$|f′(x)| \leq c|f(x)|$ for all $x \in (0,1)$ and $f(0)=0$. Show that $f(x) = 0$ for all $x \in [0,1]$. Let $f : [0,1] → \mathbb R$ be a real-valued continuous function which is differentiable on $(0,1)$ and satisfies $f(0) = 0$. Suppose there exists a constant $c ∈ (0, 1)$ such that$|f′(x)| \leq c|f(x)|$ for all $x \in (0,1)$ and $f(0)=0$. Show that $f(x) = 0$ for all $x \in [0,1]$.
My proof - Given $f(0)=0$, let $x \in (0,1),$ then by MVT, we have for some $x_0 \in (0,x)$ s.t. $$|f(x)/x| = f'(x_0)$$ and we have $0<x_0<x$. Similarly, applying MVT again, for some $x_1$ such that $0<x_1<x_0<1$ we have
$$|f(x)|= c|f(x_0)|x_0<c|f(x_0)|<c|f'(x1)|x_1 $$
$$|f(x)|< c^2|f(x_1)|x_1<c^2|f(x_1)|.$$
Continuing this way we have
$$|f(x)|<c^n|f(x_n)|$$
and as $n \rightarrow \infty $, $|f(x)|=0$.
Is this argument correct?
 A: There is no need to iterate the MVT. 
For $x\in (0,1]$ we have 
$$|f(x)|\le \left |\frac{f(x)}{x}\right| \overset{MVT}{=} |f'(m_x)|<c |f(m_x)|\le c \max_{x\in [0,1]} |f(x)|$$ 
If the maximum on the RHS is attained at $0$, then we're done. Otherwise, it is  attained at some point $x_1\in (0,1]$, and we therefore have  $|f(x_1)| \le c |f(x_1)|$, which implies $|f(x_1)|=0$. 
A: All you need to observe is this: Suppose not. Let $y \in (0,1)$ be such that $f(y) \not = 0$. Then by MVT there is an $x \in (0,y)$ s.t. $|f'(x)| = \frac{|f(y)|}{y} > 0$. Thus $|f'(x)|$ is strictly positive, and by the continuity of $f$ on $(0,1)$, it follows that $f(x)$ is bounded. Thus $|\frac{f'(x)}{f(x)}|$ is strictly greater than 0 (or is $\infty$). So take $c =|\frac{f'(x)}{2f(x)}|$ if $f(x)$ satisfies $f(x) \not = 0$ and take $c=1$ if $f(x)$ satisfies $f(x)=0$, it will follow that $f'(x) > cf(x)$ for such $x$, which contradicts the assumption.  
A: The MVT yields an $x_0$, such that $|f(x)/x|=|f^{\prime}(x_0)|$. It is not quite clear to me how your construction continues at the "similarly" part. You would need to elaborate on how you construct these constants.
Alternatively, note that $|f^{\prime}(x)|\le c|f(x)|$ for all $c\in(0,1)$ implies that $|f^{\prime}(x)|=0$ for all $x\in(0,1)$ (take a sequence $(c_n)_n$ in $(0,1)$ such that $c_n\rightarrow0$ and take limits in the inequality). Thus $f$ is constant on $[0,1]$ and you may conclude from $f(0)=0$ that $f(x)=0$ for all $x\in[0,1]$.
A: If $|f'(x)| \leq c|f(x)|$ for all $c \in (0,1)$, then $|f'(x)| = 0$, implying $f$ is constant. Since $f(0) = 0$, $f$ is zero everywhere.
