The question goes as follows:
Suppose $f: (a,b)\to \mathbb{R}$ is twice-differentiable on $(a, b)$, and suppose that $$ |f''\left( x \right) |\le |f'\left( x \right) |+|f\left( x \right) | $$ for all $x\in (a,b)$.
Suppose there exists an $x_0\in (a,b)$ such that $f(x_0) = f'(x_0) = 0$.
Prove that $f(x) \equiv 0$ for all $x \in (a,b)$.
I tried using the trick below:
$$ \begin{align} \displaystyle \frac{f'\left( x_0+\delta \right) ^2}{2}&=\int_{x_0}^{x_0+\delta}{f'f''dx} \\ &\le \int_{x_0}^{x_0+\delta}{ff'dx}+\int_{x_0}^{x_0+\delta}{\left( f' \right) ^2dx} \\ &=\frac{f\left( x_0+\delta \right) ^2}{2}+\delta \cdot f'\left( x_0+\delta \right) ^2 \\ &\le f'\left( x_0+\delta \right) ^2\cdot \left( \frac{\delta ^2}{2}+\delta \right), \end{align} $$ which is impossible when $\delta \to 0$ if $f'(x_0+\delta)\neq 0$. However, as one can see, there are too many restrictions imposed.
The hint says that one should first prove $f(x)\equiv 0 \,\,(\forall x\in (a,b), |x-x_0|<\frac 12)$, and then prove that $\forall k\in \mathbb{Z}^{+}, |f''(x)|\leq \alpha (1+|x-x_0|)^k\cdot |x-x_0|^k$, where $\alpha := \max\limits_{x\in (a,b)} |f'(x)| +\max\limits_{x\in (a,b)}|f(x)|$. But I can't understand this hint.
