About mean value theorem Suppose $f''(x)$ exists on the interval $(-1,1)$,$f(0)=f'(0)=0$,and the inequality $|f''(x)|\leqslant|f(x)|+|f'(x)|$ holds on $(-1,1)$; How to prove that $f(x)=0$ on $(-\delta,\delta)$ for some $\delta>0$? Thanks for help.
 A: Suppose that the conclusion fails. Then for all $\delta>0$, there exists a point $x_\delta\in(-\delta,\delta)$ such that $f(x_\delta)\neq0$. Thus $$m_1:=\max\{|f(x)|:x\in[-\delta,\delta]\}>0.$$ Define $$m_2:=\max\{|f'(x)|:x\in[-\delta,\delta]\}\geq0.$$ Then $\kappa^2:=m_1+m_2>0$. (Both $f,f'$ are continuous, so we can use max rather than sup, but this isn't crucial.) We can write $$|f''(x)|\leq \kappa^2 \quad \hbox{for all}\quad x\in[-\delta,\delta],$$ and so 
$$ - \kappa^2\leq f''(x)\leq \kappa^2\quad \hbox{for all}\quad x\in[-\delta,\delta].$$
Now let $x\in(0,\delta]$. By the mean value theorem, there is a point $c_x\in(0,x)$ such that 
$$f''(c_x)=\frac{f'(x)-f'(0)}{x-0}=\frac{f'(x)}{x}.$$ Thus
$$ - \kappa^2 \leq \frac{f'(x)}{x} \leq \kappa^2 \quad \hbox{for all}\quad x\in (0,\delta],$$
so that 
$$ - \kappa^2x \leq f'(x) \leq \kappa^2 x \quad \hbox{for all}\quad x\in (0,\delta].$$
Integrate this inequality (applying the fundamental theorem of calculus) to get 
$$ -\kappa^2\frac{x^2}{2} \leq f(x) \leq \kappa^2\frac{x^2}{2}\quad \hbox{for all}\quad x\in(0,\delta].$$
Now repeat this process, with $-\delta<x<0$ (integration is over $[x,0]$). Combining the results yields $$|f'(x)|\leq \kappa^2|x|, \quad \hbox{for all}\quad x\in [-\delta,\delta],$$
and 
$$|f(x)|\leq \kappa^2\frac{|x|^2}{2}, \quad \hbox{for all}\quad x\in [-\delta,\delta].$$ Take a maximum through both inequalities and add the results to obtain
$$\kappa^2 \leq \kappa^2(\delta +\frac{\delta^2}{2}).$$
Since $\kappa^2>0$, we can divide through by this term to obtain the contradiction that 
$$1 \leq \delta+\frac{\delta^2}{2}$$
for arbitrarily small $\delta$.
