This seems a bit trivial but there is no explicit proof for the statement nor can I prove it directly myself.
Let $f: X \to \mathbb{R}$ be $\alpha$-Holder continuous for some $\alpha \in (0,1)$, where $X \subset \mathbb{R}^n$ is some compact region.
Then, let us consider the quotient \begin{equation} F(x,y):=\frac{\lvert f(x)-f(y) \rvert}{\lvert x-y \rvert^{\alpha}} \end{equation} defined on $X \times X - \Bigl( x=y \mid x,y \in X \Bigr)$.
Now, I wonder if this $F(x,y)$ is "jointly" continuous with respec to $(x,y)$ on $X \times X - \Bigl( x=y \mid x,y \in X \Bigr)$.
Moreover, is it possible in general to continuously extend $F(x,y)$ to whole of $X \times X$?
Could anyone please explain for me?
Usually you would just say something along the lines of "$F$ is continuous by the standard theorems on operations on continuous functions", but of course that requires seeing which functions those are... or sweeping that issue under the rug...
I'll try to show that $F$ is continuous in a (more) proper way below:
Define the following four functions: $$G : (r,s) \in \mathbb{R} \times \left(\mathbb{R}^n \setminus \{0\}\right) \longmapsto \displaystyle\frac{|s|}{|t|^\alpha} \in \mathbb{R}$$ $$H : ((u, v), (u', v')) \in \big((\mathbb{R} \times X) \times (\mathbb{R} \times X)\big) \setminus \left\{((u,v),(u',v')) \mid v = v'\right\} \longmapsto (u - u', v - v') \in \mathbb{R} \times \left(\mathbb{R}^n \setminus \{0\}\right)$$ $$I : x \in X \longmapsto (f(x), x) \in \mathbb{R} \times X$$ $$J : (x,y) \in \left(X \times X\right) \setminus \left\{(x,x) \mid x \in X\right\} \longmapsto (I(x), I(y)) \in \big((\mathbb{R} \times X) \times (\mathbb{R} \times X)\big) \setminus \left\{((u,v),(u',v')) \mid v = v'\right\}$$ (Very long expressions but you gotta do what you gotta do.)
We have $F = G \circ H \circ J$ with $G$ and $H$ continuous, thus it suffices to show that $J$ is continuous.
To do this this, it suffices to see that $I$ is continuous as its components are continuous, and thus that $J$ is continuous because its own components are also continuous.
Hence, $F$ is continuous.
I do not know whether $F$ can be extended to $X \times X$. It would be possible if $F$ happens to be uniformly continuous thanks to the extension theorem for uniformly continuous functions in metric spaces, as $\left(X \times X\right) \setminus \left\{(x,x) \mid x \in X\right\}$ is dense in $X \times X$ and $\mathbb{R}$ is complete, but I don't know if there's a characterisation for which $f$s have this property of $F$ being uniformly continuous.