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Suppose that $f$ satisfies $a$-Holder 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 $\delta$.

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 $\delta$.

Suppose that $f$ satisfies $a$-Holder 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 $\delta$.

Suppose that $f$ satisfies $a$-Holder 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 $\delta$.

Suppose that $f$ satisfies $a$-Holder condition on $[0,1]$ (0<a<1). Fix $0<b<a$. For and $x\in [0,1]$ 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 $\delta$.

Suppose that $f$ satisfies $a$-Holder 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 $\delta$.