IWM, Volume 13(2007), Issue 3, Problem 307


Find all functions f:\mathbb{R}\to \mathbb{R} such that \forall x,y,z\in\mathbb {R}\;f((x+z)(y+z))=(f(x)+f(z))(f(y)+f(z))\;(1).

My solution.

If there is exists C\in\mathbb{R} such that f(x)=C\forall x\in\mathbb{R} then by (1) we have C=4C^2, so C=0 or \dfrac{1}{4}. If there is not such C then choose x=y=z=0 in (1) we have f(0)=0 or \dfrac{1}{4}
(A)If f(0)=0

Setting z=0 in (1) we see that f(xy)=f(x)f(y)\;\forall x,y\in\mathbb{R}\;\ (2)

In (2) if put x=y=\sqrt{t}(t\geq 0) we have f(x)\geq 0\forall x\geq 0\; (3)

So in (2) choose x=y=1 then f(1)=0 or 1.

(A-1)If f(1)=0

In (2) choose y=1 then f(x)=0\forall x\in\mathbb{R} contradiction!

(A-2)If f(1)=1

In (1) put x=y=1-z we have (f(z)+f(1-z))^2=1\forall z\in\mathbb{R}\; (4) So in (1) choose y=1-z we see that  f(x+z)=(f(x)+f(z))(f(1-z)+f(z))\forall x,z\in\mathbb{R}

\Rightarrow f^2(x+z)=(f(x)+f(z))^2\forall x,z\in\mathbb{R}(\text{by (4)})

\Rightarrow f(x+z)=f(x)+f(z)\forall x,z\geq 0(\text{by (3)})\; (5)

Now, in (1) choose y=-z then (f(x)+f(z))(f(z)+f(-z))=0\forall x,z\in\mathbb{R}\; (6)

By f isn’t const, from (6) we have f(z)+f(-z)=0\forall z\in\mathbb{R}\; (7)

From (5) and (7) we have f(x+y)=f(x)+f(y)\forall x,y\in\mathbb{R}\; (8)

By (2),(8) and f isn’t const we have f(x)=x\forall x\in\mathbb{R}.

(B)If f(0)=\dfrac{1}{4}

In (1) choose y=z=0 then f(x)=\dfrac{1}{4}\forall x\in\mathbb{R}, contradiction!

Now, by checking direct (1) we have answer:
f(x)=0\forall x\in\mathbb{R}\; , f(x)=\dfrac{1}{4}\forall x\in\mathbb{R}\; , f(x)=x\forall x\in\mathbb{R}.

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