# How to find $f(x)$ when $f\left(\frac{x+y}{x-y}\right)\ge\frac{f(x)-f(y)}{f(x)+f(y)}$

We have the function $f(x)$ continuous on $(-\infty,0)\cup (0,+\infty)$ with $f(1)=-1$ $\displaystyle\lim_{x\to 0}xf(x)=-1$ and $f'(1)$ exsits.

For any $x,y\in\mathbb R$ we have $$f\left(\dfrac{x+y}{x-y}\right)\ge\dfrac{f(x)-f(y)}{f(x)+f(y)}.$$

Show that:$$f(x)=-\dfrac{1}{x}$$

My idea: Let $y=x-1$. Then $$f(2x-1)\ge \dfrac{f(x)-f(x-1)}{f(x)+f(x-1)}.$$ Now let $x=3$, then we have $$f(3)\ge\dfrac{f(2)-f(1)}{f(2)+f(1)}$$

PS: I don't want prove $f(x)=-\dfrac{1}{x}$ satisfies these conditions. I want to show that no other solutions exist.

## 1 Answer

It is unclear if the functional inequality $$\tag{\star}{f\left(\frac{x+y}{x-y}\right)}\ge \frac{f(x)-f(y)}{f(x)+f(y)}$$ should be read as implicitly stating that for all $x,y\in\mathbb R\setminus\{0\}$ with $x\ne y$, we have $f(x)+f(y)\ne0$; or maybe the inequality should be treated as meaningless in such a case. In what follows, the latter, more conservative approach is used, i.e. instances of $(\star)$ are considered only where everything is defined.

Lemma 1. There exists $c<1$ such that $|f(x)|<c$ for $|x|$ sufficiently big.

Proof. Pick $y\ne 0$ with $f(y)\ne 0$. Then for $|x|\gg |y|$, the left hand side of $(\star)$ is $\approx f(1)=-1$ by continuity. Hence the right hand side $\frac{f(x)-f(y)}{f(x)+f(y)}$ must be negative or undefined, hence $|f(x)|<|f(y)|$ of $f(x)=-f(y)$. Starting with $y=1$, $f(y)=-1$, this shows $-1< f(x)\le 1$ for large $|x|$. Then by picking $y$ with $-1<f(y)\le 1$, we see that $-1\le f(x)<1$ and hence $-1<f(x)<1$ for $|x|$ large enough. Finally, pick $y$ with $c:=|f(y)|<1$ to conclude $|f(x)|<c$ for $|x|$ large enough. $_\square$

Lemma 2. $f(x)\ne 0$ for all $x\ne 0$.

Proof. Assume $f(y)=0$ for some $y$. Then for any $x$ with $f(x)\ne0$ (and hence $x\ne y$) we have $f(\frac{x+y}{x-y})\ge\frac{f(x)}{f(x)}= 1$. If $y$ is not an interior point of $f^{-1}(0)$, we can let $x\to y$ while keeping $f(x)\ne 0$. As this gives us a contradiction to Lemma 1, we conclude that $f^{-1}(0)$ is open. As $f$ is continouus, $f^{-1}(0)$ must also be closed, hence must contain a connected component of $\mathbb R\setminus\{0\}$ either completely or be disjoint to it. From $\lim_{x\to0}xf(x)\ne0$ we conclude that $f$ cannot be constantly zero on either of the conncted components $(-\infty,0)$, $(0,\infty)$. The claim follows. $_\square$

Lemma 3. $xf(x)<0$ for all $x\ne 0$.

Proof. From $\lim_{x\to 0}xf(x)=-1$ the claim follows for $x\approx 0$. By continuity and Lemma 2, it follows for all $x\ne 0$. $_\square$

Lemma 4. $f$ is odd.

Proof. For $y\ne 0$ let $x=-qy$ and consider the limit behaviour of $(\star)$ as $q\to 1^+$. Then $\frac{x+y}{x-y}=\frac{q-1}{q+1}\to0^+$, hence $\frac{x+y}{x-y}f\left(\frac{x+y}{x-y}\right)\to -1$ and $f\left(\frac{x+y}{x-y}\right)\to -\infty$. As $f(x)-f(y)\to f(-y)+f(y)$, the numerator remains bounded for $x\approx- y$ and we conclude $f(x)+f(y)\to 0$, so that the claim follows by continuity. $_\square$

Lemma 5. If $|u|>1$ then $f(u)=-\frac1u$.

Proof. For $u$ with $|u|>1$, there exist $x,y>0$ with $u=\frac{x+y}{x-y}$. Then for any $t>0$ $$f(u)=f\left(\frac{tx+ty}{tx-ty}\right)\ge \frac{f(tx)-f(ty)}{f(tx)+f(ty)} =\frac{tf(tx)-tf(ty)}{tf(tx)+tf(ty)}$$ (where $f(tx)+f(ty)\ne 0$ follows from Lemma 3). As $t\to 0$, we have $tf(tx)\to -\frac1x$ etc., hence $$f(u)\ge\frac{-\frac1x-\frac1y}{-\frac1x+\frac1y} =-\frac1u.$$ Since $f$ is odd, we also have $$-f(u)=f(-u)\ge -\frac1{-u}=\frac1u$$ and hence the claim. $_\square$

Theorem. $f(u)=-\frac1u$ for all $u\ne 0$.

Proof. The case $|u|>1$ is handled by Lemma 5, the case $|u|=1$ by continuity. For $u$ with $0<|u|<1$, there exist $x,y$ with $x\ne y$ and $u=\frac{x+y}{x-y}$. For $t \gg 0$ we have $f(tx)=-\frac1{tx}$, $f(ty)=-\frac1{ty}$ from Lemma 5. Then $$f(u)\ge\frac{f(tx)-f(ty)}{f(tx)+f(ty)} =\frac{-\frac1{tx}+\frac1{ty}}{-\frac1{tx}-\frac1{ty}}=-\frac1u.$$ Again, using oddness we also have $-f(u)=f(-u)\ge -\frac1{-u}=\frac1u$ and ultimately $f(u)=-\frac1u$. $_\square$

Remark: While the continuity of $f$ was essential, the above did not make use of the existence of $f'(1)$.

• It seems that not only the existence of $f'(1)$ is not needed. The condition $\lim_{x\to0}xf(x)=-1$ can also be loosened to: $\lim_{x\to0}xf(x)$ exists and is nonzero. – Hagen von Eitzen Nov 25 '14 at 16:05