# Proving $f(x)>0$ on $\mathbb{R}^+$ given that $f(0)=0$ and $f'(x)>f(x)$

Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be differentiable, with the properties

(1) $\ f(0)=0$, and

(2) $\ f'(x)>f(x)$ for all $x\in\mathbb{R}.$

Prove that $f(x)>0$ for all $x\in\mathbb{R}^+.$

It's clear that one can get that $f$ is positive on some interval, by utilizing the $\epsilon$-$\delta$ definition of the derivative (using $\epsilon=f'(0)/2>f(0)/2=0$) for all positive $x$ in some $\delta$-neighborhood centered at $0$. How would one get this over all positive reals? I've tried to proceed by contradiction, but that hasn't yielded much for me thus far, using things like the mean value or creating auxiliary functions. Maybe someone with a bit more wit could direct me down the right path?

• I think the condition $f\neq0$ is necessary or $f'>f$! Oct 2, 2017 at 2:31
• I've edited the body. I'm assuming you meant $>$ rather than $\geq$. Oct 2, 2017 at 2:37
• Typo. Should've been strict. Oct 2, 2017 at 2:37
• Thanks. Beat me by 10 seconds! Oct 2, 2017 at 2:38

## 2 Answers

Set $g(x) = e^{-x}f(x)$. One then has $g(0) = 0$ and $g'(x) = -e^{-x}f(x) + e^{-x}f'(x) = e^{-x}(f'(x)-f(x)) > 0$ for all $x$.

Thus, $g(x) > g(0)$ for all $x \in \mathbb{R}^+$, or $f(x) > 0$ for all $x\in \mathbb{R}^+$.

• Very nice. Thanks! Oct 2, 2017 at 2:26
• You are welcome.
– GAVD
Oct 2, 2017 at 2:26
• Very nice solution. Oct 2, 2017 at 7:03

$$f'(x)>f(x)\Longleftrightarrow e^{-x}(f'(x)-f(x))>0\Longleftrightarrow \frac{d}{dx}(e^{-x}f(x))>0$$ $$\Longleftrightarrow 0\le\int_0^x\frac{d}{dt}(e^{-t}f(t)) dt =e^{-x}f(x)- f(0) =e^{-x}f(x)$$ that is $$e^{-x}f(x)\ge 0\Longleftrightarrow f(x)\ge 0$$

But $f'(x)>f(x)\ge0$ which implies that $f'(x)>0$ which means $f$ is strictly increasing Therfore, $$0<x\implies 0=f(0)<f(x)$$

That is $f(x)>0$ on $(0,\infty)$