Prove: if $f(0)=0$ and $f'(0)=0$ then $f''(0)\geq 0$ 
let $f$ be a nonnegative and differentiable twice in the interval $[-1,1]$
Prove: if $f(0)=0$ and $f'(0)=0$ then $f''(0)\geq 0$

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*Are all the assumptions on $f$ necessary for the result to hold ?


*what can be said if $f''(0)= 0$ ?

Looking at the taylor polynomial and lagrange remainder we get:
$$f(x)=f(0)+f'(0)x+\frac{f''(c)x^2}{2}$$
$$f(x)=\frac{f''(c)x^2}{2}$$
Because the function is nonnegative and $\frac{x^2}{2}\geq 0$ so $f''(c)\geq 0$
For 1., all the data is needed but I can not find a valid reason.
For 2., can we conclude that the function the null function?

Your proof is enough when $f''$ is continuous.

Here's a way without the continuity assumption.
This Taylor expansion $f(x) = f(0) + xf'(0) + \frac{x^2}{2}f''(0) + o(x^2)$ yields here: $$f(x) = \frac{x^2}{2}f''(0) + o(x^2)$$
Either $f''(0)=0$ and we're good, otherwise the previous equality rewrites as 
$\displaystyle \lim_{x\to 0}\frac{2f(x)}{f''(0) x^2}= 1$
Consequently, $f(x)$ and $f''(0) x^2$ must share the same sign on a neighborhood of $0$.
Since $f\geq 0$ and $x^2\geq 0$ that implies $f''(0)\geq 0$

When $f''(0)=0$, $f(x)=o(x^2)$. I don't see anything more you could say.
