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Is there a function $f:\mathbb{R}^+\to \mathbb{R}^+$ such that:

1) $f$ continuously differentialble at $(0,\infty)$

2) $f(0)=0$

3) $f$ and its derivative $f'$ are both non-decreasing

4) there exists a positive constant $c<1$ such that for all $x>0$: $xf'(x)\leq\ cf(x)$

The only function I think of is the zero function. Do those conditions imply automatically that $f=0$ ?

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    What about $f(x)=exp(x/2)-1$ ? – 2017-02-02
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    @LJSilver It doesn't seem true that $\frac{x}{2}e^{x/2}\leq ce^{x/2}$ for all $x$ (condition 4). – 2017-02-02
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    Sorry I didnt see the x premultiplying the f' – 2017-02-02
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    So yes, f(x)=cx does the job, as Doug M said – 2017-02-02
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    @LJSilver actually, no, that is why I removed it. $f(x) = cx \implies xf'(x) = cx > cf(x) = c^2 x$ when $c< 1$ – 2017-02-02
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    Sorry again. Too tired to think properly. So I d say yes thats the zero one. In fact f(x) is C^1. So take any positive x. By the mean value theorem there exists a y in (0,x) such that f(x)/x=f'(y). But f' is non decreasing and hence f'(y)\le f'(x). It follows that f(x)/x \le f'(x), i.e. xf'(x) \ge f(x) that violates 4 as long as f(x) and f'(x) are not null.. sound correct to you guys, at least the intuition? – 2017-02-02
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    @LJSilver that is what I was thinking. – 2017-02-03

1 Answers 1

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Over any interval $(0,a)$

$f(a) \le \int_0^a \max(f'(x)|x\in(0,a)) dx =$

Since $f'(x)$ non-decreasing $\max (f'(x)|x\in(0,a)) = f'(a)$

$f(a) \le af'(a)$

but we have the restriction.

$xf'(x) \le c f(x)$

and $c<1$

$f(x) \le xf'(x) \le c f(x) < f(x)$

Indeed the $0$ function is the only function that meets all the criteria.