Suppose that $f$ satisfies $a$-Hölder condition on $[0,1]$ ($0<a<1$). Fix $0<b<a$. For any $x\in [0,1]$ and sufficiently small $\delta>0$, is the following inequality established? $$ \int_0^\delta t^{-1-b}|f(x-t)-f(x)| \, dt \leq C\delta^{-b}\sup_{0<h\leq \delta}|f(x)-f(x-h)|, $$ where $C$ does not depend on $x$ and $\delta$.


1 Answer 1


$\newcommand\ep\varepsilon\newcommand{\de}{\delta}\newcommand{\De}{\Delta}$This inequality does not hold in general, even if the function $f$ is nondecreasing, which will be assumed henceforth.

Indeed, then the inequality in question is $L\ll R$, where \begin{equation*} L:=\int_0^\de dt\, t^{-1-b}[f(x)-f(x-t)],\quad R:=\de^{-b}[f(x)-f(x-\de)], \end{equation*} and $A\ll B$ means that $A\le CB$ for some real constant $C>0$ depending only on $f$. We also write $A\asymp B$ to mean that $A\ll B\ll A$. We assume throughout that $0<\de\le x\le1$.

Note that \begin{equation*} R=r(f;x,\de):=\de^{-b}\int_{x-\de}^x df(s) \end{equation*} and \begin{equation*} \begin{aligned} L&=\int_0^\de dt\, t^{-1-b}\int_{x-t}^x df(s) \\ &=\int_{x-\de}^x df(s)\int_{x-s}^x dt\, t^{-1-b} \\ &\asymp\int_{x-\de}^x df(s)\,[(x-s)^{-b}-x^{-b}] \\ &=l(f;x,\de)-(x/\de)^{-b}r(f;x,\de), \end{aligned} \end{equation*} where \begin{equation*} l(f;x,\de):=\int_{x-\de}^x df(s)\,(x-s)^{-b}. \end{equation*} So, the inequality in question, $L\ll R$, can be rewritten as \begin{equation*} l(f;x,\de)\ll r(f;x,\de). \tag{1}\label{1} \end{equation*} Since $(x-s)^{-b}\ge\de^{-b}$ for $s\in(x-\de,x)$, we see that the inequality $l(f;x,\de)\ge r(f;x,\de)$, opposite to \eqref{1}, actually holds. Moreover, since the factor $(x-s)^{-b}$ explodes to $\infty$ as $s\uparrow x$, \eqref{1} seems unlikely to hold.

So, to disprove \eqref{1}, it seems reasonable to let $f$ grow faster in a left neighborhood of $x$. Indeed, for $\ep\in(0,\de)$, let the function $f_{x,\ep}$ be defined by the conditions $f_{x,\ep}(0)=0$ and \begin{equation*} df_{x,\ep}(s)=ds\,(x-s)^{a-1}\,1(x-\ep\le s<x). \end{equation*} Then the function $f_{x,\ep}$ is $a$-Hölder-continuous, \begin{equation*} l(f_{x,\ep};x,\de)\asymp\ep^{a-b},\quad r(f_{x,\ep};x,\de)\asymp\de^{-b}\ep^a, \tag{2}\label{2} \end{equation*} so that $l(f_{x,\ep};x,\de)>>r(f_{x,\ep};x,\de)$ if $\ep<<\de$; we write $A<< B$ or, equivalently, $B>>A$ to mean that $A=o(B)$. (Do not confuse $<<$ and $>>$ with $\ll$ and $\gg$.) This is not quite a counterexample to \eqref{1}, though, since the function $f_{x,\ep}$ depends on the varying $\ep$.

However, it is not hard to modify this idea to get a genuine counterexample to \eqref{1}. To do so, in what follows let $f$ be the function defined by the conditions $f(0)=0$ and \begin{equation*} df(s)=\sum_{j=2}^\infty df_{2^{-j},2^{-j}/j}(s). \end{equation*}

Then $f$ is nondecreasing (obviously) and $a$-Hölder-continuous. Indeed (cf. \eqref{2}), \begin{equation*} \int_{2^{-j-1}}^{2^{-j}} df(s) =\int_0^1 df_{2^{-j},2^{-j}/j}(s)\asymp(2^{-j}/j)^a \tag{3}\label{3} \end{equation*} and $\sum_j(2^{-j}/j)^a<\infty$, so that $f$ is continuous at $0$ and hence on $[0,1]$. Also, $f$ is constant on $[2^{-2},1]$. So, to check that $f$ is $a$-Hölder-continuous on $[0,1]$, it suffices to check that $f$ is $a$-Hölder-continuous on $(0,2^{-2}]$.

To do so, suppose that $0<x<y\le2^{-2}$. Then there exist integers $j$ and $k$ such that $2\le k\le j$, $x\in\De_j:=(2^{-j-1},2^{-j}]$, and $y\in\De_k$.

If $k=j$, then \begin{equation*} 0\le f(y)-f(x)=f_{2^{-j},2^{-j}/j}(y)-f_{2^{-j},2^{-j}/j}(x) \\ =\frac{g_j(x)^a-g_j(y)^a}a\le\frac{(y-x)^a}a, \end{equation*} where $g_j(x):=2^{-j}-\max(x,2^{-j}-2^{-j}/j)$, so that $f$ is $a$-Hölder-continuous on $\De_j$.

If $k=j-1$, then the intervals $\De_j$ and $\De_k$ are adjacent to each other. Since $f$ is continuous on $[0,1]$ and $a$-Hölder-continuous on each of the intervals $\De_j$ and $\De_k$, we see that $f$ is $a$-Hölder-continuous on $\De_j\cup\De_k$. So, \begin{equation*} 0\le f(y)-f(x)\ll(y-x)^a \tag{4}\label{4} \end{equation*} if $k\ge j-1$.

If now $k\le j-2$, then, in view of \eqref{3}, \begin{equation*} 0\le f(y)-f(x)\le\sum_{i=k}^j\int_{2^{-i-1}}^{2^{-i}} df(s)\ll(2^{-k})^a, \end{equation*} whereas $y-x>2^{-k-1}-2^{-j}\asymp2^{-k-1}\asymp2^{-k}$. Thus, \eqref{4} holds whenever $0<x<y\le2^{-2}$, which completes the verification that $f$ is $a$-Hölder-continuous on $[0,1]$.

Finally, using \eqref{2} with $x=2^{-j}$, $\de=2^{-j-1}$, $\ep=2^{-j}/j$, and $j\to\infty$, we get $\ep<<\de$, \begin{equation*} l(f;x,\de)=l(f_{x,\ep};x,\de)\asymp\ep^{a-b}, \end{equation*} and \begin{equation*} r(f;x,\de)=r(f_{x,\ep};x,\de)\asymp\de^{-b}\ep^a, \end{equation*} so that $l(f;x,\de)>>r(f;x,\de)$, as claimed. $\quad\Box$

Here is the graph of $f$ for $a=1/2$:

enter image description here


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.