Prove that $f(x)\equiv0$ on $\left[0,1\right]$ if $f(0)=0$ and $|f'(x)|\le|f(x)|$ Let $f(x)$ differentiable on $\left[0,1\right]$ such that $f(0) = 0$.  Also, assuming that $\forall x\in \left[0,1\right]:\left|f'(x)\right| \le \left|f(x)\right|$. Prove that $f(x)\equiv 0$  
What I did so far:
Using LMVT:
$$\frac{f(x)-f(0)}{x-0} = \frac{f(x)}{x} = f'(c)$$ where $c\in \left(0,x\right)$
$$\left|f('c)\right| = \left|\frac{f(x)}{x}\right|.$$  
Since, $\left|x\right| \in (0,1)$
$$\left|\frac{f(x)}{x}\right| \ge \left|f(x)\right| \ge \left|f'(x)\right|$$
So, $$\left|f('c)\right| \ge \left|f'(x)\right|$$
But I'm not sure how is it helping me. What am I missing?  
 A: Let $x_0 \in [0,1]$ with $f(x_0) = 0$. We will show that $f(x) = 0$ for all $x \in [x_0, x_0 + \varepsilon]$ for some $\varepsilon > 0$ (if $x_0 < 1$). This is sufficient to prove your claim because you can then consider $\gamma := \sup \{x \in [0,1] \mid f|_{[0,x]} \equiv 0\}$, show $\gamma = 1$ and deduce $f \equiv 0$.
Because $f$ is continuous and $f(x_0) = 0$, there is some $\varepsilon_1 \in (0, \frac{1}{2})$ with $|f(x)| \leq \frac{1}{2}$ for all $x \in [x_0 , x_0 + \varepsilon]$.
We will now show inductively that actually $|f(x)| \leq \left(\frac12\right)^n$ for all $x \in [x_0, x_0 + \varepsilon]$. For $n=1$ this follows by our choice of $\varepsilon$.
For the inductive step note that $|f'(x)| \leq |f(x)| \leq \left(\frac12\right)^n$ for all $x \in [x_0, x_0 + \varepsilon]$. Using the mean-value theorem, it is now easy to see that
$$|f(x)| = |f(x) - f(x_0)| = |f'(\xi)| \cdot |x - x_0| \leq \left(\frac{1}{2}\right)^n \cdot |x - x_0| \leq \left(\frac{1}{2}\right)^{n+1}$$
for $x \in [x_0, x_0 + \varepsilon]$, because $\varepsilon \leq \frac{1}{2}$.
In the limit $n\rightarrow \infty$, we get our claim.
A: Assume $f$ is not identically zero on $[0,1]$. Since $f$ is continuous on $[0,1]$, there is a maximum point $b\in[0,1]$ such that $|f(b)|\geq |f(x)|$ for all $x\in [0,1]$. If possible, we will choose $b<1$. In other words, either $b<1$, or $b=1$ and $|f(x)|<|f(b)|$ for all $x\in[0,b)$. Without loss of generality, we may assume that $f(b)>0$. (Otherwise consider the function $-f(x)$.)
Use the mean value theorem on the interval $[0,b]$. This yields a point $c\in (0,b)$ such that $f'(c)={f(b)-f(0)\over b-0}={f(b)\over b}$. There are now two cases:
(i) If $b<1$ then $f'(c)={f(b)\over b}>f(b)\geq |f(c)|$, which contradicts $|f'(x)|\leq |f(c)|$.
(ii) If $b=1$ then $f'(c)=f(b)>|f(c)|$, again contradicting $|f'(x)|\leq |f(c)|$.
A: This is similar to an older question 
Should $f(x) \equiv 0$?.
Consider $g(x)=\mathrm e^{-x}f(x)$. Then $g'(x)=\mathrm e^{-x}\left(f'(x)-f(x)\right)$. Consider a fixed $x\in[0,1]$. If $f(x)\geq0$ then $g(x)\geq0$ and since $\lvert f(x)\rvert>\lvert f'(x)\rvert$, we have $g'(x)\leq0$. Conversely, if $f(x)\leq0$ then $g(x)\leq0$ and $g'(x)\geq0$. So we have, for all $x\in[0,1], g(x)g'(x)\leq0$. And consequently $h(x)=\frac12g(x)^2$ is decreasing on $[0,1]$. As $h(0)=0$ and $h(x)\geq0$ we conclude that $h(x)=0$ for all $x\in[0,1]$ and thus $f(x)=0$ for all $x\in[0,1]$.
A: You can use the fact that $|f'(c)| \leq |f(c)|$ and since $f$ is differentiable it is also continuoues, therefore you could get the equality $|f(x)| = |\frac{f(x)} x |$
which means $f(x)=0$
A: $\displaystyle \left |\frac{f(x)}x \right | \le |f(c)| \implies |f(x)| \le |f(c)| $ where $0<c<x$. This implies $\displaystyle |f(x)| \le \left| \lim_{\epsilon \to 0^+}f(\epsilon)\right| = 0$
To show that $c_n \to 0$, either there exists finitely $c \in (0, x)$ such that $f'(c) = f(x)/x$ or there and interval $(a,b)$ where  $f'(c) = \mathrm{constant}$, second case, $c = \inf\{(a,b)\}$.
A: Solution 1: Suppose by contradiction that there exists $x_0 \in (0,1]$ such that $f(x_0) \neq 0$. Because $f$ is continuous, there exist $a<b$ such that $f$ is nonzero on $[a,b]$; without loss of generality, suppose it is positive (otherwise consider $-f$). In particular, you have
$$b-a \geq \int_a^b \frac{f'(x)}{f(x)}dx = \ln(f(b)) - \ln(f(a)).$$
Now, according to mean value theorem, there exists $c \in (a,b) \subset (0,1)$ such that
$$1 < \frac{1}{c} = \frac{\ln(f(b))- \ln(f(a))}{b-a} \leq 1,$$
a contradiction.
Solution 2: Let $x_m \in [0,1]$ be such that $|f(x_m)|= \max\limits_{x \in [0,1]} |f(x)|$. Without loss of generality, we may suppose $f(x_m) \geq 0$ (otherwise consider $-f$). By mean value theorem, for all $x<x_m$, there exists $c \in (x,x_m)$ such that
$$f(x_m) \geq |f(c)| \geq |f'(c)| = \left| \frac{f(x_m)-f(x)}{x_m-x} \right| = \frac{f(x_m)-f(x)}{x_m-x}.$$
Therefore,
$$f(x_m) \leq \frac{f(x)}{1-x-x_m} = \frac{f(x)}{x} \cdot \frac{x}{1-x-x_m}. \hspace{1cm} (1)$$
On the other hand, from Taylor's expansion
$$f(x) = \underset{=0}{\underbrace{ f(0) }} + x \underset{=0}{\underbrace{ f'(0) }} + o(x),$$
hence $\frac{f(x)}{x}=o(1)$. When $x \to 0$, $(1)$ gives $0 \leq f(x_m) \leq 0$. Consequently, $f \equiv 0$.
A: From the application of the LMVT we get $$f(x)=xf'(\theta x),\ $$ for some $\theta\in [0,1]$. So, $$|f(x)|=x|f'(\theta x)|\le x|f(\theta x)|$$ So, by repeated application of this inequality $n$ times you get $$|f(x)|\le \prod_{k=1}^n \theta_k \cdot x^nf\left(x\prod_{k=1}^n \theta_k \right)$$ Since $x,\ \theta_i,\ i=1,2,\cdots ,\ n\in [0,1]$, as $n\to \infty$ for any $x\in [0,1]$, $f(x)=0$.
