Find all functions $f\colon \mathbb R \to \mathbb R$ such that $f(f(x)(y + 1)) = x[f(y) + 1]$, $\forall x, y \in \mathbb R$.

Question

Find all functions $f\colon \mathbb R \to \mathbb R$ such that $$\large f(f(x)(y + 1)) = x[f(y) + 1], \forall x, y \in \mathbb R$$

Let $P(x, y)$ be the assertion $f(f(x)(y + 1)) = x[f(y) + 1]$.

For $P(0, -1)$ and $P(x, 0)$, we respectively have that $f(0) = 0$ and $f^2(x) = x, \forall x \in \mathbb R$.

This implies that $f(x)$ is a bijective function, (although I am not sure why).

For $P(x, x - 1)$, we have that $f(xf(x)) = x[f(x - 1) + 1], \forall x \in \mathbb R$

Then... I don't know what to do next.

Well, this is a duplicate of this problem I've just found out. https://math.stackexchange.com/questions/2051991/functional-equation-f-mathbbr-rightarrow-mathbbr-fx-1-fy — Lê Thành Đạt, Jan 04 '20 at 14:31
in your question you have a whole part [.] though, but I suspect doesn't make problem much harder? — govordovsky, Jan 04 '20 at 15:02
Does this answer your question? Functional equation: $f: \mathbb{R} \rightarrow \mathbb{R}$, $f((x + 1) f(y)) = y (f(x) + 1)$ — govordovsky, Jan 04 '20 at 15:30

score 1 · Accepted Answer · answered Jan 04 '20 at 15:15

1

To answer your question about the bijection: if $f(f(x)) = x$ then $f$ is certainly surjective (since $x \mapsto x$ is surjective) and it is also injective (since if $f(x) = f(y)$ then $x = f(f(x)) = f(f(y)) = y$).

answered Jan 04 '20 at 15:15

Patrick Stevens

36,135

1

Thanks for that, although you should have written this as a comment, not an answer, but still, thanks for the explanation. – Lê Thành Đạt Jan 04 '20 at 15:18

Find all functions $f\colon \mathbb R \to \mathbb R$ such that $f(f(x)(y + 1)) = x[f(y) + 1]$, $\forall x, y \in \mathbb R$.

1 Answers1