# Estimate an improper integral

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

$$\newcommand\ep\varepsilon\newcommand{\de}{\delta}\newcommand{\De}{\Delta}$$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 $$\begin{equation*} df_{x,\ep}(s)=ds\,(x-s)^{a-1}\,1(x-\ep\le s Then the function $$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 $$0. Then there exist integers $$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 $$0, which completes the verification that $$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 $$a=1/2$$: 