit arguments, would be appreciated.

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  • $\begingroup$ Since Tf is always something x f I don't see how you can change the support of the measure. E.g., suppose f(x) non zero only on (1/2, 3/4), then the same is true of Tf, TTF etc. In particular, it will not converge to a delta function at 0 or 1. $\endgroup$
    – mike
    Commented Oct 8, 2022 at 6:03
  • $\begingroup$ @mike Thank you for considering my question. I am not sure I understand your comment. For example, suppose $f_n$ is defined on $(0,1)$ as follows: $f_n(x)=n-1$ if $x\in (1-1/(2n),1+1/(2n))$ and equals $1/(n-1)$ otherwise. I think $f_n$ is a density on $(0,1)$ with full support and converges weakly to the delta function at $1/2$. But I may well be confused. $\endgroup$
    – Not_Dustin
    Commented Oct 8, 2022 at 15:44
  • $\begingroup$ I had thought that $\delta_0, \delta_1$ in 'strong result' 1 & 2 referred to a delta function at 0 or 1 respect., but in light of your reply I now assume it only means 2 possibly different delta functions. $\endgroup$
    – mike
    Commented Oct 10, 2022 at 11:50

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