Hoelder continuity of $\frac yx$ for $x\in (0,1)$ and $0<y<x^2$

Question

I would like to see a proof that the function $$ f(x,y) = \frac yx $$ is Hoelder continuous with exponent $\frac 12$ on the region $$ D:= \{ (x,y): \ x\in (0,1), \ 0<y<x^2 \}. $$ That is, I am looking for a proof of $$ \left| \frac{y_1}{x_1} - \frac{y_2}{x_2}\right| \le c \left( \sqrt{|x_1-x_2|} + \sqrt{|y_1-y_2|} \right)\quad \forall (x_1,y_1),(x_2,y_2)\in D. $$ Numerical experiments suggests that this inequality is valid for $c=1$. I could not find a proof, all attempts went to nowhere.

Answer 1

Assume without loss of generality that $x_1 \le x_2$. Then $$ \frac{y_1}{x_1} - \frac{y_2}{x_2} = \frac{y_1 x_2 - y_2 x_1}{x_1 x_2} = \frac{y_1(x_2-x_1)}{x_1 x_2}-\frac{y_2 - y_1}{x_2} \, . $$ The first term can be estimated as $$ \left | \frac{y_1(x_2-x_1)}{x_1 x_2}\right| \le \frac{x_1^2}{x_1x_2}|x_1 - x_2| \le |x_1 - x_2| \le \sqrt{|x_1 - x_2|} $$ because $|x_1 - x_2| \le 1$.

The second term can be estimated as $$ \left| \frac{y_2 - y_1}{x_2}\right| = \frac{\sqrt{|y_2 - y_1|}}{x_2} \sqrt{|y_2 - y_1|} \le \frac{\sqrt{\max(y_1, y_2)}}{x_2}\sqrt{|y_2 - y_1|} \\ \le \frac{\max(x_1, x_2)}{x_2}\sqrt{|y_2 - y_1|} \le \sqrt{|y_2 - y_1|} \, . $$

Combining these results we get $$ \left| \frac{y_1}{x_1} - \frac{y_2}{x_2}\right| \le \sqrt{|x_1-x_2|} + \sqrt{|y_1-y_2|} \, , $$ which is the desired estimate with the constant $c=1$.