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Let $f$ be a differentiable Function on $(0,\infty)$ such that $\lim_{x\to \infty}\frac {f(x)}{x}$=0.

I have to show that $\liminf_{x\to \infty}|f'(x)|=0$.

I had thought to apply the mean value theorem for the function $g(x)= \frac {f(x)}{x}$ over the interval $(x,2x)$ so that $$ \frac{\frac {f(2x)}{2x}-\frac {f(x)}{x}}{x}=g'(c)=\frac{cf'(c)-f(c)}{c^2} $$ for $c\in (x,2x)$. Then $$ \frac {f(2x)-2f(x)}{2}=\frac{cf'(c)-f(c)}{c^2}=\frac {f'(c)}{c}-\frac {f(c)}{c^2} $$ How to proceed further I have no idea. Is my way is right? Or do I have to choose another way? Any help will be appreciated

egreg
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Curious student
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  • "How to Proceed Further" I am not sure what it is exactly you think you "proceeded" with so far, but anyway... What happens if $$\liminf|f'(x)|\geqslant\epsilon$$ for some positive $\epsilon$? This has some easy, rather drastic, consequences on $f'$, hence on $f$. Please state the ones you can think about. (Oh, and by the way, are you sure the hypothesis is that $f$ is differentiable, or that $f$ is continuously differentiable?) – Did May 06 '18 at 12:22
  • First of all Sorry Sir I have not such intention to Write . As per your suggestion,I think if lim inf of that that derivative is positive then that function must goes to Infinity. because function is increasing . – Curious student May 06 '18 at 12:33
  • @Did What would you want continuous derivatives for? – Hagen von Eitzen May 06 '18 at 12:33
  • Function is given to differentible on that interval .Sir what will happen if this function is continously differentible ? – Curious student May 06 '18 at 12:35
  • It seems a duplicate of https://math.stackexchange.com/questions/298832/if-lim-x-to-infty-fx-l-then-lim-x-to-infty-frac-fxx – rtybase May 06 '18 at 12:36
  • @HagenvonEitzen OK, not needed. – Did May 06 '18 at 12:37

1 Answers1

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Let $\epsilon>0$ be given. Then there exists $M_\epsilon>0$ such that $\left|\frac{f(x)}{x}\right|<\frac\epsilon3$ for all $x>M_\epsilon$. So for such $x$, the Mean Value Theorem tells us that there exists $\xi\in(x,2x)$ such that $$x|f'(\xi)|=|f(2x)-f(x)|\le|f(2x)|+|f(x)|<\frac\epsilon3\cdot 2 x+\frac\epsilon3\cdot x=\epsilon x.$$ Therefore $|f'(\xi)|<\epsilon$ for some $\xi \in(x,2x)$ whenever $x>M_\epsilon$. This makes $\liminf|f'(x)|\le \epsilon$. As this holds for all $\epsilon>0$, we get $\liminf |f'(x)|=0$.

Hagen von Eitzen
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