Suppose $f$ is a continuous function on $\mathbb{R}$. $0<a<1$. $B(x,r)$ is open ball centered at $x$ with radius $r$. Is it true that $$ \varlimsup_{r\rightarrow 0} \frac{|f(x+r)-f(x)|}{|r|^\alpha} \leq C \varliminf_{r \rightarrow 0^+}\frac{\sup_{x_1,x_2\in B(x,r)} |f(x_1)-f(x_2)|}{r^a} $$ for some positive constant $C$?
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The claim does not hold in general. I shall give a counterexample. I interprete the ball $B(x,r)$ as the interval $(x-r,x+r)$. My example will not be continuous, but one can replace the jumps with linear pieces of fast growing slopes. You will get the drift.
Define $$ f(t)=\begin{cases} 0&t\le 0,\\ e^{-a(n-1)^2}&e^{-n^2}< t\le e^{-(n-1)^2},\ n\in{\mathbb N}\\ 1&t> 1. \end{cases} $$ Let $$ h(r)=\sup_{|x_1|,|x_2|<r}|f(x_1)-f(x_2)| =\sup_{0\le t<r}f(t). $$ Our claim is that $$ \liminf_{r\searrow 0}\frac{h(r)}{r^a}\le 1, $$ whereas $$ \limsup_{r\searrow 0}\frac{f(r)}{r^a}=\infty. $$ Note that $$ h(r)=\sup_{0\le t<r}f(t)=f(r). $$ Let $f^+(r)=\lim_{t\searrow r}f(t)$. Then \begin{align*} \liminf_{r\searrow 0}\frac{h(r)}{r^a} &\le \lim_{n\to\infty} \frac{f(e^{-n^2})}{e^{-an^2}}=1 \end{align*} and \begin{align*} \limsup_{r\searrow 0}\frac{f(r)}{r^a} \ge \lim_n\frac{f^+(e^{-n^2})}{e^{-an^2}}= \lim_n\frac{e^{-a(n-1)^2}}{e^{-an^2}}=\lim_ne^{a(2n-1)}=\infty. \end{align*}