Given differentiable $f$ with $f(0)=0$, $f''(0)>0$ prove there exists $x > 0$ such that $f(2x) > 2f(x)$ Let $f : \mathbb R \to\mathbb R$ be a differentiable function such that $f(0) = 0$, and $f''(0)$ exists and is positive. Prove that there exists $x > 0$ such that $f(2x) > 2f(x)$.
By Taylor's theorem with Lagrange's reminder I can write
$$f(x)=xf'(0)+x^2f''(0)/2+x^3f'''(y)/6$$
with $0<y<x$, so if I write $g(x)=f(2x)-2f(x)$ I have
$$g(x)=x^2f''(0)+x^3f'''(y).$$
If $f'''$is bounded in $(0,h)$ for some $h>0$, then
I have $g(x)/x^2$ tends to $f''(0)>0$ as $x$ tends to 0, and by the limit definition $g(x)>0$ for some $x$ small enough, and we're done. However we don't have conditions for $f'''(x)$ to be bounded near 0, so I don't know how to get $g(x)>0$ near 0.
 A: Let $g(x)=f(2x)-2f(x)$, so $g(0)=0$ and $g'(x)=2f'(2x)-2f'(x)$ so $g'(0)=0$. Finally, $g(x)=\frac{g''(0)}{2}x^2+o(x^2)$ which in turn gives $g(0)>0$ for $x \in (0,\varepsilon)$ since it behaves like a degree two polynomial with positive degree two coefficient close to zero. That gives you the answer and you don't need to assume any more differentiability for $g$.
A: At $x=0$, $f(2x)$ and $2f(x)$ are both $0$ and equal.
At $x=0$, their derivative are both $2f'(0)$ and equal.
At $x=0$, their derivative is increasing as $f"(0)>0$.
The double derivative of the former is $4f''(0)$, is larger than that of the latter, $2f''(0)$.
Hence at a point "close" to 0, the derivative of the former increases faster than the latter, and as they both start from the same point, the former has a value larger than the latter for a point "close" to 0.
To be more "rigorous", you may follow the same procedure by defining a function $g(x)=f(2x)-2f(x)$ and show that it is positive at some point(but I would prefer the more logical approach as above :) )
A: I'll assume the following: $f$ is differentiable in a neighborhood of $0,$ and $f''(0)>0.$ We do not need the existence of $f''$ anywhere else and we certainly don't need the existence of $f'''$ anywhere.
From Taylor we then have
$$f(x) = 0+ f'(0)x + f''(0)x^2/2 +o(x^2).$$
Thus
$$f(2x)= 2f'(0)x + 2f''(0)x^2 + o(x^2),$$
whereas
$$2f(x) = 2f'(0)x +f''(0)x^2 + o(x^2).$$
Subtracting, we get
$$f(2x) - f(2x) = f''(0)x^2 + + o(x^2).$$
Because $f''(0)>0,$ the last expression will be positive for small nonzero $x.$
