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A Hölder norm of square root of a $C^2$ function

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Background

I am reading a proof of the Calabi-Yau theorem from these notes. In page 15 he claims the following statement without proof (calling it elementary): Let $M$ be a compact manifold. There exists some $C$ that depends only on $M$ such that$$||\sqrt{h}||_{1,0}\leq C(1+||h||_{1,1}),\qquad \forall h\in C^2(M), h\geq 0 .$$

The author says this follows from a one real variable analysis. I found this stack-exchange question showing the one-dimensional case if $h>\epsilon>0$ (bounding with the $C^2$ norm, though). The answerer claims it can be proven for $h\geq 0$ by approximations, which I am failing to see too.

In the Calabi-Yau proof, the author then goes on to use it with $||\sqrt[n-1]{h}_{1,0}||$ instead of $||\sqrt[2]{h}||_{1,0}$.

My Questions

  1. How do you prove this in the one-dimensional case with $h\geq 0$ general?
  2. Is the $C^{2,0}$ norm bounded by the $C^{1,1}$ norm? I doubt this is true...
  3. How to conclude this for the multiple dimensional (compact) case?
  4. What about the fact this result is used for a root that is not the square root?

My attempt

Following the guidance, assuming this is true for all $h\geq \epsilon>0$, take $h_\epsilon=h+\epsilon$, and then, clearly$$ \sup{\sqrt{h_\epsilon}}\leq \sup\sqrt{h}+\sqrt \epsilon,$$so $\sqrt{h_\epsilon}\xrightarrow{C^1}\sqrt{h}$. For the Lipschitz bound, I do:\begin{align*}|\sqrt{h_\epsilon(x)}-\sqrt{h_\epsilon(y)}-(\sqrt{h(x)}-\sqrt{h(y)})| &= \left\vert\int_0^\epsilon\frac{\partial}{\partial \eta}\left({\sqrt{h_\eta(x)}}-{\sqrt{h_\eta(y)}}\right)d\eta\right\vert \\&\leq \frac12\int_0^{\epsilon} \left\vert\frac{1}{\sqrt{h_\eta(x)}}-\frac{1}{\sqrt{h_\eta(y)}}\right\vert d\eta \\&\leq C(1+||h||_{1,1})|x-y|\int_0^\epsilon\frac1{\sqrt{h_\eta(x)h_\eta(y)}}d\eta.\end{align*}This is almost good, except that the last integral is not integrable. In fact, writing this, I see that such an approach will not work when at a point $x$ such that $h(x)=0$, because then the LHS is independent of $x$.

I am quite disappointed someone writes such a statement that is impossible to verify in such a dismissing manner.


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