# Solution of the functional equation $f(x) + f(y) =f\left(x\sqrt{1-y^2 }+y\sqrt{1-x^2 }\right)$

If $$f(x) + f(y) =f\left(x\sqrt{1-y^2 }+y\sqrt{1-x^2 }\right)\text,$$ prove that $$f(4x^3 -3x) + 3f(x) =0\text.$$

I started by substituting $$x = y$$, in the expression and I get $$2f(x)=f\left(2x\sqrt{1-x^2}\right)\text.$$ Then, I also substitute $$y = \sqrt{(1-x^2)}$$ and I get $$f(x) +f\left(x\sqrt{1-x^2}\right) = 0\text.$$ How shall I proceed further and solve this problem?

• Do you want $f(x)+f(y) =f(x\sqrt{1-y^2}+y\sqrt{1-x^2})$? Jun 7 '18 at 18:47
• @LordSharktheUnknown Sorry, I wrote it wrong. I have edited it now,
– ShiS
Jun 7 '18 at 18:49
• So $f$ should be $C\arcsin$? Jun 7 '18 at 18:50
• Yes, that satisfies the equation. But how shall I prove this in general.Or how will I get this solution by solving the equation forward
– ShiS
Jun 7 '18 at 18:51
• @LordSharktheUnknown Ineterestingly, $\arcsin$ does not work as $f(0)+f(0)=f(0)$ implies $f(0)=0$ and then $f(1)+f(1)=f(0)$ implies $f(1)=0$. In fact, transporting the equation back to angles, one gets a (local form of) Cuachy's functional equation. Therefore the only continuous solution is $f(x)=0$. Jun 7 '18 at 20:12

I'll assume the arguments of $f$ are always between $-1$ and $1$. Then $$2f(x)=f(2x\sqrt{1-x^2})$$ and $$3f(x)=f(x)+f(2x\sqrt{1-x^2})=f(y)$$ where $$\begin{split} y &= x\sqrt{1-4x^2(1-x^2)}+\sqrt{1-x^2}2x\sqrt{1-x^2}\\ &= x\sqrt{1-4x^2+4x^4}+2x\left(1-x^2\right)\\ &= x\left(1-2x^2\right)+2x\left(1-x^2\right)=3x-4x^3. \end{split}$$ But $f(y)+f(-y)=f(0)=f(0)+f(0)$ and so $3f(x)=f(-y)$.
Easy way to prove that for every real number $x$ with $- 1 \le x \le 1$, $f ( x ) = 0$ (using some steps already mentioned in the comments, but repeated here to be complete):
Let $x = y = 0$ in $$f ( x ) + f ( y ) = f \Big( x \sqrt { 1 - y ^ 2 } + y \sqrt { 1 - x ^ 2 } \Big) \tag { * } \label { * }$$ and you get $f ( 0 ) = 0$. Again, let $x = y = 1$ in \eqref{ * } and you get $f ( 1 ) = 0$. Thus, letting $y = 1$ in \eqref{ * } you'll have $f ( x ) = f \big( \sqrt { 1 - x ^ 2 } \big)$. This also shows that $f ( | x | ) = f ( x )$. Now, substituting $| x |$ for $x$ and $\sqrt { 1 - x ^ 2 }$ for $y$ in \eqref{ * }, you get $f ( x ) = 0$.