Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a function such that $f'(x)$ is continuous and $|f'(x)|\le|f(x)|$ for all $x\in\mathbb{R}$ 
Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a function such that $f'(x)$ is continuous and $|f'(x)|\le|f(x)|$ for all $x\in\mathbb{R}$. If $f(0)=0$, find the maximum value of $f(5)$.

$f'(x)=f(x)$ is true when $f(x)=ke^x$. 
$f(x)=0$ satisfies the condition. So $f(5)=0$ which is also the correct answer. But is there any method other than substitution?
 A: The inequality indicates
$$
\left|\frac{\mathrm{d}}{\mathrm{d}x}\log\left|f(x)\right|\,\right|\le1
$$
If $\log\left|f(5)\right|=a$ is finite, then $\log\left|f(0)\right|\ge a-5$. Since $f(0)=0$, we must have $f(5)=0$.
A: Chose $N\in\mathbb N$ arbitrarily. Then $f$ is continuous in $[0,N]$ and therefore attains its maximum and minimum, so  $\exists M>0:|f(x)|\leq M,\,\forall x\in [0,N]$.
For each $x\in[0,N]$ you have
 $$f(x)=\int\limits_{0}^{x}{f'(t)dt}\Rightarrow |f(x)|\leq \int\limits_{0}^{x}{|f(t)|dt}\leq \int\limits_{0}^{x}{Mdt}=Mx$$
Take the inequality $|f(t)|\leq Mt,\forall t\in [0,N]$ and integrate it again in $[0,x]$ for arbitrary $x\in [0,N]$ to get 
$$|f(x)|\leq \int\limits_{0}^{x}{|f(t)|dt}\leq \int\limits_{0}^{x}{Mtdt}=\frac{Mx^2}{2}$$
and continue the same way. You get that $|f(x)|\leq \frac{Mx^n}{n!},\forall n\in\mathbb N$ which means that $|f(x)|\equiv 0,\,\forall x\in [0,N]$ because $\lim\limits_{n\to\infty}\frac{x^n}{n!}\to 0,\,\forall x\in\mathbb R$. Because $N$ was arbitrarily big it follows that $f(x)\equiv 0,\,\forall x\in \mathbb R$ and this is the only function satisfying the conditions in the question $\Rightarrow $ the maximum value at $x=5$ is the only possible value $0$.
