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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.

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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 $\delta$.

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You have to be a bit careful if $f''$ is not continuous. I'm not sure about the validity of the fundamental theorem of calculus in this case, but you can replace its usage by the mean value theorem. – Florian Dec 5 '12 at 17:02
The hypothesis that $f'$ is differentiable seems to be enough to allow use of the fundamental theorem - see what wikipedia calls the second fundamental theorem of calculus - the proof of which is essentially the mvt patch that you're suggestion would lead to. – user12477 Dec 5 '12 at 17:13
I'm not quite convinced; $f''$ can become unbounded and non-Lebesgue integrable (OK, in this case it can't; I'm only arguing about a general fundamental theorem of calculus without further hypotheses on $f'$ or $f''$. And the Wikipedia page states Riemann-integrability as an assumption). – Florian Dec 5 '12 at 17:32
I share your lack of conviction! I missed the bit about Riemann integrability being a hypothesis... – user12477 Dec 5 '12 at 17:50
I've edited the answer to avoid the illegal use of the fundamental theorem. Thanks @Florian. – user12477 Dec 5 '12 at 18:07

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