What is the solution of the functional equation $f(xy)=f(x)f(y)$ for all $x$ and $y$ in $I$ where $f$ is a real-valued mapping with domain the unit closed interval $I$?
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1Why have you used "Information Theory" in the title? – Jaideep Khare Jan 25 '18 at 12:26
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$f(x)=x^a$ for some $a>0$. But you need some regularity assumption on $f$, e.g. continuity. – Yurii Savchuk Jan 25 '18 at 12:33
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For example, $f(x)=1$ for all $x$ , or $f(x)=1$ for all $x\ne0$, and $f(0)=0$. – kimchi lover Jan 25 '18 at 12:42
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1See the "$f(xy)=f(x)f(y)$" section under the canonical answer to Overview of basic facts about Cauchy functional equation. – dxiv Jan 26 '18 at 07:23
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It is not full solution but it can help you solve it fully (if I solve it fully I will edit this).
Take $y=x$, then $f(x^2)=f(x)^2$.
Take the derivative of both sides.
$$f'(x^2) \cdot 2x = 2f(x) \cdot f'(x)$$
$$f'(x^2) \cdot x=f(x) \cdot f'(x)$$
1) If $f'(x) = 0$, then $f(x) = C$, where $C$ is some real number.
Then from the functional equation we get $C = C^2$. Solving this equation we get $C=0$ or $C=1$. Then $f(x) = 0$ or $f(x) = 1$.
2) $f'(x) \ne 0$
Then we can get such a differential equation:
$$f(x)=\frac{f'(x^2) \cdot x}{f'(x)}$$
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But it was not given that $f$ is differentiable. If it is continuous, which is a less demanding requirement, then one can prove that it's a power function, see the quoted answer to the duplicate question. – dxiv Jan 26 '18 at 07:21
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