An inequality from derivatives Let $f:\;\mathbb{R}\to\mathbb{R}$ be differentiable with $f(0)=0$ and $f''(0)$ exists and is positive. I would like to show that there exists $x>0$ such that $f(2x)>2f(x)$.
My try: 
$$\begin{aligned}f''(x)&=\frac{1}{h}\Big(f'(x+h)-f'(x)\Big)+\frac{o(h)}{h}\\&=\frac{1}{h^2}\Big(f(x+2h)-2f(x+h)+f(x)+o(h)\Big)+\frac{o(h)}{h}\end{aligned}$$
At $0:$
$$\begin{aligned}f''(0)&=\frac{1}{h^2}\Big(f(2h)-2f(h)\Big)+\frac{o(h)}{h^2}+\frac{o(h)}{h}\end{aligned}$$
Therefore $\lim_{h\to 0}\frac{1}{h^2}\Big(f(2h)-2f(h)\Big)>0$ so there must exist $\epsilon>0$ such that $\forall x\in (0,\epsilon),\;\frac{1}{x^2}\Big(f(2x)-2f(x)\Big)>0$ which gives  $f(2x)>2f(x)$.
The problem is I don't know how to deal with the $\frac{o(h)}{h^2}$ term (I don't know how to argue that it vanishes). Can I make this more rigorous?
 A: I finally have the proof for the general case.


*

*Let me restate clearly the hypothesis.


$f: \mathbb R \rightarrow \mathbb R$ is differentiable over $\mathbb R$ and twice differentiable at $0$, such that $f(0)=0$, and $f''(0)>0$.


*

*Let $g:\mathbb R \rightarrow \mathbb R$ defined as follows: $$\forall x,  g(x) =f(2x)-2f(x)$$


Then $$\forall x, g'(x) =2(f'(2x)-f'(x))$$
Notice that $g'(0)=0$
Let's see what $g''(0)$ looks like.
$$g''(0)=2\lim_{x \rightarrow 0} \frac{f'(2x)-f'(x)}{x}=2\left( \lim_{x \rightarrow 0} \left(2\frac{f'(2x)-f'(0)}{2x} -\frac{f'(x)-f'(0)}{x}\right)\right)$$
Hence $$g''(0)=2f''(0)$$


*

*By the definition of limit,
there exists $\delta>0$ such that$$x\in]-\delta,\delta[ \Rightarrow \|\frac{g'(x)}{x} -2f''(0) \| < f''(0)$$


and equivalently
$$x\in]-\delta,\delta[ \Rightarrow f''(0)< \frac{g'(x)}{x} < 3f''(0)$$
Hence 
$$x\in[0,\delta[ \Rightarrow 0<f''(0)< g'(x)$$
Hence
$$x\in[0,\delta[ \Rightarrow 0< g'(x)$$


*

*This means that $g$ is strictly increasing over $[0,\delta[$


Since $g(0)=0$ and by definition of $g$
$$\forall x \in ]0,\delta[, f(2x)>2f(x)$$.
This is even slightly stronger than what you requested.
A: Under the given assumptions there are numbers $a$ and $b$ with $b>0$ such that
$$f(x)=a x+b x^2+o(x^2)\qquad(x\to0)\ .$$
It follows that
$$f(2x)-2f(x)=2b x^2\bigl(1+o(1)\bigr)\qquad(x\to0)\ ,$$
which implies that $$f(2x)-2f(x)>0\qquad(0<x<h)$$ for some $h>0$.
