Solve the functional equation $4f(x)=f(2x)$. As for now I know that one solution is $f(x)=cx^2$, where c is a constant value.
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Is the function assumed differentiable ? – servabat Dec 06 '14 at 14:43
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1Show us your efforts or ideas. And also edit the question. In its current version, the wording is too vague. – Dec 06 '14 at 14:43
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If $f$ is not necessarily continuous, then there are many solutions. It can, for instance, be defined as $$ f(x) = \cases{x^2 & if $x$ is rational\\ 3x^2 & if $x$ is a rational multiple of $\pi$.\\ -5x^2 & if $x$ is a rational multiple of $e$\\ x^2\ln 2 & if $x$ is a rational multiple of $\sqrt{2}$\\ 0 & otherwise} $$ and even the rational numbers (and thus each of the posts above) can be further subdivided into more and more subsets of the real line that will never interfere with each other through the functional equation, and therefore can have their own definition.
Arthur
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At least for any prime fraction you can even chose a different value of $f(\frac1p)$ independent of the other values. – AlexR Dec 06 '14 at 14:52
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@AlexR I believe your statements holds for any odd (positive) number $p$, and not just primes, since the functional equation only relates the function value at $x$ and $2x$. – Arthur Dec 06 '14 at 14:54
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It seems to me that we can even use rationals with odd numerator and denominator (in reduced form, ofc) – AlexR Dec 06 '14 at 14:56