Derivative bounded and functional value 0 Let $f:[0,1]\to \mathbb{R}$ be a real-valued continuous function which is differentiable on $(0,1)$ and satisfies $f(0)=0$.Suppose there exists a constant $c\in (0,1)$ such that $|f'(x)|\le c|f(x)|$ for all $x\in (0,1)$. Show that $f(x)=0$ for all $x\in [0,1]$
Do we need to use taylor or mean value 
 A: Hint: Observe
\begin{align}
e^{-cx}|f(x)|=&\ \left|\int^x_0 \frac{d}{ds}\left(e^{-cs}f(s) \right)\ ds\right| \leq \int^x_0 |e^{-cs}f'(s)-ce^{-cs}f(s)|\ ds\\
\leq&\ 2c\int^x_0 e^{-cs}|f(s)|\ ds.
\end{align}
This reduces the problem down to showing if
\begin{align}
F(x) \leq 2c\int^x_0 F(s)\ ds
\end{align}
for $F\geq 0$ and $F(0) = 0$ then $F(x) \equiv 0$. 
Additional Hint:

 $\frac{d}{dx}\left(e^{-2cx} \int^x_0F(s)\ ds \right)\leq 0$

A: Putting $x=0$
shows that
$f'(0) = 0$.
$f(x)
=\int_0^x f'(t) dt
$
so
$\begin{array}\\
|f(x)|
&=|\int_0^x f'(t) dt|\\
&\le \int_0^x |f'(t)| dt\\
&\le \int_0^x |cf(t)| dt\\
&= c\int_0^x |f(t)| dt\\
&\le cx\max_0^x |f(t)|
\end{array}
$
Let
$t(x)$ be such that
$|f(t(x))|
=\max_0^x |f(t)|
$.
Putting $t(x)$ for $x$,
$|f(t(x))|
\le ct(x)\max_0^x |f(t)|
= ct(x)|f(t(x))|
$
which is a contradiction
(since
$0 \le ct(x)
\lt 1$)
unless
$f(t(x)) = 0$.
Therefore
$\max_0^x |f(t)|
=0$
for all $x$,
so $f(x) = 0$
for all $x$.
A: Here is a proof by contradiction using directly the MVT without any integration:
Consider the function first on smaller intervals $I_{\epsilon}=[0,1-\epsilon]$.


*

*$f$ is continuous $\Rightarrow \exists x_{\epsilon} \in I_{\epsilon}:\; M_{\epsilon} := |f(x_{\epsilon})|= \max_{x \in I_{\epsilon}} |f(x)|$.

*Assume $M_{\epsilon}>0 \Rightarrow x_{\epsilon} \in (0,1-\epsilon]$


$$\color{blue}{M_{\epsilon}} \color{blue}{\stackrel{x_{\epsilon} \in I_{\epsilon}}{<}} \frac{M_{\epsilon}}{x_{\epsilon}} =\frac{|f(x_{\epsilon})|}{x_{\epsilon}} = \left|  \frac{f(x_{\epsilon}) - f(0)}{x_{\epsilon}-0}\right| \stackrel{MVT}{=} |f'(\xi_{x_{\epsilon}}) | \leq c|f(\xi_{x_{\epsilon}})|\stackrel{c \in (0,1)}{\leq}|f(\xi_{x_{\epsilon}})| \leq \color{blue}{M_{\epsilon}}$$
This contradiction shows, that $M_{\epsilon} = 0 \Rightarrow f(x) = 0$ on each interval $I_{\epsilon}$. With continuity we get $f(x) = 0$ on $[0,1]$.
A: solve by using mean value theorem :
By lagrange mean value theorem $\frac{f(x)-f(0)}{x-0} =f'(x_1)$ where $0<x_1<x$
$|f(x)|\le cx|f(x_1)|\le c^2xx_1|f(x_2)|\le ...\le c^nxx_1...x_{n-1}|f(x_n)|$, where $0<x_n<x_{n-1}<...<x_1<x$.
So $|f(x)|\le (cx)^n |f(x_n)|$.
Now $f$ is continuous on $x\in [0,1]$ .so $f$ is bounded.
it follows that $f(x)=0$ since $c\in (0,1)$ 
Can it be done by this way??
