# Theorem concerning Linear Cauchy functional equation $f ( a x + b y + c ) = p f ( x ) + q f ( y ) + r$

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 $$f ( 0 ) = 0$$.

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.
• Could you please tell that how can I prove $f (0)=0$ ???? Dec 1 '20 at 10:47
• @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 $0$. Dec 1 '20 at 11:35