Prove or disprove: A differentiable function $f$ is always non-negative with this condition

Question

I want to prove that a differentiable function $f: [0,\infty) \rightarrow \mathbb{R} $, which satisfies the following condition is always non-negative:

Assume $f(0)=0$ and whenever $f(a)=0$, then $f'(a) \geq 0$ and $f''(a) \neq 0$.

Note that if I change the condition to this: $f(0)=0$ and whenever $f(a)=0$, then $f'(a) > 0$, then it's always non-negative. But in the condition I have $f'(a) \geq 0$ and that makes the problem a bit more complicated. Therefore I have added the second derivative condition $f''(a) \neq 0$.

So if this condition doesn't make it non-negative, what could be a slight modification on the condition which makes it non-negative? (e.g. $f'(a) \geq 0$ and $f''(a) > 0 ?$)

Answer 1

As Dieter Kadelka noted, the condition "$f(0)=0$ and whenever $f(a)=0$, then $f'(a)\ge0$ and $f''(a)\ne0$" is not enough for $f\ge0$.

On a positive note, we will have $f\ge0$ if the following condition (say $C_m$) holds for some $m\in\{1,2,\dots,\infty\}$:

Suppose that a function $f\colon[0,\infty)\to\mathbb{R}$ with $f(0)=0$ is continuous and has derivatives $f^{(j)}$ on $[0,\infty)$ of all natural orders $j<m$. (The derivatives $f^{(j)}(0)$ are understood as the right derivatives.) Suppose also that for any real $a\in[0,\infty)$ such that $f(a)=0$ there is a natural $k<m$ such that $f^{(j)}(a)=0$ for $j\in\{0,\dots,k-1\}$ and $f^{(k)}(a)>0$.

Indeed, suppose the contrary: the set $$E:=\{x\in[0,\infty)\colon f(x)<0\}$$ is nonempty. Let $a:=\inf E$. Then $a\in[0,\infty)$ and $f(a)=0$. So, there is a natural $k<n$ such that $f^{(j)}(a)=0$ for $j\in\{0,\dots,k-1\}$ and $f^{(k)}(a)>0$. So, by downward induction on $j$, in a right neighborhood of $a$ we have $f^{(j)}>0$ for all $j\in\{0,\dots,k-1\}$. In particular, $f>0$ in a right neighborhood of $a$, so that $\inf E>a$, which contradicts the definition of $a$.

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The condition $C_1$ will never hold, because $f(0)=0$ but there is no natural $k<1$.

The second part of the condition $C_2$ means that for any real $a\in[0,\infty)$ such that $f(a)=0$ we have $f'(a)>0$.

The second part of the condition $C_3$ means that for any real $a\in[0,\infty)$ such that $f(a)=0$ we have either $f'(a)>0$ or $f'(a)=0<f''(a)$.