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The statement of the problem is as follows, Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be differentiable, and $f''(0)$ exists and is greater than $0$. We also have $f(0) = 0$. Prove that there exists an $x > 0$ such that $f(2x) > 2f(x)$. I have tried doing this based on the sequential definition of limits and the definition of the derivative but I have only been able to show that there exists $x > 0$ such that $f(2x) > f(x)$.

EDIT: There was an additional condition I forgot to include, $f(0) = 0$

Peter Booth
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1 Answers1

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\begin{align} f(x) = f(0) + f'(0) x + \frac{f''(0)}{2} x^2 + R(x) \end{align}

We have $f(0) = 0$ and $a = f''(0) > 0$ and some $b = f'(0)$. Thus

\begin{align} \Delta(x) = f(2x) - 2 f(x) &= 2 b x + 2 a x^2 + R(2x) - 2 b x - a x^2 - 2 R(x) \\ &= a x^2 + R(2x) - 2 R(x) \end{align} where $$ \lim_{x \to 0} \frac{R(x)}{x^3} = 0 $$ This means the rests decrease faster than positive $ax^2$ when we make $x$ smaller (but keep $x > 0$), so there must be a $x > 0$ with $\Delta(x) > 0$.

mvw
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