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I am finding this functional equation from a past high school mathematics competition rather tricky.

Find all continuous functions $f:\mathbb{R}\rightarrow\mathbb{R}^+$ such that:

$f\big(x+f(x)\big)=f(x)\quad \forall x\in\mathbb{R}$

and

$f(2012)=2012$.

(One must prove that the trivial constant solution is the only solution.)

Mohsen Shahriari
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goonfiend
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  • I have no idea how to do it, but here's maybe an idea: if you can show that $f(x)=2012$ for all rational numbers $x$, you can deduce that it holds for all real numbers,too. Of course, you probably knew that already. – goblin GONE May 17 '13 at 15:22
  • Yep. So I had a couple of approaches in mind, none of which worked out (at least for me right now).

    1. Using some sort of number theoretic argument to prove that $f(x)=2012$ for a dense subset of the rationals and using continuity to finish.

    2. Let $S$ be the set of $x$ for which $f(x)=2012$ and using some clever use of the intermediate value theorem to arrive at a contradiction if $S$ is not all of $\mathbb{R}$.

    One possibly useful result is that $g(x)=x+f(x)$ can be easily shown to be injective (in fact monotonically increasing).

     – goonfiend May 17 '13 at 15:30
  • Does the function have to be differentiable? – Quark May 17 '13 at 15:45
  • No, we cannot assume differentiability. – goonfiend May 17 '13 at 15:48
  • @Eric Jablow: Why do you delete your answer? – 23rd May 17 '13 at 16:46
  • Presumably because the problem we pointed out was correct, Landscape! – not all wrong May 17 '13 at 22:36
  • Is this written correctly? I am a bit confused. If the solution is the constant solution why do I get contradictions? For example $1=f(1)=f(1+f(1))=f(1+1)=f(2)=2$. – Derek Luna Nov 04 '20 at 12:11
  • @DerekLuna You are taking $f(x)=x$, not $f(x)=c$. – goonfiend Nov 04 '20 at 13:57

1 Answers1

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By induction, $f(x+nf(x))=f(x)$. Hence as you observe, each $x+nf(x)$ is injective, since $$x+nf(x)=y+nf(y) \implies f(\cdots)=f(\cdots) \implies f(x) = f(y) \implies x + \cdot = y + \cdot$$ and therefore monotonically increasing, since $x+nf(x)\to\infty$ as $x\to\infty$.

Now for $x<y$, we have $x+nf(x)<y+nf(y)$ for all $n$ and so $f(y)-f(x)\ge0$.

But $f(x+nf(x))=f(x)$ and $f(x)>0$ implies that for all $x$ there exist points arbitrarily far away which take the same $f$ value, and hence as $f$ is non-decreasing, $f$ must be constant.

not all wrong
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