Is there $\varepsilon \in (0, 1)$ such that $\sup_{t \in [0, \varepsilon]} [\ell_t]_\beta < \infty$?

Question

$ \newcommand{\bR}{\mathbb{R}} \newcommand{\bT}{\mathbb{T}} \newcommand{\bN}{\mathbb{N}} \newcommand{\bP}{\mathbb{P}} \newcommand{\bE}{\mathbb{E}} \newcommand{\bF}{\mathbb{F}} \newcommand{\bD}{\mathbb{D}} %%%%%%%%%%%%%%%%%%%%%%%%%%%% \newcommand{\cA}{\mathcal{A}} \newcommand{\cB}{\mathcal{B}} \newcommand{\cD}{\mathcal{D}} \newcommand{\cE}{\mathcal{E}} \newcommand{\cF}{\mathcal{F}} \newcommand{\cG}{\mathcal{G}} %%%%%%%%%%%%%%%%%%%%%%%%%%%% \newcommand{\sD}{\mathscr{D}} \newcommand{\sE}{\mathscr{E}} \newcommand{\sG}{\mathscr{G}} \newcommand{\sH}{\mathscr{H}} \newcommand{\sK}{\mathscr{K}} \newcommand{\sP}{\mathscr{P}} %%%%%%%%%%%%%%%%%%%%%%%%%%%% \newcommand{\eps}{\varepsilon} \newcommand{\diff}{\mathop{}\!\mathrm{d}} \newcommand{\andd}{\quad \text{and} \quad} \newcommand{\qtext}{\quad\text} $ Let $\sP_1 (\bR^d)$ be the space of Borel probability measures on $\bR^d$ with finite first moment. We endow $\sP_1 (\bP^d)$ with the Wasserstein metric $W_1$. We consider $$ \mu :[0, 1] \to \sP_1 (\bP^d), \, t \mapsto \mu_t. $$

We assume that each $\mu_t$ admits a probability density function (p.d.f.) denoted by $\ell_t$. We fix $\alpha \in (0, 1), \beta \in (0, 1)$ and $C>0$. We assume that for $t \in [0, 1]$: \begin{align} W_1 (\mu_0, \mu_t) &\le C t^\alpha, \\ \|\ell_t\|_\infty &\le C, \\ [\ell_t]_\beta :=\sup_{x \neq y} \frac{|\ell_t (x) - \ell_t (y)|}{|x-y|^\beta} &< \infty. \end{align}

Is there $\eps \in (0, 1)$ such that $\sup_{t \in [0, \eps]} [\ell_t]_\beta < \infty$?

Thank you so much for your elaboration!

Answer 1

Pick a sequence of functions $g_n$ where (i) $\|g_n\|_\beta=n$ and (ii) $\|g_n-\mathbf 1\|_\infty\le 3^{-n}$. For example $$ g_n(x)=\begin{cases} 1+n(a_n-x)^\beta&\text{if $0\le x\le a_n$};\\ 1&\text{if $a_n\le x\le 1-a_n$;}\\ 1-n(x-(1-a_n))^\beta&\text{if $1-a_n\le x\le 1$}, \end{cases} $$ where $a_n=(n3^n)^{-1/\beta}$.

Notice that condition (ii) implies $W_1(g_n,\mathbf 1)\le 3^{-n}$.

Now define a family of functions $f_t$ by

$$ f_t= \begin{cases} \frac {t-3^{-n}}{3^{-n}}g_n + \frac{2\times 3^{-n}-t}{3^{-n}}\mathbf 1 &\text{if $3^{-n}\le t\le 2\times 3^{-n}$;}\\ \frac {3^{-(n-1)}-t}{3^{-n}}g_n + \frac {t-2\times 3^{-n}}{3^{-n}}\mathbf 1 &\text{if $2\times 3^{-n}\le t\le 3^{-(n-1)}$.} \end{cases} $$ In words, $f_n$ is equal to $\mathbf 1$ at $3^{-n}$ and $3^{-(n-1)}$ and linearly interpolates to $g_n$ at $2\times 3^{-n}$.

It's easy to check from (ii) that $t\mapsto f_t$ is Lipschitz in the $W_1$ distance. And $\|f_{2\times 3^{-n}}\|=n$. Each $f_t$ has a finite Hölder norm, but there is no uniform upper bound on any non-trivial interval containing $t=0$.