0
$\begingroup$

I was reading a book on functional equation by Titu Anderescu and came across the following functional equation.

Linear Cauchy Equation

This functional equation is an extension of both Cauchy's and Jensen's equation. It has the form $$ f ( a x + b y + c ) = p f ( x ) + q f ( y ) + r \text , \tag {*} \label {eqn} $$ where $ a $, $ b $, $ c $, $ p $, $ q $ and $ r $ are real constants and $ a b \ne 0 $.

Then we have the following Theorem.

Theorem. Let $ f : \mathbb R \to \mathbb R $ be a nonconstant continuous function satisfying \eqref{eqn}. Then $ f ( x ) = s x + t $ for some constants $ s \ne 0 , t \in \mathbb R $ and $ p = a $, $ q = b $, $ r = s c + t ( 1 - a - b ) $.

Proof. setting $$ ( x , y ) = \left( - \frac c a , 0 \right) , \left( \frac { u - c } a , 0 \right) , \left( - \frac c a , \frac v b \right) , \left( \frac { u - c } a , \frac v b \right) $$ in \eqref{eqn} gives respectively $$ f ( 0 ) = p f \left( - \frac c a \right) + q f ( 0 ) + r \text , $$ $$ f ( u ) = p f \left( \frac { u - c } a \right) + q f ( 0 ) + r \text , $$ $$ f ( v ) = p f \left( - \frac c a \right) + q f \left( \frac v b \right) + r \text , $$ $$ f ( u + v ) = p f \left( \frac { u - c } a \right) + q f \left( \frac v b \right) + r \text . $$ Hence $ \require {color} \color {red} f ( u + v ) = f ( u ) + f ( v ) - f ( 0 ) $ for all $ u , v \in \mathbb R $.
Thus $ g ( u ) = f ( u ) - f ( 0 ) $ is a continuous additive function and we conclude that $ g ( x ) = s x $ for some constant $ s \in \mathbb R $, i.e. $ f ( x ) = s x + t $, where $ t = f ( 0 ) $ and $ s \ne 0 $. Now plugging $ f ( x ) $ in \eqref{eqn} shows that $ p = a $, $ q = b $ and $ r = s c + t ( 1 - a - b ) $.

I am unable to understand that how the author has concluded $ f ( u + v ) = f ( v ) + f ( u ) + f ( 0 ) $, as for that to happen, $ 2 \Big( p f \left( - \frac c a \right) + q f (0) + r \Big) = 2 f ( 0 ) $ should be equal to $ 0 $, but the author has not proved anything like that in the theorem.

It would be highly helpful if I am provided with a pedantic proof that how $ f ( 0 ) = 0 $.

$\endgroup$
1
$\begingroup$

Actually you get $f(u+v)=f(u)+f(v)+tf(0)$ for some $t$ [ $t=1-2q$]. Put $u=v=0$ to get $(2+t)f(0)=f(0)$ If $f(0) \neq 0$ then this gives $t=-1$ so $f(u+v)=f(u)+f(v)-f(0)$. If $g(u)=f(u)-g(0)$ then we get $g(u+v)=g(u)+g(v)$. So in both the cases $f(0)=0$ and $f(0) \neq 0$ we can reduce the given functional equation to Cauchy's equation.

$\endgroup$
9
  • $\begingroup$ That the author has done already $\endgroup$
    – user832197
    Dec 1 '20 at 10:46
  • $\begingroup$ Could you please tell that how can I prove $f (0)=0$ ???? $\endgroup$
    – user832197
    Dec 1 '20 at 10:47
  • 1
    $\begingroup$ @user832197 You cannot prove that $f(0)=0$ That is why you consider $g(x)=f(x)-f(0)$ The final conclusion is $f(x)=sx+t$ and $f(0)=t$ need not be $0$. $\endgroup$ Dec 1 '20 at 11:35
  • 1
    $\begingroup$ I don't understand why you are asking this. In the question you said 'for that to happen...' so you already know the answer to your last question. You have to just add the first three equations and conpare it with the last equation. $\endgroup$ Dec 1 '20 at 12:01
  • 1
    $\begingroup$ @user832197 I have edited my answer. I hope everything is clear now. $\endgroup$ Dec 3 '20 at 5:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.