I am reading a paper and it uses the following fact:
Suppose $f_1,f_2:X\to (0,\infty)$ where $(X,d)$ is a bounded metric space, and $\gamma\in(0,1)$.
Suppose that $|\log{f_1}|_\gamma\geq |\log{f_2}|_\gamma$ where $|f|_\gamma:=\sup\limits_{x\neq y}\frac{|f(x)-f(y)|}{d(x,y)^\gamma}$
(alternatively, $|f|_\gamma$ is the smallest $C\geq 0$ such that $|f(x)-f(y)|\leq C d(x,y)^\gamma$ $\;\;\forall$ $x,y\in X$).
Then $|\log{(f_1+f_2)}|_\gamma\leq |\log{f_1}|_\gamma$.
I have been stuck on showing this for hours! If somebody could help that would be great!
1. For $a, b,c,d$ four positive real numbers, we have $$ \frac{a+b}{c+d} \leq \max\big\{ \frac{a}{c} , \frac{b}{d} \big\} \qquad (\#) $$
2. For $x,y\in X$, we have $$ \mathbf{J}_{x,y}:=\big\vert \log( f_1(x) + f_2(x) ) - \log( f_1(y) + f_2(y) ) \big\vert = \big\vert \log \frac{ f_1(x) + f_2(x)}{ f_1(y) + f_2(y)} \big\vert. $$ We can assume $ \frac{ f_1(x) + f_2(x)}{ f_1(y) + f_2(y)} \geq 1$, otherwise replacing the positions of x, y.
Then, we deduce from (#) that $$ \mathbf{J}_{x,y} \leq \max\Big\{ \log \frac{f_1(x)}{f_1(y)}, \log \frac{f_2(x)}{f_2(y)} \Big\} \leq |\log f_1|_\gamma d(x,y)^\gamma. $$ In the last step, we used the assumption that $|\log f_1|_\gamma \geq |\log f_2|_\gamma$. Hence, the proof is over.