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Let $f:\mathbb{R}→\mathbb{R}$ such that: $$f(xy)=xf(y)+yf(x), f(x+y)=f(x^{2021})+f(y^{2021}), \forall x, y\in \mathbb{R}$$ Calculate $f(\sqrt{2020})$.

So far I found out that $f(x)$ is an additive function that has the form of $f(x)=ax\log_c(x)$, with $a, c$ is constant, $a\neq 0$, $c>0$. I figured it out from the first equation. Now I don't know what to do with the second equation, neither as how to calculate.

Lalit Tolani
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Trường Hưng Nguyễn
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  • Do you have any additional condition? Do you know if the function is continuous, derivable...? – Miguel Sep 04 '21 at 13:49
  • No, I don't know anything more than that. – Trường Hưng Nguyễn Sep 04 '21 at 14:03
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    If your deduction that $f(x) = ax\log_c(x)$ is correct, then such an $f$ as above does not exist. First, since $\log_c(x) = \frac{\log(x)}{\log(c)}$, you can reduce the form of $f$ to $f(x) = ax\log(x)$. Now, the second equation implies $f(2x) = 2f(x^{2021})$. Hence, we have $2ax\log(2x) = 2\cdot ax^{2021}\log(x^{2021})$, which should hold for any $x$, which is obviously not the case (except when $a=0$, and $f\equiv 0$ indeed is a solution). – amsmath Sep 04 '21 at 14:13
  • How do you get your conclusion $f(X)=a x \log_c x$? – Miguel Sep 04 '21 at 14:40
  • https://math.stackexchange.com/questions/2091212/proving-f1-exist-for-f-satisfying-fxy-xfyyfx – Trường Hưng Nguyễn Sep 04 '21 at 14:45
  • @TrườngHưngNguyễn The answer there assumes that $f$ is continuous, which you do not have here. However, you can solve the problem without any additional assumption on $f$. See my answer. – amsmath Sep 04 '21 at 14:47
  • If my assumption is wrong, then how can I prove that the function is additive? The original problem has 2 sub-problems: prove it's additive on $\mathbb{R}$, and calculate $f(\sqrt{2020})$. – Trường Hưng Nguyễn Sep 04 '21 at 14:56

1 Answers1

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For $x\neq 0$ set $g(x) := \frac {f(x)}x$. Then the first equation shows that $g(xy) = g(x)+g(y)$, whenever $x,y\neq 0$. In particular, $g(2x) = g(2) + g(x)$ and $g(x^n) = ng(x)$ for $n\in\mathbb N$. The second equation shows that \begin{align} g(2) + g(x) &= g(2x) = \frac{f(2x)}{2x} = \frac{2f(x^{2021})}{2x} = \frac{f(x^{2021})}{x^{2021}}\cdot x^{2020} = x^{2020}g(x^{2021})\\ &= 2021x^{2020}g(x). \end{align} Plugging in $x=2021^{-(2020)^{-1}}$ in this equation gives $g(2) = 0$ and therefore $f(x)=0$ for $x\neq 0$. The first equation also yields $f(0)=0$. Therefore, $f\equiv 0$ is the only solution.

amsmath
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