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.
About mean value theorem
1 Answers
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 $-\delta
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0I was thinking of using the direct approach, thinking that f(0) = f'(0) = 0 implies that$f(x)$looks like a smiley face (x^2) or a sad face (-x^2) or like x^3 or -x^3 in some neighbourhood of x = 0. But I think x^4*sin(1/x) is a counterexample. But I'm not too sure – 2012-12-07