differential inequality implies zero function Let $f$ be twice continuously differentiable on $(-1,1)$, and $f(0)=f'(0)=0$, $\quad|f''(x)|\leq |f(x)|+|f'(x)|$. Show that $f=0$ in some neighborhood of $0$.
How can we deduce something from this $|f''(x)|\leq |f(x)|+|f'(x)|$? I have no idea... Also, argue by contradiction, I could not deduce further...
 A: Since $|f''(0)| \le |f(0)|+|f'(0)| = 0$, we have $0 = f''(0) = \displaystyle\lim_{x \to 0}\dfrac{f'(x)-f'(0)}{x-0} = \lim_{x \to 0}\dfrac{f'(x)}{x}$.
Hence, there exists a $\delta \in (0,1)$ such that $\left|\dfrac{f'(x)}{x}\right| < 1$, i.e. $|f'(x)| < x$ for $x \in (0,\delta)$. 
Then, $|f(x)| = \left|\displaystyle\int_{0}^{x}f'(t)\,dt\right| \le \displaystyle\int_{0}^{x}|f'(t)|\,dt \le \displaystyle\int_{0}^{x}t\,dt = \dfrac{1}{2} x^2$ for $x \in (0,\delta)$. 
Hence, $|f''(x)| \le |f(x)|+|f'(x)| \le x(1+\frac{1}{2}x) \le (1+\dfrac{1}{2}\delta)x \le (1+\delta)x$ for $x \in (0,\delta)$. 
Now, suppose for some $n \in \mathbb{N}$, we have $|f''(x)| \le (1+\delta)^nx^n$ for all $x \in (0,\delta)$. 
Then, for $x \in (0,\delta)$ we have: 
$|f'(x)| = \left|\displaystyle\int_{0}^{x}f''(t)\,dt\right| \le \displaystyle\int_{0}^{x}|f''(t)|\,dt \le \displaystyle\int_{0}^{x}(1+\delta)^n t^n\,dt = \dfrac{(1+\delta)^n}{n+1} x^{n+1}$
and, 
$|f(x)| = \left|\displaystyle\int_{0}^{x}f'(t)\,dt\right| \le \displaystyle\int_{0}^{x}|f'(t)|\,dt \le \displaystyle\int_{0}^{x}\dfrac{(1+\delta)^n}{n+1} t^{n+1}\,dt = \dfrac{(1+\delta)^n}{(n+1)(n+2)}x^{n+2}$. 
Hence, $|f''(x)| \le |f(x)|+|f'(x)| = \dfrac{(1+\delta)^n}{(n+1)(n+2)}x^{n+2}+\dfrac{(1+\delta)^n}{n+1} x^{n+1}$ $= \dfrac{(1+\delta)^n}{n+1}\left(\dfrac{1}{n+2}x+1\right) x^{n+1} \le (1+\delta)^n(1+\delta)x^{n+1} = (1+\delta)^{n+1}x^{n+1}$ for all $x \in (0,\delta)$. 
Therefore, for any positive integer $n$, we have $|f''(x)| \le (1+\delta)^{n}x^{n}$ for all $x \in (0,\delta)$. 
You can get a similar bound for $x \in (-\delta,0)$. Do you see how to proceed from here?
