# If $f$ is a continuous function from $R \rightarrow R$ and $f(x)=f(x+f(x))$ then prove that $f$ is constant.

If $f$ is a continuous function from $R \rightarrow R$ and $f(x)=f(x+f(x))$ then prove that $f$ is constant.

I could prove that $f(x)=f(x+f(x))=..=f(x+nf(x))$ after $n$ iterations.Then , how will I proceed?

• Your $f(x) = f(x+nf(x))$ doesn't seem right if you mean $nf(x)$ as in $n * f(x)$ – Aneesh Apr 17 '16 at 15:40
• We have $f(x+f(x))=f(x+f(x)+f(x+f(x)))=f(x+f(x)+f(x))=f(x+2f(x))$.Do this thing similarly for n times ,, you should get the above relation – Legend Killer Apr 17 '16 at 15:43
• Ah okay, you're right, I didn't see that – Aneesh Apr 17 '16 at 15:45
• See the answer to this question. – Andrew Apr 17 '16 at 16:12
• Can you prove rigourously that f is injective? – Legend Killer Apr 17 '16 at 16:22

If you follow Andrew's link to An elementary functional equation. you will find a solution for the case all the values of $f$ are non-negative. A slight variation of the argument works in the general case.
Start like there: Sharkos proves that $h_n(x) = x + n f(x)$ is an injective function of $x$ for all integers $n \ge 1$. Then observe that a continuous function can only be injective if it is strictly monotonic (either strictly increasing or strictly decreasing). Now consider the infinitely many values of $n$. There must be either infinitely many values of $n$ for which $h_n$ is strictly increasing or infinitely many values of $n$ for which $h_n$ is strictly decreasing.
As in Sharkos's proof in the old thread, if there are infinitely many values of $n$ for which $h_n$ is strictly increasing, we conclude that $f$ itself is non-decreasing. Now suppose f is not constant. Then there must exist two points $a$ and $b$ with $f(a) = A < f(b) = B$ and moreover either $A, B > 0$ or $A, B < 0$. (Prove that!) However, $f(a + nA) = A$ and $f(b + mB) = B$ for all integers $n, m \ge 1$. Choosing either very large $n$ or very large $m$, depending on the sign of $A$ and $B$, we can contradict the fact that $f$ is non-decreasing.
If instead there are infinitely many values of $n$ for which $h_n$ is strictly decreasing, then $f$ itself is non-increasing, by a similar argument to Sharko's. Then the proof proceeds like in the previous paragraph.