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This is a question from a sample paper of a entrance exam. I did this type of problems using MVT but this time i dont have any hint how to solve this The question is "let $ f :[0,1]- R $ is differentiable and $ f(0)=0. $ Given that there exist a $c\in (0,1)$ such that $ |f'(x)| \leq c|f(x)|$ for all $ x \in [0,1]. $ Then prove that $ f(x) =0$ for all $x\in [0,1].$

I was thinking if i can prove this by contradiction. That is if I assume $ f(x)>0 $ or $ f(x)<0 $ then by MVT i can arrive at a contradiction. But that's where i am stuck now since $c\in (0,1) $ then $c<1$ then $|f'(x)|<|f(x)|$ but i dont know what to do after that. Next i have some idea if i can show that $f'(x) = 0$ for all $x $ then $ f $ is a constant function and since $f(0)=0 $ then the whole function is zero . That's why i need some hints to slove this problem.

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4 Answers 4

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Hint: suppose the maximum absolute value of $f$ is attained at some $y\in [0,1]$ (by the extreme value theorem). Then \begin{equation} \vert f(y)\vert=\bigg\vert \int_0^y f'(x)dx\bigg\vert\leq \int_0^y \vert f'(x)\vert dx\leq c\int_0^y \vert f(x)\vert dx\leq c\cdot y\cdot \vert f(y)\vert. \end{equation}

What happens if $\vert f(y)\vert\neq 0$?

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  • $\begingroup$ Thank you sir. But i have a question. How can you look at a sum like this and think of this kind of approach which is awesome. Can you suggest me any book of real analysis from where can i learn this kind of approach? $\endgroup$ Apr 13, 2020 at 4:17
  • $\begingroup$ I’m glad you appreciate it! I don’t really have book suggestions; these sorts of things are more experience than anything else as opposed to a technique in some book. Personally, my thought process was just if you want to relate a derivative and the original function, the fundamental theorem of calculus is probably relevant. Hope this helps! $\endgroup$
    – J.G
    Apr 13, 2020 at 4:33
  • $\begingroup$ $c<1$ is also not necessary in this proof after a small modification, you just end up having to repeat the argument on a few intervals of length less than $1/c$. $\endgroup$ Apr 13, 2020 at 5:46
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As I mentioned in a comment, I'm not sure what is wrong with the answer of Eeyore Ho. Here is an elaboration.

Considering instead the differentiable function $g:=f^2=|f|^2$, we note that the assumptions imply $$ g' \le 2c g$$

One can now use Gronwall's inequality to finish, or indeed just follow the proof of Gronwall's inequality. Consider the function

$$ h(x) = g(x)e^{-2cx}$$ then $$ h'(x) = (g'(x)-2c g(x))e^{2cx} \le 0$$ So $h$ is non-increasing. But $h(0) = 0 $ and $h\ge 0$, so $h(x)=0$ for all $x$; so $g(x)=0$ for all $x$, so $f(x)=0$ for all $x$.

PS this doesn't require $c\in(0,1)$.

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I will use the mean value theorem.

Let $a\in [0,1]$ we have by MVT $a_1\in [0,a]$ such that $\frac{f(a)- f(0)}{a} = f'(a_1)\leq c f(a_1)$ thus $f(a) \leq ac f(a_1)$

Again for $a_1 \in [0,a_1]$ we have $a_2\in [0,a_1]$ such that $\frac{f(a_1)-f(0)}{a_1-0} = f'(a_2) \leq c f(a_2)$ combine the first and the second we get

$f(a) \leq ac f(a_1) \leq aa_1c^2f(a_2)$ (note that $c\in (0,1)$ also $a,a_1 \leq 1$ ) So $f(a) \leq c^2 f(a_1)$ keep going we get

$f(a) \leq c^n f(a_n)$

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Let $g(x)=f^{2}(x),\, g(x)\geq 0,$

$$\left |g^{'}(x)\right |\leq 2cg(x)$$ Prove $g(x)=0$ for all $x\in[0,1]. $ And then you can get $ f(x)=0$ for all $x\in[0,1]. $

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