## Spring 2018, problem 56

Let $\mathbb{Q}_+$ denote the set of all positive rational numbers and define $f:\mathbb{Q}_+\to \mathbb{R}$ to be a function such that the following two inequalities hold

$$f(xy)\le f(x)f(y), \quad \quad f(x)+f(y)\le f(x+y).$$

Show that if $f$ has a fixed point with value strictly greater than $1$, that $f(x)=x$ for all $x\in \mathbb{Q}_+$. Show by example that the requirement that on the value of the fixed point is necessary.

### Comments

Your answer is wrong. You never made use of the fact that fixed point is greater than 1. You just used the fact it is positive which is not sufficient. Take f(x)= 2x^2 fixed point is 1/2 and satisfies both conditions but it is not identity map.

I found the answer and it is little more complicated but makes assumption that the fixed point is an integer greater than 1. I am trying to fix this but seems difficult

In fact I did not solve the problem: as specified I just listed some "first results"

First results.$f$ being defined only on $\mathbb Q_+$, the choice $f(x):=\lambda x$ gives a counterexample for any real $\lambda\not\in(0,1)$.

From now on we assume the existence of a fixed point. This implies that $f$ vanishes nowhere. In fact the existence of $x_0$ such that $f(x_0)=0$ would imply $f(x)\le0\ \forall x$ (because $f(x)=f(x_0*\frac x{x_0})\le f(x_0)f(\frac x{x_0})=0$); but on the fixed point $f$ must be positive.

From $f(x)=f(1*x)\le f(1)*f(x)$ we get $f(x)*[1-f(1)]\le0$; more generally in this inequality we can replace the number 1 by any $m\in\mathbb N$ because

$(i)\quad\qquad mf(x)=f(x)+f(x)+\cdots+f(x)\le f(x+x+\cdots+x)=f(mx)\le f(m)f(x)$

that we can rewrite in the form

$(ii)\quad\qquad[m-f(m)]*f(x)\le0$

Let us show that $(ii)$ implies $f(x)>0$ for any $x$. In fact $f$ is positive in the fixed point; if $f$ could also take negative values we would get not only

$(iii)\quad\qquad m=f(m)$

but also that all inequalities in $(i)$ are equalities; in particular we have $mf(x)=f(mx)$, property that justifies the last "=" of the next line: $$m*f(\frac xm)=f(\frac xm)+…+f(\frac xm)\le f[\frac xm+…+\frac xm]=f[m*\frac xm]=mf(\frac xm)$$ Now, using again the previous argument (inequalities that are in fact equalities) we get $m*f(\frac xm)=f(x)$, say $f(\frac xm)=\frac{f(x)}m$. Choosing $x=n$ (an integral value) and using $(iii)$, we end up with $f(\frac nm)=\frac nm$, say $f(x)=x\ \forall x\in\mathbb Q_+$ (of course this contradict the change of sign for $f$).

Now from $f(x)=f(1*x)\le f(1)*f(x)$ with $x=1$ we get $f(1)\ge1$; and more generally, from $(i)$, we get a weaker form of $(iii)$, say $f(m)\ge m\ \forall m\in\mathbb N$.