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I need to show that if $f$ is a $C^1$ function with $f( 0 ) = 0$ and $f'( x ) > f( x )$ for all $x \in \mathbb{ R }$ then $f( x ) > 0$ for all $x > 0$.

I think I need to show that $f( x ) < 0$ for some $x$ leads to a contradiction ( $f( x ) \neq 0$ for some $x \neq 0$ follows ). I know that if there is at least one point $x_0$ such that $f( x_0 ) < 0$ then by the mean value theorem, I can find infinitely many. So $f \to \infty$ on $( 0, x_0 )$.

I'm not sure how I can exactly reach the contradiction from this knowledge.

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    Consider the function $g(x) = e^{-x}f(x)$. What can you say about its derivative?2012-11-30

2 Answers 2

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Take $g(x)= e^{-x}f(x)$ then observe that $g'(x)= e^{-x}( f'(x)-f(x))>0 \forall x>0$

So $g$ is increasing on $[0,\infty)$ which implies $g(x)\geq g(0)=0 \forall x \geq 0$

ADDED You can see $g$ is increasing from mean value theorem. $\mathrm{for} \, x >0 \exists \xi \in (0, x) \,\frac{g(x)-g(0)}{x}= g'(\xi) >0 $

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    @Carlos Glad I could help:-)2012-11-30
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Assume there is an $x_0>0$ such that $f(x_0)=0$. Since $f(0)=0$, we know $f'(0)>0$ by hypothesis, and we can also apply Rolle's Theorem, which says that there exists a $c\in(0,x_0)$ such that $f'(c)=0$. But $f'(c)>f(c)$ so $f$ must be negative for all such values of $c$. This is a contradiction since $f$ must have been increasing at $x=0$.

I hope this helps.

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    Since $f$ is increasing at $0$ and $f(0)=0$, this means there is a neighborhood containing $0$ where f> 0 except at $0$. In order for $f(x_0)=0$ at any point, the function must come back down since it is $C^1$. Thus, 0=f'(c)>f(c)>0 for some $c$, a contradiction.2012-12-01