Estimate an improper integral 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 $\delta$.
 A: $\newcommand\ep\varepsilon\newcommand{\de}{\delta}\newcommand{\De}{\Delta}$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 $a=1/2$:

