Determination of $f(x)$

Question

A continuous function $f$ satisfies $f(x+f(x))=f(x)$ for all $x\in[0,1]$. prove that $f(x)$ is constant.

$f(f(0))=f(0)$. I am unable to proceed further.

@GHD Here's my issue with $f$ being any constant function. If that is the case, then $x+f(x)$ may not be in the domain of $f(x)$ when $x$ is and vice versa. — John Lou, Jan 19 '18 at 14:49
@john lou domain of f(x) is R(all real) it satisfies this functional equation when x is in[0,1] — Siddhartha Raja, Jan 19 '18 at 14:51
@GhD If $f$ is defined on all $\mathbb R$ but the functional equation only holds for $x\in [0,1]$ then $f$ does not need to be a constant. Suppose $f(x)=1$ on $[-\infty,2]$ but $f(x)=3-x$ for $x>2$ — lulu, Jan 19 '18 at 14:54
@GhD, ah true. In any case, I see that your posted counterexample is effectively the same as mine. I think the problem is meant to require that everything stays in $[0,1]$. — lulu, Jan 19 '18 at 14:56
@SiddharthaRaja As you can see, you need to clarify the domain of your function and the region in which the functional equation is to hold. Look at the counterexamples you have been given. To exclude those you need to clarify your conditions. — lulu, Jan 19 '18 at 15:00

score 1 · Accepted Answer · answered Jan 19 '18 at 14:55

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Put $f(x)=0$ when $x\in\mathbb R_+\cup\{0\}$ and put $f(x)=x$ when $x\in\mathbb R_-$. This is not constant and satisfy youR question.

answered Jan 19 '18 at 14:55

GhD

@TheSimpliFire: but if the functional equation does not hold outside of the given interval for $x$, then your comment does not apply... – abiessu Jan 19 '18 at 14:58
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@TheSimpliFire The problem is very poorly worded. The OP insists that the functional equation is only meant to hold for $x\in [0,1]$ and the function here does indeed satisfy the functional equation there. – lulu Jan 19 '18 at 14:58

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