If $f''(x) \gt 0$ for $x\in(a,b)$, and $f(a)=f(b)=0$, then $0 \gt f(x)$ for all $x \in (a,b)$. I find a word in my calculus exercise book. It says that:

If $f''(x) \gt 0$ for $x\in(a,b)$, and $f(a)=f(b)=0$, then $0 \gt f(x)$ for all $x \in (a,b)$.

I try to prove this. First, since $f''(x) \gt 0$, so $f'(x)$ is increasing in interval $(a,b)$. However, I don't know how to continue the proof. I want to use IVT, but the three kinds of IVT need that $f(x)$ also continious in $[a,b]$, in this statement, I can not prove $f(x)$ is continious at point $a$ or $b$.
In my opinion, if we want to prove that $0 \gt f(x)$, I try to prove that $0 \gt\frac{f(x)-f(a)}{x-a}  \forall x \in (a,b)$ and use Lagerange theorem, but I can't prove that continuity at $a$ otr $b$.
In fact, I think that maybe this statement is false. Can anyone tell me if this statement right and if it is right how to prove it?
ps: I construct this counterexample. (I can't be sure about the correctness.)

 A: The condition $f(a)=f(b)=0$ means nothing without continuity at $a$ and $b$.  A counter-example would be $f(x)=e^{x}$ for $a <x<b$ and $0$ for $x=a$ and $x=b$.
Assuming continuity at $a$ and $b$ we can prove this as follows.
$f$ is a convex function so $f(x)=f(ta+(1-t)b) \leq tf(a)+(1-t)f(b)=0$ where $t=\frac {b-x} {b-a}$. So $f(x) \leq 0$. It remains to prove strict inequality. If $f(x)=0$ for some $x \in (a,b)$ then MVT shows that $f'(s)=0$ and $f'(t)=0$ for some points $s \in (a,x)$ and $t \in (x,b)$.  But then $f''(z)=0$ for some $z$ between $s$ and $t$, contradicting the hypothesis.
A: Suppose $f(u)\geq 0$, $u\in (a,b)$.
First case: $f(u)=0$. There exists $v_1\in (a,u), v_2\in (u,b)$ with $f'(v_1)=f'(v_2)=0$ implies there exists $w\in (v_1,v_2)$ such that $f"(w)=0$.
Case 2. $f(u)>0$, there exist $v_1\in (a,u), v_2\in (u,b)$ with $f(v_1)=f(v_2)>0$ implies there exists $w\in (v_1,v_2)$ with $f'(w)=0$. Since $f">0$, it implies that for every $x<w$, $f'(x)\leq f'(w)=0$, implies that $f$ decreases on $[a,w]$, we deduce that $f(a)\geq f(v_1)$ since $v_1<w$, contradiction.
