I have been spending times on the following question for 2hrs, and I couldn't get any kind of proper proof yet.
Q: Let $f$ be continuous on $[0,\infty)$ and differentiable on $(0,\infty)$. If $f(0)=0$ and $|f'(x)|\leq |f(x)|$ for all $x>0$, then $f(x)=0$ for all $x\geq 0$.
What I guessed are as follows:
First trial: Tried to use the MVT to show the contradiction when we assume that there exists a real $k$ s.t $f(k)\neq0$... but I failed.
Second trial: I saw that suppose $|f'(x)|= |f(x)|$ , we set a function $g(x)=\frac{f(x)}{e^x}$, then $f(x)=0$ for all $x\geq0$. But this is not the exact one since the original question involves inequality ($\because |f'(x)|\leq |f(x)|$)
Third trial: I tried to divide the domain into $[0,1]$ and $[1,\infty)$ and show that $f(x)=0$ for each of the intervals... Somehow I think I can prove in this way (for the interval $[0,1]$, used the property of compact set $\to$ so that there exists maxima & minima (supposed to be all 0 for this $f$ I guess)... and so on and for the interval $[1,\infty)$, I had no idea.
Will there be any suggestions on this question? I would like to know if what I was missing or doing some error on my trials. Need your help.
Here is a hint for the interval $[0,\frac 12]$. Let $M = \max_{[0,1/2]} |f|$ and choose $x \in [0,\frac 12]$ with the property that $|f(x)| = M$. If $M = 0$ you have that $f$ is identically $0$ on $[0,\frac 12]$. Otherwise $x > 0$.
Now apply the Mean Value Theorem. There is a point $c \in (0,x)$ satisfying $$\frac{f(x) - f(0)}{x-0} = f'(c).$$ Under the hypotheses of the problem this gives you $M = |f(x)| = |x||f'(c)| \le |x||f(c)| \le \frac 12 M.$ This implies $M = 0$.
The conclusion is that $f$ is constant on $[0,\frac 12]$. Can you extend this to the rest of the line?