$f(y+zf(x))=f(y)+xf(z) $ with $x,y,z\in \mathbb{R}$

Question

Find all $f\colon\mathbb{R} \rightarrow \mathbb{R}$ such that

$f(y+zf(x))=f(y)+xf(z)$ with $x,y,z\in \mathbb{R}$

When I study functional equation, I have some difficult in solving above equation. Can anyone help me please?

Answer 1

By applying $z = 0$, we have $f(0) = 0$.

$f(x) = 0$ for all $x\in\mathbb{R}$ is clearly a solution.

Assume that there exists $t\in\mathbb{R}$ such that $f(t)\neq 0$. Since $f(y+zf(t)) = f(y) + tf(z)$, $f$ is surjective.

Let $m, n\in\mathbb{R}$ such that $f(m) = f(n)$. $f(y) + mf(t) = f(y+tf(m)) = f(y+tf(n)) = f(y) + nf(t)$, so $m = n$. Hence $f$ is bijective.

By applying $y = 0$, we have $f(zf(x)) = xf(z)$. By changing the roles of $x$ and $z$, we have $f(f(zf(x))) = f(xf(z)) = zf(x)$. So for all $l\in\mathbb{R}$, by applying $z = \frac{l}{f(t)}, x = t$, we obtain $f(f(l)) = l$.

By applying $z = 1$ in $f(zf(t)) = tf(z)$, we have $f(1) = 1$.

So $f$ satisfies $f(y + z) = f(y)+ f(z)$.

By applying $y = 0, z = f(m)$, we obtain $f(f(x)f(m)) = f(x)f(m)$ for all $x, m\in\mathbb{R}$. Since $f$ is bijective, $f(xy) = f(x)f(y)$ for all $x, y\in\mathbb{R}$.

Therefore $f$ is an automorphsim of field $\mathbb{R}$ ! And the identity is the only automorphism of field $\mathbb{R}$.

It can be demonstrated by the following steps :

1. Show that $f(x) = x$ for all $x\in\mathbb{Q}$.
2. Show that $f(\mathbb{R}_{+})\subset\mathbb{R}_{+}$. (use the fact that $x\geq 0$ iff x is a square)
3. Deduce that $f$ is strictly increasing.
4. For all $x\in\mathbb{R}\setminus\mathbb{Q}$, if $x\neq f(x)$, choose a rational number between $x$ and $f(x)$.