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I have a question that occurred to me and has been bothering me, because maybe graphically it seems obvious but I don't know how to get there. It has to do with the distribution function and monotone rearrangment.

Given a bounded function $f\colon [a,b] \rightarrow \mathbb{R}$, the (right-continuous) distribution of $f$ is the function $D_{f}$ definded by

\begin{align} D_{f}(y):= \mu(\{t\in I\colon f(t)\leq y \}),\; \forall y\in [\inf f, \sup f]. \end{align}

$\mu$ denotes de Lebesgue-measure. The (left-continuous) rearrangment of $f$, denoted by $f^{\ast}$, is the function: \begin{align} f^{\ast}(t):= \inf\{y\in [\inf f, \sup f] \colon D_{f}(y)\geq t \},\; \forall t\in [0,b-a]. \end{align}

It turns out that $f^{\ast}$ is the left inverse function of $D_{f}$, moreover it is the generalized inverse. They are both monotone increasing functions. Here is my question:

Let $t'\in [0,b-a]$ be a point such that $f^{\ast}$ is discontinuous in $t'$, this means $D_{f}$ has a constant segment over $[y_{j},y_{j+1})$ of value $t'$, so

\begin{align} t' = D_{f}(y_{j}) = D_{f}(y), \; \forall y\in [y_{j},y_{j+1}). \end{align}

How can I proof that: \begin{align} \lim_{t\to t'^{+}} f^{\ast}(t) = y_{j+1} \end{align}

Clearly: \begin{align} \lim_{t\to t'^{-}} f^{\ast}(t) = f^{\ast}(t') = f^{\ast}(D_{f}(y_{j}))=y_{j}. \end{align}

I would appreciate your help very much! If I am not being clear on something or you need more details please let me know.

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    $\begingroup$ Cross-posted to Math.SE $\endgroup$
    – Glorfindel
    Nov 5, 2021 at 7:58
  • $\begingroup$ Do you have a response to the answer below? $\endgroup$ Nov 10, 2021 at 2:46

1 Answer 1

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$\newcommand{\ep}{\varepsilon}$Let \begin{equation*} c:=\inf f,\quad d:=\sup f,\quad I:=[a,b],\quad F:=D_f, \end{equation*} so that \begin{equation*} F(y)=\mu(\{t\in I\colon f(t)\le y \})\quad \forall y\in[c,d]. \end{equation*} Also introduce the set \begin{equation*} E_t:=\{y\in[c,d]\colon F(y)\ge t\}\quad \forall t\in[0,b-a]. \end{equation*} Since $F$ is nondecreasing and right-continuous, for all $t\in[0,b-a]$ \begin{equation*} f^*(t)=\inf E_t=\min E_t,\quad E_t=[f^*(t),d], \end{equation*} and, for all $y\in[c,d]$, \begin{equation*} F(y)\ge t\iff y\ge f^*(t). \tag{1} \end{equation*}

Now take any $t'\in[0,b-a]$ such that $f^*$ is discontinuous at $t'$. Since $f^*$ is nondecreasing and left-continuous, we have $f^*(t'+)>f^*(t')$. So, $t'<b-a$, and for some real $\ep>0$ and all $t\in(t',b-a]$ we have $f^*(t)>\ep+f^*(t')$; so, by (1), $F(\ep+f^*(t'))<t$ for all $t\in(t',b-a]$ and hence $F(\ep+f^*(t'))\le t'$. On the other hand, again by (1), $F(f^*(t'))\ge t'$. Since $F$ is nondecreasing, it follows that $F(y)=t'$ for all $y\in[f^*(t'),\ep+f^*(t')]$. So, introducing
\begin{equation*} Y:=\{y\in[c,d]\colon F(y)=t'\}, \end{equation*} \begin{equation*} y_1:=\inf Y,\quad y_2:=\sup Y, \end{equation*} we do have $y_1<y_2$ and $F(y)=t'$ for all $y\in[y_1,y_2)$.

Moreover, for any $y\in[y_1,y_2)$ and any $t\in(t',b-a]$, we have $F(y)=t'<t$ and hence, once again by (1), $y<f^*(t)$. Letting here $y\uparrow y_2$ and $t\downarrow t'$, we get $f^*(t'+)\ge y_2$.

Suppose now that $f^*(t'+)\ne y_2$. Then for some real $\ep>0$ we have $f^*(t'+)>\ep+y_2$. So, for all $t\in(t',b-a]$ we have $f^*(t)>\ep+y_2$, whence, once again by(1), $t>F(\ep+y_2)$. So, $t'\ge F(\ep+y_2)\ge F(y_1)=t'$ and hence $F(\ep+y_2)=t'$, which contradicts the definition $y_2:=\sup Y$.

Thus, indeed $f^*(t'+)=y_2$.

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