2
$\begingroup$

$g:[0,1]\to[0,1]$ continuously differentiable and increasing such that for all integers $t>0$ and for all $r\in(0,1)$, $g(r^{t+1})>g(r)\cdot g(r^t)$. Does this imply that for all $r\in(0,1)$, and $m,n \in N$, $g(r^{m+n})>g(r^m)\cdot g(r^n)$? If not, what can be a counter example?

This is the last part of a larger question that I had asked before here: https://math.stackexchange.com/questions/810277/a-unsolved-puzzle-from-number-theory-functional-inequalities. Overcoming this would help me solve the whole question.

Edit: I missed this out, when I posted initially: $g(0)=0$ and $g(1)=1$

Improve this question
$\endgroup$
3
  • $\begingroup$ Just replacing like that does not get very far, unless $(m/n)=t$ for an integer t. Do you see my point @PietroMajer ? $\endgroup$
    – Juanito
    Commented May 27, 2014 at 23:24
  • $\begingroup$ When you write $g:[0,1]\to[0,1]$, you mean $g(0)=0$ and $g(1)=1$, or could it be that $g(0)>0$ and $g(1)<1$ (comparing with your MSE post)? $\endgroup$
    – username
    Commented May 28, 2014 at 17:11
  • $\begingroup$ @AthanagorWurlitzer, thank you for your comment. I have also thanked you on the MSE website. I missed out those two conditions, and have edited the post. $\endgroup$
    – Juanito
    Commented May 28, 2014 at 20:09

1 Answer 1

Reset to default
1
$\begingroup$

$\let\eps\varepsilon$It is easier to consider a function $f(x)=-\log g(e^{-x})$. Then $f\colon (0,\infty)\to(0,\infty)$ is still continuously differentiable and increasing (the bordering conditions are almost not in the game), and it satisfies $$ f((t+1)x)<f(tx)+f(x) \qquad (*) $$ for all $x\in(0,+\infty)$ and positive integer $t$ (I do not know why you set $t>1$...).

Now let us consruct such an $f$ violating the desired condition. Firstly, we set $f(x)=x+\eps$ for a small $\eps>0$, so that it satisfies $(*)$. Next, we perturb $f$ in a small neighborhood of $x=5$ so that now $f(5)$ is slighly greater than $f(2)+f(3)$, thus violating the desired condition. Surely, now it also violates $(*)$, but this happens only for $x$ being in a small neighborhood of $5/2$, or for $x$ being almost in $(0,5/3)$ (or, recursively from $5/2$, for $x$ around $5/4$ and smaller ones). Thus, now it is easy to perturb $f$ in a neighborhood of $5/2$, and then simply to increase it on, say, $(0,\,1.9)$ so that it becomes, say, $1.1x+\eps$ on $(0,\,1.7)$. The details can be simply recovered.

Improve this answer
$\endgroup$
7
  • $\begingroup$ Hello @Ilya, thanks for your reply. I think this should work out. TO get back g(x) from a function $f(x)=-\log g(e^{-x})=x+\eps$ I could replace $x$ by $-logx$, and I end up getting $g(x)=x.e^{-\eps}$. Similarly, $f(x)=1.1x+\eps$ on $(0,\,1.7)$ would give $g(x)=x^{1.1}.e^{-\eps}$. This function would not be differentiable at the kink. I think I missed including the conditions $g(0)=0$ and $g(1)=1$ in my initial post, and I apologize for that. But, given this extra information, $g(1)=1.e^{-\eps} \neq 1.$ $\endgroup$
    – Juanito
    Commented May 28, 2014 at 21:39
  • $\begingroup$ @Juanito: Are you sure then that there are no other misses? I am particularly interested in $t>1$ (in your present formulation, you don't say that $g(r^2)>g(r)^2$ --- is it what you wanted?). $\endgroup$
    – Ilya Bogdanov
    Commented May 29, 2014 at 5:22
  • $\begingroup$ you are absolutely correct, I also require $g(r^2)>g(r)^2$, thus, the correct condition should be $t>0$. $g(r^2)>g(r)^2$ is also implied by $m=n=1$ on the right hand side. $\endgroup$
    – Juanito
    Commented May 29, 2014 at 15:29
  • $\begingroup$ If I use $f(x)=-\log g(e^{-x})$, I would get $g(x)=e^{-f(-\log x)}$. Given the boundary conditions $g(0)=0$ and $g(1)=1$, I would require, the candidate function $f$ to satisfy $f(0)=0$ and $f(\infty)=\infty $. I was thinking of following the steps you recommended by using $f(x)=x^{1/2}$. This satisfies the boundary conditions and $ f((t+1)x)<f(tx)+f(x) $ for all $x \in(0,+\infty)$ and positive integer $t$. I perturb the function at $5,5/2$ and in $(1,1.9)$ Someone mentioned using bump functions to smooth the slope discontinuities. I hope using a bump fn dont violate $ f((t+1)x)<f(tx)+f(x)$. $\endgroup$
    – Juanito
    Commented May 30, 2014 at 5:45
  • $\begingroup$ Is there a way to ensure that the bump fn introduced does not violate $ f((t+1)x)<f(tx)+f(x)$? $\endgroup$
    – Juanito
    Commented May 30, 2014 at 5:55

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .