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Find all continuous functions $f:\mathbb R \rightarrow \mathbb R$ such that

1) $\forall x \in \mathbb R$ $$f(-x)=-f(x)$$

2) $\forall x \in \mathbb R$ $$f(x+1)=f(x)+1$$

3$\forall x \in \mathbb R/ \{0\}$ $$f\left(\frac1x\right)=\frac1{x^2}f(x)$$

My work so far:

1) $f(0)=0$

2) $\forall x \in \mathbb Z$ $$f(x+m)=f(x)+m$$

3) I see that the answer $f(x)=x$, but can not rove.

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    $\begingroup$ I think that $2$ and $3$ imply $f(x)=x$ for $x\in\mathbb Q$, so that we may use continuity to show $f(x)=x$ for $x\in\mathbb R$. $\endgroup$ – Simply Beautiful Art Oct 6 '16 at 14:42
  • $\begingroup$ @SimpleArt: $f(-x)=-f(x)$. I am sorry. I edited. $\endgroup$ – Roman83 Oct 6 '16 at 14:46
  • $\begingroup$ @SimpleArt this should be an answer ;) $\endgroup$ – Scientifica Oct 6 '16 at 14:47
  • $\begingroup$ @Scientifica Working on making it. $\endgroup$ – Simply Beautiful Art Oct 6 '16 at 14:48
  • $\begingroup$ @SimpleArt Oh I didn't mean a complete answer, but a hint-type answer. If the OP manages to solve it, your answer will be accepted ;) $\endgroup$ – Scientifica Oct 6 '16 at 14:54
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Let $S=\{\,x\in\Bbb R\mid f(x)=x\,\}$. By the given conditions, $x\in S$ implies $-x\in S$, $x+1\in S$, and (if $x\ne 0)$ $\frac 1x\in S$. Also, $(1)$ shows $0\in S$. Define the injective map $\Bbb Q\to \Bbb N^2$, $\frac ab\mapsto (b,|3a-1|)$ (where $\frac ab$ is in lowest terms and $b>0$). The lexical order on $\Bbb N^2$ is a well-order, and this induces a well-order $\prec$ on $\Bbb Q$.

Assume $\Bbb Q\setminus S$ is not empty. Then there is a $\prec$-smallest rational number $x$ with $x\notin S$. Assume $x<0$. Then $-x\prec x$, hence $-x\in S$ and so $x\in S$, contradiction. Assume $x\ge1$. Then we have $x-1\prec x$, hence $x-1\in S$ and so $x\in S$, contradiction. Clearly $x\ne 0$ as we know $0\in S$. Hence $0<x<1$. But then $\frac 1x\prec x$, hence $\frac 1x\in S$ and so $x\in S$, again a contradiction.

We conclude $\Bbb Q\subset S$. By continuity of $f$, $ S$ is closed, i.e., $S=\Bbb R$ and so $$f(x)=x. $$

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  • $\begingroup$ Wow, that was a beautiful proof by contradiction. $\endgroup$ – Simply Beautiful Art Oct 6 '16 at 16:20
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Get $y = f(x)$ , then i have $$f(x+1)=y+1,f(\frac{1}{x+1})=\frac{y+1}{(x+1)^2},f(\frac{-1}{x+1})=\frac{-(y+1)}{(x+1)^2}$$ $$f(\frac{x}{x+1})=f(-\frac{1}{x+1}+1)=\frac{-(y+1)}{(x+1)^2}+1=\frac{x^{2}+2x-y}{(x+1)^2}$$ $$f(\frac{x}{x+1})=f(\frac{1}{\frac{x+1}{x}})=\frac{f(\frac{x+1}{x})}{(\frac{x+1}{x})^2}=\frac{\frac{x^{2}+y}{x^2}}{(\frac{x+1}{x})^2}=\frac{x^2+y}{(x+1)^2}$$ So i have $$\frac{x^{2}+2x-y}{(x+1)^2}=\frac{x^2+y}{(x+1)^2}$$ $$x=y$$ And easy to check $f(0)=0,f(-1)=-1$ so $$f(x) \equiv x$$

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  • $\begingroup$ Because $f(x+1)=f(x)+1$ for all $x$ , here i suppose $x$ is not $0$ or $-1$ $\endgroup$ – Gankedbymom Oct 6 '16 at 15:58
  • $\begingroup$ Wait, how did you get $\frac{x^{2}+2x-y}{(x+1)^2}=\frac{x^2+y}{(x+1)^2}$? $f\left(\frac x{x+1}\right)\ne f\left(\frac{x+1}x\right)$ The first step of the third line is wrong. Once you fix that, I think your good. $\endgroup$ – Simply Beautiful Art Oct 6 '16 at 16:00
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    $\begingroup$ sorry , i fixed that $\endgroup$ – Gankedbymom Oct 7 '16 at 12:12
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For part 1) you want to consider all odd functions, for which their definition is the condition set. For example, $Sin(-x) = -Sin(x)$. Graphically, this is to say that $f$ is symmetric about the origin. Another example is $(-x)^3 = -(x^3)$

For part 2), you are looking for some function $f$ such that, for a given argument $x$, a translation by 1 unit left in the x-direction is equivalent to a translation by 1 unit up in the y-direction.

When considering part 3 as well, it does appear that, as other posts have suggested, the only function which satisfies all 3 constraints is $f(x) = x$

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  • $\begingroup$ I believe the OP is looking for all functions that satisfy ALL 3 conditions. $\endgroup$ – Scientifica Oct 6 '16 at 15:02
  • $\begingroup$ Ah okay, by misunderstanding. I have edited my post, but left it there if anyone is interested in the graphical interpretation of the transformations. Thank you for the heads up. $\endgroup$ – Pablo Oct 6 '16 at 15:04
  • $\begingroup$ Ok. No problem. $\endgroup$ – Scientifica Oct 6 '16 at 15:06

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