TITLE: To control first derivative with the function itself: $f'(x)^2\leq Cf(x)$ near where $f(x_0)=f'(x_0)=f''(x_0)=0$.
QUESTION [6 upvotes]: Let $f$ be a compactly supported nonnegative $C^2$ function. 
I want to show that there exists $C$, such that for all $x\in \mathbb R$, we have $f'(x)^2\leq  C f(x) $ by showing that for every point $x\in \mathbb R$, we can find a neighborhood $U$, that on $U$ we can find such a $C$.
However I have a trouble to find such a $U$ and $C$ when $f, f', f''$ all vanish at x. Any advice would be appreciated, thanks.

REPLY [4 votes]: There is an $M>0$ with $$|f''(x)|\leq M\qquad\forall\>x\in{\mathbb R}\ .\tag{1}$$ 
Claim: $\quad f'^2(x)\leq 2M \>f(x)\quad\forall x\in{\mathbb R}$.
Proof. $\ $ When $f(x)=0$ at some point $x\in{\mathbb R}$ then $f'(x)=0$ as well, since otherwise $f$ would assume negative values in some points.
Therefore it is enough to show that $f(0)=:y_0>0$ and $f'(0)=-m_0$ with
$$m_0^2>2My_0\tag{2}$$
is impossible. Assume to the contrary that these conditions hold. From $(1)$ we then get
$$f'(x)\leq-m_0+Mx\qquad(x\geq0)\tag{3}$$
and therefore
$$f(x)=y_0+\int_0^x f'(t)\>dt\leq y_0+\int_0^x(-m_0+Mt)\>dt=y_0-m_0x+{1\over2}Mx^2\qquad(x\geq0)\ .$$
Putting $x:={m_0\over M}$ here it follows from $(2)$  that
$$f\left(m_0\over M\right)\leq y_0-{m_0^2\over 2M}<0\ .$$
From this we conclude that $f(\xi)=0$ for a $\xi$ with $0<\xi<{m_0\over M}$. At the same time, from $(3)$ it would  follow that $f'(\xi)<0$, which is impossible.