Spring 2018, problem 56

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

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

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

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 $z$.

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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.

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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.

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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 $\{y=x\}$.