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.


4 years ago

First results. A almost complete solution in my second post

$f$ 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


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 $f(m)\ge m\ \forall m\in\mathbb N$.

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.

chorgeshashank1729 4 years ago

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

chorgeshashank1729 4 years ago

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

Claudio 4 years ago
3 years ago

Let $\varphi>1$ be a fixed point. We will prove that $f(x)\equiv x$ under the assumption:

$(\star)\qquad\qquad\varphi\in\mathbb N$

As a first consequence of $(\star)$ we have:

$(\bullet)\qquad\qquad f(\varphi*x)=\varphi*f(x)\quad\forall x\in\mathbb Q_+$

In fact, using both functional inequalities on $f$ we have: $$f(\varphi*x)\le f(\varphi)*f(x)=\varphi*f(x)= f(x)+\dots+f(x)\le f(x+\dots x)=f(\varphi*x)$$ thus all inequalities are in fact equalities.

We are now in measure of prove the existence of infinitely many fixed points, namely:

$(\star\star)\qquad\qquad f\left(\varphi^z\right)=\varphi^z\qquad\forall z\in\mathbb Z$

Such a property will easely follow from suitable choices of $x$ in $(\bullet)$:

  • $x=1$ gives $f(1)=1$; in particular $(\star\star)$ holds true for $z=0$;

  • $x=\frac1\varphi$ gives $f(1)=\varphi*f\left(\varphi^{-1}\right)$ thus (because of $f(1)=1$) property $(\star\star)$ holds true also for $z=-1$;

  • $x=\varphi^n$ with $n>0$ gives $f\left(\varphi^{n+1}\right)= \varphi*f\left(\varphi^n\right)$ thus, by induction, $(\star\star)$ holds true for any $z>0$;

  • finally choosing $x=\varphi^{-n}$ in $(\star)$ we get by induction that $(\star\star)$ holds true for any negative value of $z$.


Remarks (i) As I proved in my previous post, the existence of a fixed point implies $f(x)>0\ \forall x$

(ii) Our proof will show that the only function $f:\mathbb Q_+\to\mathbb R_+$ satisfying $(\star\star)$ and the second functional inequality:

$(\star\star\star)\qquad\qquad f(u)+f(v)\le f(u+v)\qquad u,v>0$

is the identity function, say $f(x)\equiv x$; in particular the first functional inequality on $f$ is no more needed.

(iii) the relevance of the assumption $(\star)$ was pointed out by chorgeshashank1729 in his post.


From now on we will denote by $f$ a positive function on $\mathbb Q_+$ satisfying $(\star\star)$ and $(\star\star\star)$; and we will prove that $f(x)\equiv x$.

Let us summarize the proof. For any $z\in\mathbb Z$ we denote by $S_z$ the vertical strip of points $(x,y)$ such that $\varphi^z\le x\le\varphi^{z+1}$; remark that the union of these strips cover the whole half-plane $x>0$; thus, denoting by $\mathbb G(f)$ the graph of $f$ and by $\mathbb G_z(f)$ the intersection of $\mathbb G(f)$ with the strip $S_z$, we have $\mathbb G(f)=\bigcup_z\mathbb G_z(f)$. In each strip we want to build a family of parallelograms $\mathbb P_z(\lambda)$, depending on a parameter $\lambda\in[0,1[$, in such a way that $\mathbb G_z(f)\subseteq\mathbb P_z(\lambda)$.

Let $D_z$ be the intersection of the strip $S_z$ with the first bissectrice $\{y=x\}$; of course the vertices of the segment are in $\mathbb G(f)$, and we want to build $\mathbb P_z(\lambda)$ "around" $D_z$. More precisely, for any $\lambda$ we define the our parallelograms $\mathbb P_z(\lambda)$ by imposing that one of the diagonals is $D_s$; furthermore two parallel sides will be vertical, being part of the boundaries of the strip; and the other two sides have $\lambda$ as slope.

How to prove that such $\mathbb P_z(\lambda)$ contain the whole of $\mathbb G_z$? E.g. for $\lambda=0$ the parallelogram $\mathbb P_z(0)$ is the square with diagonal $D_z$; and the wanted property $\mathbb G_z(f)\subseteq\mathbb P_z(0)$ follows because $f$ is increasing and the vertices of $D_z$ belong to the graph.

Joining the South-East vertices of thes squares we get a stright line, say $y=\mu*x$ with $\mu:=\frac1\varphi$; the graph $\mathbb G(f)$ being contained in a family of parallelograms that are all above such a line, from $(\star\star\star)$ we get $$f(u)-f(v)\ge\mu(u-v)\qquad\qquad(u>v>0)$$ From the weaker $f(x)-f(y)\ge0$ we derived $\mathbb P_z(0)\supseteq\mathbb G_z(f)$; now let us show that the stronger bound on $f$ implies $\mathbb P_z(\mu)\supseteq\mathbb G_z(f)$.
In fact, starting from $v=\varphi^z$ and moving to a $u$ on the right, the new inequality ensures that we must remain above $y=u+\mu(x-u)$; while starting from $u=\varphi^{z+1}$ and moving to a $v$ on the left, we must remain below $y=u+\mu(x-u)$.

Remark The strategy can easely be followed on a picture describing some of the strips and the corresponding parallelograms for the initial steps of $\lambda$; however, figures being not allowed on the site, I suggest the reader to draw by yourself the picture in order to better understand the proof.


Our argument applying to the general case, we can conclude:

Claim For any $\lambda\in[0,1[$, if $\mathbb G_z\subseteq\mathbb P_z(\lambda)$ then, for a suitable $\mu>\lambda$, it is $\mathbb G_z\subseteq\mathbb P_z(\mu)$.

Starting from $\lambda=0$ and proceeding in such a way, the upper bound for the values of $\lambda$ is 1; thus for any $z$ the parallelogram $\mathbb P_z(\lambda)$ can be stretched towards the diagonal segment $D_z$; in other words no point of the graph can exist outside the bissectrice $\{y=x\}$.