Consider the function $H: \mathbb{R}^2 \to \mathbb{R}$ defined by
$$H(x,y) = \frac{-1}{2} \left[ a - x^2 + \sqrt{a^2 + x^4 -2ax^2 + 4y^2}\right]^{\frac{1}{2}}$$
where $a > 0$ is fixed.
This function is clearly well defined and $C^{\infty}$. One can easily see that $H$ is lipschitz on any ball centered at $(0,0)$ with radius $R$ since $|\nabla H|$ is bounded there. I am having trouble computing the Lipschitz constant $C_R$. I want to find how it depends on $R$. Does it go to infinity as $R$ goes to infinity? If so, at what rate?
I computed the derivative $|\nabla H|$ and tried to find a bound for it, but it gets very messy. I get an upper bound for $C_R$ but I think it's very weak, and there probably is a way to get a sharper upper bound on $C_R$. Any help is appreciated.
I do not know if it is helpful, but the proposed formula resembles the $\sinh^{-1}$ function.
Precisely speaking, if $z > 0$, one has that \begin{align*} \sinh^{-1}\left(\frac{w}{z}\right) = \ln\left(\frac{w + \sqrt{w^{2} + z^{2}}}{z}\right) \end{align*}
At your case, one has that \begin{align*} a - x^{2} + \sqrt{a^{2} + x^{4} - 2ax^{2} + 4y^{2}} & = a - x^{2} + \sqrt{(a-x^{2})^{2} + (2y)^{2}}\\\\ & = \left(\frac{a - x^{2} + \sqrt{(a-x^{2})^{2} + (2y)^{2}}}{2y}\right)\times 2y\\\\ & = \exp\left(\sinh^{-1}\left(\frac{a-x^{2}}{2y}\right)\right)\times 2y \end{align*}
Hopefully this helps!