# How to show that $f(x)=0$ for all $x\in [0,1]?$

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.

• $c\in(0,1)$ is written as $c\in(0,1)$, for example. Please format your question. Consult this: math.meta.stackexchange.com/questions/5020/… Apr 13 '20 at 3:30
• A more general version is discussed here. Also see this thread. Apr 17 '20 at 2:24

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

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

• 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? Apr 13 '20 at 4:17
• 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!
– J.G
Apr 13 '20 at 4:33
• $c<1$ 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 $1/c$. Apr 13 '20 at 5:46

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 $$c\in(0,1)$$.

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

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

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 $$x\in[0,1].$$