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?

Another approach is via the method of contradiction. Assume that $f''(0) < 0$. Then the function $f'$ is strictly decreasing at $0$ which means that there is a neighborhood $I$ of $0$ such that if $x \in I, x < 0$ then $f'(x) > f'(0) = 0$ and if $x \in I, x > 0$ then $f'(x) < f'(0) = 0$. We can obviously choose $I$ of the form $(-h, h)$ and hence observing sign of $f'$ in $(-h, h)$ we see that $f$ is strictly increasing in $(-h, 0]$ and strictly decreasing in $[0, h]$ and since $f(0) = 0$ it follows that $f(x) < 0$ for all $x \in I$ except $x = 0$. And this is contrary to the hypotheses that $f$ is non-negative. Thus we obtain the desired contradiction.
BTW the above argument can be replaced by the following concise argument. Since $f'(0) = 0, f''(0) < 0$ the point $0$ is a local strict maximum of $f$ and hence $f(x) < f(0) = 0$ for all $x$ in some neighborhood of $I$ of $0$ except $x = 0$. And this contradicts that $f$ is non-negative.
The argument in first paragraph actually shows how the vanishing of derivative and sign of second derivative guarantee local maxima/minima and if the reader is well aware of the proof of second derivative test for maxima/minima, it is preferable to adopt the argument in second paragraph.

As can be seen from the arguments above we don't need continuity of $f''$. And we can avoid difficult theorems like Taylor or L'Hospital. The argument used in first paragraph leads to one of the simpler proofs of Taylor's Theorem with Peano's form of remainder.
Update: The condition $f' (0)=0$ is superfluous here. The function is non-negative and $f(0)=0$ and hence $0$ is a point of local minima so that $f'(0)=0$.
