# Solve functional equation $f(x)+f(y)=f(x\sqrt{1-y^2}+y\sqrt{1-x^2})$ [duplicate]

Abstract

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:

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}

Approach

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 $$f=0$$.

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