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For $x>0$, $f(x)>0$ and continuous. Also $$f(x)=xf\bigg(\frac{1}{x}\bigg).$$ The function $f(x)=x \ \text{for} \ x\in(0,1]$ and $f(x)=1 \ \text{for} \ x>1$ certainly satisfies the above conditions. Is this the unique function satisfying the above conditions? Or what is another example function?

Thanks

vnd
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    How about arbitrary $f$ for $0 < x <1$, $f(1) = 1$, and $f(x) = x f(x^{-1})$ for $x > 1$? – Hans Engler Oct 05 '21 at 22:16
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    For the record, the functions mentioned by @HansEngler are all the solutions to the functional equation (with the exception that $ f ( 1 ) $ can be chosen arbitrarily). And those solutions are continuous iff the function chosen on $ ( 0 , 1 ) $ is continuous and its limit when $ x \to 1 ^ - $ exists, and is equal to $ f ( 1 ) $. – Mohsen Shahriari Oct 08 '21 at 19:08
  • Another class of solutions: $f(x)=C\sqrt{x}.$ – md2perpe Jan 20 '22 at 09:00

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\begin{align} &f(x)=xf\Big(\frac 1 x\Big), f: \mathbb{R^{+}} \to \mathbb{R^{+}}\text{ and Continuous.} \\ \ \\ &\text{if } f \equiv c: \\ &c=xc \Rightarrow c=0, \text{ Contradiction.(By $f: \mathbb{R^+} \to \mathbb{R^+}$.)} \\ \therefore \; & f \not\equiv c. \\ \end{align} This shows that the function isn't constant...

I just thought about another form of this function and found another one.

enter image description here

I think these 2 functions satisfy the F.E...

I thought very shortly, so nevermind this one.

RDK
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