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If $\dfrac{f(x)}{x^2}$monotone increasing function on $x\in (0,+\infty)$,and there exist constant $M$,such $f(x)<M,\forall x\in (0,+\infty)$,then Find the $M_{min}$

If we let $g(x)=\dfrac{f(x)}{x^2}$,then for any $x,y>0(x<y)$,we have $g(x)<g(y)$ or $$\dfrac{f(x)}{x^2}<\dfrac{f(y)}{y^2}$$ but I don't have any idea how to start proving it, Thanks

2 Answers2

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EDIT: It is not needed to assume $f$ continuous.

If $M_{min}>0$ then $\forall \epsilon>0: \exists x>0: f(x)>M_{min}-\epsilon$. Then for small enough $\epsilon>0$ such that $M-\epsilon>0$ we have $\exists x>0: f(x)>M-\epsilon>0$. But then $\frac{f(x)}{x^2}<\frac{M}{x^2}\to 0^+$ when $x\to\infty$ which means that $\frac{f(x)}{x^2}$ can not be monotonically increasing. Therefore $M_{min}\leq 0$. Also if we take $f(x)=-\epsilon,\,\epsilon>0$ then $\frac{-\epsilon}{x^2}$ is monotonically increasing in $(0,\infty)$ and it follows that $M_{min}$ can not be negative (because we can take $\epsilon$ as small as we want). So $M_{min}=0$. The last means that from the assumptions in the question we can only infer that $M_{min}\leq 0$. But we can not infer that $M_{min}<0$ because if we assume it, then there are functions $f(x)$ for which $\frac{f(x)}{x^2}$ is monotonically increasing and $f(x)>M_{min}$.

Svetoslav
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  • It is a good practice for the downvoter to tell what he thinks is not OK with the answer. – Svetoslav Oct 06 '15 at 14:47
  • I did not downvote, but can you elaborate on what you mean by 'because we can take $\epsilon$ as small as we want' – porridgemathematics Oct 06 '15 at 14:52
  • It means that if we assume that from the assumptions in the question it follows that $M_{min}<0$, then we can find immediately a counterexample with $\epsilon>0$ and $f(x)=-\epsilon$ such that $M_{min}<-\epsilon$. Then obviously $f(x)$ is not less than $M_{min}$ and so we can not conclude that $M_{min}<0$ – Svetoslav Oct 06 '15 at 14:56
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    How can we immediately find such a counterexample? – porridgemathematics Oct 06 '15 at 15:01
  • @moorish The counter example is $f(x)=-\epsilon$ for $M_{min}<-\epsilon$. – Svetoslav Oct 06 '15 at 15:06
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    How about $f(x) = \frac{-1}{x} - 70$, clearly $f(x) < -70 $ and $\frac{f(x)}{x^2}$ is monotonically increasing, $-70 < 0$, so how can zero be $M_{min}$? – porridgemathematics Oct 06 '15 at 15:08
  • If $m<0$, $-m$ is larger than zero, so that $f(x) = \frac{-m}{2} > 0 > m$. If you say $m = M_{min} < 0$ $f(x)$ still needs to be less than $M_{min}$, but your 'counter example' of $f(x) = \frac{-m}{2}$ is positive, and your $m = M_{min}$ is negative, this is not a 'counter example' you are just choosing an invalid $f(x)$ as per the assumptions of the question. – porridgemathematics Oct 06 '15 at 15:21
  • @moorish I had a typo in my previous comment. Here is what I meant: In other words, the question asks if we have only that $f(x)<M$ for some $M$ and $\frac{f(x)}{x^2}$ monotonically increasing then what can we infer for the minimal constant $M_{min}$. Clearly, we can not infer that $M_{min}=m<0$, because if we assume that really $m<0$ then starting from the function $f(x)=\frac{m}{2}$ which is less than for example $M=1$ we arrive at a contradiction – Svetoslav Oct 06 '15 at 15:24
  • Do you agree that $f(x) < m$ is an assumption we MUST make? – porridgemathematics Oct 06 '15 at 15:26
  • I think the question is as I paraphrased it, because if it was: Let $f(x)$ is a fixed function, for which you know that $f(x)<M$ and $\frac{f(x)}{x^2}$ is monotone increasing. Then what is the $M_{min}$ ?

    Then $M_{min}$ could be anything, because $f(x)$ could be $-1$, could be $-10000$

     – Svetoslav Oct 06 '15 at 15:27
  • Yes, $M_{min}$ does not exist – porridgemathematics Oct 06 '15 at 15:29
  • But it exists, if you ask the question as : Given the assumptions on $f$, what $M_{min}$ can you infer ?

    Then $M_{min}$ that you can infer is $0$.

     – Svetoslav Oct 06 '15 at 15:30
  • What do you mean given the assumptions on $f$, there is no assumption on $f$ apart from the fact that $f$ is bounded above. – porridgemathematics Oct 06 '15 at 15:31
  • We are to find the least possible number that can bound $f$ above, such that $\frac{f(x)}{x^2}$ is monotonically increasing in $x>0$ – porridgemathematics Oct 06 '15 at 15:32
  • And of course, we can bound $f$ above by as small numbers as we like, and still have $\frac{f(x)}{x^2}$ monotonically increasing, just take $f(x) = \frac{-1}{x} - K$ where $K>0$, this ensures that $f(x) < -K $ – porridgemathematics Oct 06 '15 at 15:33
  • I mean that if you ask the question the other way (not the one to which I answered) then the question is not well posed. Because you can not find the $M_{min}$ given only the assumptions on $f$ – Svetoslav Oct 06 '15 at 15:37
  • What are your assumptions on $f$ apart from the fact that $f$ must be bounded above, and please enlighten as to what your interpretation of the question is. – porridgemathematics Oct 06 '15 at 15:38
  • @moorish Firstly, $M_{min}$ can not be $-\infty$ whatever the question is, because there is no real valued function, for which $f(x)<-\infty$. Secondly, the question is not well posed if you imagine the following situation: I have in mind some function, which is bounded above by some constant $M$ and for which $\frac{f(x)}{x^2}$ is monotonic increasing and I ask you to guess what is $M_{min}$, meaning what is $\sup {f(x)}$. But you can not answer me, because I may have in mind the function $f(x)=-1$, or maybe $f(x)=-100$, or even $f(x)=-100000000$. In each of these cases $M_{min}=-1,-100,$ etc – Svetoslav Oct 06 '15 at 15:45
  • Yes, I did not say $M_{min}$ existed the way the question is posed, which is why I asked you what your interpretation of the question was in order to make it 'answerable' – porridgemathematics Oct 06 '15 at 15:48
  • Your interpretation seems to be 'Calculate $M$ such that for all $f(x)>M$, $\frac{f(x)}{x^2}$ is NOT monotonic increasing. I do not know how you managed to interpret this from the question though. – porridgemathematics Oct 06 '15 at 15:53
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    @moorish Maybe I can state the question like this: If $\forall f:(0,\infty)\to\mathbb R$ : $f<M=const$ and $\frac{f(x)}{x^2}$ is monotonic increasing, what is $M_{min}$ ? – Svetoslav Oct 06 '15 at 16:05
  • Yes, this is how I thought you interpreted the question, then the answer is indeed 0. – porridgemathematics Oct 06 '15 at 16:07
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The function $f$ must not have an infimum. Take $f(x) := -\frac{1}{x}$, then $f(x)$ is bounded from above by $0$ and $\frac{f(x)}{x^2} = -\frac{1}{x^3}$ is monotonically increasing, but $f(x) \to - \infty$ as $x \to 0$.

Three.OneFour
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