Function Exercise asked in Csb.gov.lb Level 2

Exercise:

Consider the function $f:\mathbb{R}\to \mathbb{R}$ defined to be

         $f\left( {{x}_{1}}+{{x}_{2}} \right)=\frac{f\left( {{x}_{1}} \right)+f\left( {{x}_{2}} \right)}{1+f\left( {{x}_{1}} \right)f\left( {{x}_{2}} \right)}$  $\forall {{x}_{1}},{{x}_{2}}\in \mathbb{R}$ With $f\left( {{x}_{1}} \right)f\left( {{x}_{2}} \right)\ne -1$

1) Show that the possible values of $f\left( 0 \right)$ are $-1,0\,\And \,1$

2) Show that , $f$ is constant function whenever $f\left( 0 \right)=\pm 1$

3) Now assume that $f\left( 0 \right)=0$

    a)    Show that $f$ is odd function

    b)    Show that $f$ is different from $\pm 1$ $\forall x\in \mathbb{R}$

    c)   Show that,$\left| f\left( x \right) \right|<1\,\,,\,\,\forall x\in \mathbb{R}$

Solution:

1) First of all to obtain $f\left( 0 \right)$ we need to make ${{x}_{1}}+{{x}_{2}}=0\Leftrightarrow 0+0=0$

Hence ${{x}_{1}}=0\,\,\And \,\,{{x}_{2}}=0$ thus $f\left( 0 \right)=\frac{f\left( 0 \right)+f\left( 0 \right)}{1+{{f}^{2}}\left( 0 \right)}=\frac{2f\left( 0 \right)}{1+{{f}^{2}}\left( 0 \right)}$

$\Leftrightarrow f\left( 0 \right)+{{f}^{3}}\left( 0 \right)=2f\left( 0 \right)\Leftrightarrow f\left( 0 \right)\left[ 1+{{f}^{2}}\left( 0 \right)-2 \right]=0\Leftrightarrow f\left( 0 \right)=0\,\,or\,{{f}^{2}}\left( 0 \right)=1$

Thus $f\left( 0 \right)=0\,or\,f\left( 0 \right)=\pm 1$

2) Fix ${{x}_{2}}=0\,\,\And \,{{x}_{1}}=x$

$f\left( {{x}_{1}} \right)f\left( {{x}_{2}} \right)\ne -1\Leftrightarrow f\left( x \right)f\left( 0 \right)\ne -1\Leftrightarrow f\left( x \right)\ne \pm 1$

So $f\left( {{x}_{1}}+{{x}_{2}} \right)=f\left( x \right)=\frac{f\left( x \right)+f\left( 0 \right)}{1+f\left( x \right)f\left( 0 \right)}$

And consider two cases for $f\left( 0 \right)$

If $f\left( 0 \right)=-1\Leftrightarrow f\left( x \right)=\frac{f\left( x \right)-1}{1-f\left( x \right)}=\frac{f\left( x \right)-1}{-\left( f\left( x \right)-1 \right)}=-1$ since $f\left( x \right)\ne 1$

If $f\left( 0 \right)=1\Leftrightarrow f\left( x \right)=\frac{f\left( x \right)+1}{1+f\left( x \right)}=1$ since $f\left( x \right)\ne -1$

Thus $f$ is constant function whenever $f\left( 0 \right)=\pm 1$

3) (a) We know that $f$ is odd function iff $f\left( -x \right)=-f\left( x \right)$

We have $f\left( 0 \right)=0$ so put ${{x}_{1}}=x\,\,\And \,\,{{x}_{2}}=-x$

Thus $f\left( {{x}_{1}}+{{x}_{2}} \right)=f\left( x-x \right)=\frac{f\left( x \right)+f\left( -x \right)}{1+f\left( x \right)f\left( -x \right)}$

$\Leftrightarrow f\left( 0 \right)=0=\frac{f\left( x \right)+f\left( -x \right)}{1+f\left( x \right)f\left( -x \right)}\Leftrightarrow f\left( x \right)=-f\left( x \right)$

Therefore $f$ is odd function.

(b) as $f$ is odd function and $f\left( 0 \right)=0$ so ${{x}_{1}}=x\,\,\And \,\,{{x}_{2}}=-x$

Thus $f\left( 0 \right)=\frac{f\left( x \right)+f\left( -x \right)}{1+f\left( x \right)f\left( -x \right)}=\frac{f\left( x \right)-f\left( x \right)}{1-{{f}^{2}}\left( x \right)}$ as $f$ is odd

Hence $f$ is defined when $1-{{f}^{2}}\left( x \right)\ne 0\Leftrightarrow f\left( x \right)=\pm 1$

( c) we need now to prove $\left| f\left( x \right) \right|<1$ , $\forall x\in \mathbb{R}$

We have $f\left( {{x}_{1}}+{{x}_{2}} \right)=\frac{f\left( {{x}_{1}} \right)+f\left( {{x}_{2}} \right)}{1+f\left( {{x}_{1}} \right)f\left( {{x}_{2}} \right)}$

Put ${{x}_{1}}=\frac{x}{2}\,\,\And \,\,{{x}_{2}}=\frac{x}{2}\Leftrightarrow {{x}_{1}}+{{x}_{2}}=\frac{x}{2}+\frac{x}{2}=x$

So $f\left( x \right)=\frac{f\left( \frac{x}{2} \right)+f\left( \frac{x}{2} \right)}{1+{{f}^{2}}\left( \frac{x}{2} \right)}=\frac{2f\left( \frac{x}{2} \right)}{1+{{f}^{2}}\left( \frac{x}{2} \right)}$

$\Leftrightarrow 1-f\left( x \right)=1-\frac{2f\left( \frac{x}{2} \right)}{1+{{f}^{2}}\left( \frac{x}{2} \right)}=\frac{1+{{f}^{2}}\left( \frac{x}{x} \right)-2f\left( \frac{x}{2} \right)}{1+{{f}^{2}}\left( \frac{x}{2} \right)}=\frac{{{\left( 1-f\left( \frac{x}{2} \right) \right)}^{2}}>0}{1+{{f}^{2}}\left( \frac{x}{2} \right)>0}>0$

$\Leftrightarrow 1-f\left( x \right)>0\Leftrightarrow f\left( x \right)<1$ Also on the other hand we have:

$f$ is odd function then $f\left( -x \right)=-f\left( x \right)$ whenever $f\left( 0 \right)=0$

So $1-f\left( -x \right)=1+f\left( x \right)>0\Leftrightarrow f\left( x \right)>-1$ hence $\left| f\left( x \right) \right|<1\,\,\,\forall x\in \mathbb{R}$