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Let $~~f : [1,3] \rightarrow [1,2] ~~$ be a bijective , continuous and strictly increasing function , and let $g$ be another function such as $ ~~g: [0,+\infty[ \rightarrow \mathbb{R} ~~~~, ~~g(0) = 0 ~~$ and :

$g(x) = 2^{\textbf{-}n}f(3^{n}x) ~~~~~~~~~~~~~~~~~~~~~~~~~/~~~~~~~~ n \in \mathbb{Z} ~~~~~~,~~~~~~ 3^{n}x \in[1,3[$

prove that : $ 2g(x) = g(3x) $

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    Does the given relation hold for all $\forall n \in \mathbb{Z}\,$, or for one particular integer value $n\,$? – 2017-01-05

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Since $g(x) = 2^{-n}f(3^{n}x) $, $g(3x) = 2^{-n}f(3^{n}3x) = 2^{-n}f(3^{n+1}x) $.

But $g(x) = 2^{-n-1}f(3^{n+1}x) = \frac12\cdot 2^{-n}f(3^{n+1}x) =\frac12 g(3x) $, so $2 g(x) = g(3x) $.

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    straight method , thanks :-) – 2017-01-04
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    A bit odd that, other than the functional relation itself, none of premises were used (or needed). – 2017-01-05
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    That occurred to me also. – 2017-01-05