Let $f\in C^2(\Bbb R^m), f\geq 0$, Hessian matrix of $f$ is upper bounded by some constant $C$. Do we have $|\nabla f|\leq \alpha \sqrt{f}$ for some $\alpha$, even if the Hessian matrix is degenerate?



A property of $C^2$ functions
$\newcommand\R{\mathbb R}$Let $\R:=R$. Suppose that $|f''(x)(h,h)|\le C|h|^2$ for all $x$ and $h$ in $\R^m$ -- this is how we interpret the condition "Hessian matrix of $f$ is upper bounded by some constant $C$". Of course, here $f''(x)$ is the bilinear form that is the second derivative of $f$ at $x$, so that $f''(x)(h,k)=h^\top H(x)k$ for all $x,h,k$ in $\R^m$, where $H(x)$ is the Hessian matrix of $f$ at $x$.


Consider first the case $m=1$. Take any $x\in\R$ and any real $h>0$. Then
$$0\le f(x+h)\le f(x)+f'(x)h+Ch^2/2,$$
whence
$$f'(x)\ge-\frac{f(x)}h-\frac C2\,h=-K\sqrt{f(x)}$$
if $h=\sqrt{2f(x)/C}$,
where $K:=\sqrt{2C}$. Similarly (or by the left-right symmetry), $f'(x)\le K\sqrt{f(x)}$, and hence
$$|f'(x)|\le K\sqrt{f(x)} \label{1}\tag{1}$$
for all $x\in\R$.


Now take any natural $m$. Considering the restrictions of $f$ to all straight lines in $\R^m$, we see that \eqref{1} holds for all $x\in\R^m$, where now $|f'(x)|$ denotes the norm of the linear form $f'(x)$ that is the derivative of $f$ at $x$, so that $f'(x)(h)=h^\top\,\nabla f(x)$ for all $h\in\R^n$ and hence
$|f'(x)|=|\nabla f(x)|$.


Thus, the desired conclusion holds, in general.


