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 $C$?



Compare two limits related to Hölder condition
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\*}


