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Find all function $f:\mathbb{R}^+\rightarrow\mathbb{R}^+$ satisfying

(i) $f(xf(y))=yf(x)$ for all positive real numbers, $x, y$

(ii) $ f(x) $ is bounded function for domain interval $(1,\infty)$

Please help me check my work below. Thank you.

Let $P(x,y)$ denote $f(xf(y))=yf(x),\;\;\forall x, y \in \mathbb{R^+}$.

Let $y_1, y_2 \in \mathbb{R}^+$ such that $f(y_1)=f(y_2)$.

Consider $P(x,y_1)$ and $P(x,y_2)$ : we have

$f(xf(y_1))=f(xf(y_2))= y_1f(x) = y_2f(x)$ then $y_1=y_2$ so $f$ is injective.

Consider $P(1,1)$ : $f(f(1))=f(1) \rightarrow f(1)=1$

substitute $x=1$ in (i), we have $f(f(y))=y$.

Consider $P(x, x)$ : $f(xf(x))=xf(x)$

Let $xf(x)=z$ so $f(z)=z, \;\exists z \in \mathbb{R}^+$

If $z>1$ then $P(x,f(y)) : f(x)f(y)=f(xy)$ so $f(z)f\left(\frac{1}{z}\right)=f(1)=1$ so $f\left(\frac{1}{z}\right) = \frac{1}{z}$

then $f(z^2)=f(z)f(z)=z^2$ , $f(z^4)=f(z^2)f(z^2)=z^4$ so $f(z^{2^n})= z^{2^n}, \;\forall n \in \mathbb{N}$, contradict (ii)

If $z<1$ then $\frac{1}{z}>1$. Since $f\left(\frac{1}{z}\right) = \frac{1}{z}$ so $f\left(\frac{1}{z^2}\right) = f\left(\frac{1}{z}\right)f\left(\frac{1}{z}\right)=\frac{1}{z^2}$

then $f\left(\frac{1}{z^{2^n}}\right)=\frac{1}{z^{2^n}}$, contradict (ii)

Thus $xf(x)=1$, $f(x)=\frac{1}{x} ,\;\;\forall x \in \mathbb{R^+}$

Answer check :

(i) $f(xf(y))=\frac{y}{x}=yf(x) $

(ii) $f(x)=\frac{1}{x} < f(1)=1 ,\;\;\forall x \in (1,\infty) \;\;\blacksquare$

user403160
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    Not sure what $f(x),:,(1,\infty)$ means. Also, are we to assume that $f(x)$ is continuous? If not, then I don't believe that $f(xy)=f(x)f(y)\implies f(x)=x^c$. Take, for example, any additive non-linear function, $g(x) $ with $g(x+y)=g(x)+g(y)$. Then $f(x)=e^{g(\log x)}$ satisfies $f(xy)=f(x)f(y)$. – lulu Aug 16 '17 at 12:42
  • @lulu, I've edited requirement (ii). – user403160 Aug 16 '17 at 16:38
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    And what about continuity? Your argument seems to rely on it. – lulu Aug 16 '17 at 17:00
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    As @lulu has noted, $ f ( x y ) = f ( x ) f ( y ) $ is not sufficient to deduce $ f ( x ) = x ^ c $. You can use (ii) to deduce that. see https://math.stackexchange.com/questions/505847/real-analysis-proofs-additive-functions. – Mohsen Shahriari Aug 16 '17 at 21:43
  • @MohsenShahriari Thanks for the link; never knew that criterion. Interesting. – lulu Aug 16 '17 at 23:04
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    Related: https://math.stackexchange.com/questions/221248/find-all-positive-functions-of-a-positive-real-such-that-fxfy-yfx-and In fact, the solution works here with minor changes. – Mohsen Shahriari Aug 17 '17 at 21:48
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    The only problem with your argument is that "$ f ( x y ) = f ( x ) f ( y ) $ is not sufficient to deduce $ f ( x ) = x ^ c $". You can use the additional hypothesis that $ f $ is bounded on $ ( 1 , + \infty ) $ to get that. The idea is similar to the one which I gave a link to. – Mohsen Shahriari Aug 18 '17 at 16:54
  • @lulu For more facts about Cauchy's functional equations see https://math.stackexchange.com/questions/423492/overview-of-basic-facts-about-cauchy-functional-equation. – Mohsen Shahriari Aug 18 '17 at 16:56
  • @Mohsen Shahriari, lulu, is my edited work correct now ? – user403160 Aug 19 '17 at 03:45
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    Now it's OK. Just a typo: $ f \big( x f ( x ) \big) = x f ( x ) $ is yielded by considering $ P ( x , x ) $, not $ P \big( x , f ( x ) \big) $. – Mohsen Shahriari Aug 19 '17 at 05:38
  • @Mohsen Shahriari, lulu. Edited, thank you for your kind help. – user403160 Aug 19 '17 at 05:56

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