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Suppose I have differentiable functions $f,g: \mathbb{R} \rightarrow \mathbb{R}_{\geq 0}$ with the property that for all $x \in \mathbb{R}$ we have that

$$ f(x)- g (x) \leq f'(x)- g'(x) $$

This seems like a nice property however I am not sure what the implications are for the functions $f,g$. Any help is appreciated.

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Any help with what? Help with "investigating the condition" seems a bit broad... do you have specific conjectures? – Dirk Jan 22 '13 at 9:37
Do you really want your condition as you wrote it, or you want to have pair of absolute values there? – Martin Argerami Jan 22 '13 at 14:52
up vote 1 down vote accepted

You can use Gronwall-like techniques: for instance $$\frac{d}{dx}e^{-x}(f(x)-g(x)) = e^{-x}\left((f'(x)-g'(x))-(f(x)-g(x))\right) \geq 0,$$ hence $e^{-x}((f(x)-g(x)))$ is non-decreasing. Actually, this statement is totally equivalent to your inequality.

I do not know what kind of consequences you are seeking. For example $$ \lim_{x\to+\infty} e^{-x}(f(x)-g(x))\in\mathbb{R}\cup\{ +\infty\},\qquad\text{and}\qquad\lim_{x\to-\infty}e^{-x}(f(x)-g(x))\in\mathbb{R}\cup\{ -\infty\} $$ both exist.

Another consequence is that, if $f(a) \geq g(a)$ for some $a$, then $f(x) \geq g(x)$ for all $x \geq a$ and $$ f(x) \geq f(x) - g(x)\geq C_a \,e^x $$ where $C_a = e^{-a}(f(a)-g(a))$.

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The derivative $\frac{d}{dx}$ is linear so what you wrote really means $ f(x)- g (x) \leq \frac{d}{dx}(f(x)- g(x)) $ or $ h(x) \leq \frac{d}{dx}h(x) .$ Just consider $f(x)\equiv h(x)+g(x)$, where you implement your bondaries regarding the codomain.

Well okay, the differential equation $ \frac{d}{dx}h(x) = h(x) $ has solutions $A\ \mathbb{e}^x$ and so your $h(x)$ grows stronger.

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As Nick Kidman points out, it is equivalent to characterizing solutions to $$ h(x) \le \frac{d}{dx} h(x)$$ Let $h(x) = e^x \cdot k(x)$ for some function $k$. Now, remark that $h'(x) = e^x \cdot k(x) + e^x \cdot k'(x)$. We want: $$ e^x \cdot k(x) \le e^x \cdot k(x) + e^x \cdot k'(x)$$ $$k'(x) \ge 0 $$ Thus $h$ is a solution iff it is of the form $e^x \cdot k(x)$ for some non-decreasing function $k$. To revert back into the problem statement, $f,g$ work iff their is some non-decreasing function $\ell (x)$ such that $f(x) - g(x) = e^x \cdot \ell(x)$.

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You could choose a different letter instead of reusing $f$... – Rahul Jan 22 '13 at 10:22
Oops, did not notice that. I have editted it. – dinoboy Jan 22 '13 at 19:08

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