I have viewed this questions here, in which the author asked to prove that $f(4x^3-3x)+f(x)=0$ for all $x \in \mathbb{R}$. Scanning through the comments, I saw one trying to figure out the general solution, which suggests that \begin{align} f(x)=C\times \arcsin(x) \end{align} But no answer was proposed, so I suggest the following problem:


Find the continuous function $f: [-1,1] \to \mathbb{R}$, such that \begin{align} f(x)+f(y)=f(x\sqrt{1-y^2}+y\sqrt{1-x^2}) \end{align}


My initial idea was to use the Cauchy functional equation schemes to expand the condition suggested in the abstract. But no solution has been yielded.

Any help is appreciated


$x=y=0\;\Longrightarrow\;f(0) = 0$

$x=y=1\;\Longrightarrow\;2f(1) = f(0) = 0\;\Longrightarrow\;f(1)=0$

$x=\sin t,\,y=\cos t\;\Longrightarrow\;f(\sin t)+f(\cos t) = f(1) = 0$

$x=y=\cos t\;\Longrightarrow\;2f(\cos t) = f(2\sin t\cos t)$

$x=y=\sin t\;\Longrightarrow\;2f(\sin t) = f(2\sin t\cos t)$

Hence, $f(\sin t) = f(\cos t) = -f(\sin t)$ and so the only solution is $f=0$.

  • $\begingroup$ Seems like the continuity was not even needed... $\endgroup$
    – Sil
    Oct 22 '21 at 11:38

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