$\frac{f'(x)}{f(x}\le \frac{g'(x)}{g(x)} \Rightarrow f(x)\le g(x) \,\forall\,x\in[a,b]$

Question

Let $\,f,g:\,[a,b]\rightarrow \mathbb R$ be differentiable, postive functions with $f(a)=g(a)$ and $\frac{f'(x)}{f(x)}\le \frac{g'(x)}{g(x)}\,\forall\,x\in[a,b]$ $$Prove,\,that: \frac{f'(x)}{f(x)}\le \frac{g'(x)}{g(x)} \Rightarrow f(x)\le g(x) \,\forall\,x\in[a,b]$$

I suppose it has something to do with Rolles Theorem and/or the mean value theorem for differential equations, but I have no idea how I can constructively approach this problem. Thank you in advance for any help!

coltrane

Answer 1

$\frac{d}{dx}\log f\leq \frac{d}{dx}\log g $ implies $\log f\leq \log g$ by integrating both sides. By exponentiation, $\,f\leq g$.

If both $f$ and $g$ are negative one may apply the same argument to $-f$ and $-g$, obtaining $-f\leq -g$, hence $f\geq g$.

Answer 2

We have $$(\ln f(x))' \leq (\ln g(x))'$$ so $$ \Big(\ln g(x)/f(x)\Big)'\geq 0 = c'$$

so $$g(x) /f(x) \geq e^c$$ and thus $$g(x) \geq e^cf(x)$$ Since $f(a)=g(a)$ we get $c=0$ so $$g(x) \geq f(x)$$

Answer 3

Correct me if wrong.

A bit of detail:

$\dfrac {d}{dx} \log f(x) \le \dfrac {d}{dx} \log g(x);$

$\dfrac {d}{dx} (\log g(x) - \log f(x)) \ge 0.$

This implies that $(\log g(x) - \log f(x))$

is a monotonically increasing function :

For $b \ge x \ge a:$

$\log g(x) - \log f(x) \ge$

$\log g(a) -\log f(a) = 0,$

since $g(a)=f(a).$

Hence $ \log g(x) \ge \log f(x),$ and

finally by exponentiation :

$g(x) \ge f(x).$