find functions f such that $f(f(x))=xf(x)+1$,

let $f:R\longrightarrow R$, and $f$ is continous,and such that $f(f(x))=xf(x)+1$,

find all this $f$?

follow is my some idea:(but I don't have solution)

We have $f(f(0)) = 1$, so there is your $c = f(0)$, such that $f(c) = 1$. Assume there exists $v$ such that $f(v) = 0$. Then $f(0) = f(f(v)) = vf(v) + 1 = 1$, meaning $c=1$. Now, $f(0) = f(1) = 1$, so $1 = f(f(0)) = f(f(1)) = f(1)+1 = 2$, absurd.

So $f(x) \neq 0$ for all $x$, thus $f$ takes constant sign, being continuous. Assume now $f(x) = f(y) = t \neq 0$, so $xt+1 = f(f(x)) = f(f(y))= yt + 1$, whence $(x-y)t=0$, thus $x=y$. This means $f$ is injective, therefore monotonous, being continuous. Moreover, assume $x=f(x)$, so $x = f(x) = f(f(x)) = xf(x) + 1 = x^2+1$, thus $x^2-x+1 = 0$, but this has no real roots, so $f(x) \neq x$ for all $x$. Then either $f(x) > x$ for all $x$, or $f(x) < x$ for all $x$, since $f(x)-x$ is continuous

mark: this problem is my found,come from this probelm,if when $f:N\longrightarrow N$,and add $f(1)=1$then this problem is equivalent follow problem $$a_{n+1}=na_{n}+1,a_{1}=1$$

we can find $a_{n}=[e(n-1)!]$,

Thank you everyone can help

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today I see this problem:math.stackexchange.com/questions/408004/… – math110 Jun 1 '13 at 1:06
Your problem is really different from that one, I think. – Eric Stucky Jun 1 '13 at 1:08
Hahaha,Thank you @EricStucky – math110 Jun 1 '13 at 1:16
The recurrence of the title should be $f(x+1)=xf(x)+1$ if it is to match that of your recurrence for $a_n$ with $f(n)=a_n$. Also you might mention that $a_n=[e(n-1)!]$ gives $2$ at $n=1$, so you are really only saying this $a_n$ satisfies your recurrence provided you define $a_1=1$ separately, not using the formula. – coffeemath Jun 1 '13 at 1:35
I don't have an answer, but here is a fact that may be useful: The function is decreasing and positive, and hence must decrease to some $L \geq 0$ as $x \rightarrow -\infty$. Since $$f(x) = \frac{f(f(x)) - 1}{x},$$ we see that the LHS converges to the number $L$ while the RHS converges to 0, and so $L = 0$. – Zach L. Jun 1 '13 at 2:49

From the OP and Zach L's comment to the OP, we can continuously extend $f$ to a function on the extended real numbers by setting $f(-\infty) = 0$ and $f(+\infty) = +\infty$.

Define a sequence $a_n$ of extended real numbers for all natural numbers by $a_0 = -\infty$, and $a_{n+1} = f(a_n)$. Observe that $f$ is a bijective function $[a_n, a_{n+1}] \to [a_{n+1}, a_{n+2}]$.

Let $g$ be the inverse of $f$ (with domain the non-negative extended real numbers)

The sequence $a_n$ is monotonic and increasing, and therefore has a limit $L$ in the extended real numbers. This satisfies

\begin{align} L &= \lim_{n \to +\infty} a_n \\&= \lim_{n \to +\infty} a_{n+1} \\&= \lim_{n \to +\infty} f(a_n) \\&= f(\lim_{n \to +\infty} a_n) \\&= f(L) \end{align}

The OP has already shown that $f$ has no finite fixed points, so therefore $L = +\infty$.

This means the intervals $[a_n, a_{n+1}]$ cover the entire range $[-\infty, +\infty)$.

The fact that $f(f(x)) = x f(x) + 1$ means that the value of $f$ on $[a_{n+1}, a_{n+2}]$ is determined by its values on $[a_n, a_{n+1}]$ (by considering $x \in [a_n, a_{n+1}]$).

Therefore, $f$ is completely determined by its values on $[a_0, a_1] = [-\infty, 0]$.

Conversely, I assert that if you choose any continuous, monotonically increasing function $f_0$ on $[-\infty, 0]$ such that $f_0(-\infty) = 0$ and $0 < f_0(0) < 1$, then we have a n increasing sequence (converging to $+\infty$) recursively defined by

• $a_0 = -\infty$
• $a_1 = 0$
• $a_2 = f_0(0)$
• $a_{n+2} = a_{n+1} a_n + 1$

and a sequence of invertible functions $f_n : [a_n, a_{n+1}] \to [a_{n+1}, a_{n+2}]$ recursively defined by

• $f_{n+1}(x) = f_n^{-1}(x) x + 1$

and then the function

$$f(x) = \begin{cases} f_n(x) & x \in [a_n, a_{n+1}] \\ +\infty & x = +\infty \end{cases}$$

is (the continuous extension to the extended real numbers of) a solution to the problem.

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While there are many continuous solutions, there is probably a unique analytic solution. – Hurkyl Jun 1 '13 at 9:25

The definition implies that we can take $x$ from $f(f(x))$. That means that there exists a function that is the inverse of $f$. Let's call that function $g$ with $g(f(x)) = x$.

We would have:

$f(f(x)) = g(f(x))f(x) +1$

Right now we only have defined the case $f(f(x))$. Since we have to find functions that satisfy that case, but there are no restriction upon other cases, we can safely define the general case $f(x)$ as:

$f(x) = g(x)x +1$

By simple algebra, we can define $g$ as:

$g(x) = (f(x) - 1)/x$

Now, suppose we have $f(z_1) = z_2$, then we can use $g(f(x)) = x$ to find $g(z_2)$, and use that to find $f(z_2)$. We can repeat this method indefinitely, since OP proved that $f(x) \not= x$, for all $x$. We still have to find a suitable values for $z_1$ and $z_2$

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Your functional equation $f(x) = g(x)x + 1$ is only valid for $x$ in the range of $f$. You cannot use it to deduce $f(0)=1$ since $0$ is not in the range of $f$ (proved by the OP already). – Erick Wong Jun 1 '13 at 4:12
You are absolutely right Erick. I edited it so that it might be at least partially useful. – jmj Jun 1 '13 at 6:24