Let $f$ be a bounded and continuous function, $0<a < 1$. $U(x,r)$ is the neighborhood of $x$ with diameter $r$. Can we prove the following equation of two limits
$$ \lim\_{r\rightarrow 0} \sup\_{y,z \in U(x,r)} \frac{|f(y)-f(z)|}{|y-z|^a} = \lim\_{r\rightarrow 0} \sup\_{y,z \in U(x,r)} \frac{|f(y)-f(z)|}{r^a}$$
whenever they are finite or infinite?



A limit problem
This equality is not true. E.g., if $f(u)=u$, $x=0$, and $a=1$, then the left-hand side of the equality is $1$, whereas its right-hand side is $2$.




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The OP has now switched the meaning of $r$ from radius to diameter. The equality is still not true. E.g., if $f(u)=|u|$, $x=0$, and $a=1$, then the left-hand side of the equality is $1$, whereas its right-hand side is $1/2$.


