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I wish to prove/disprove that there exists a unique solution to the functional equation $$xyF(xy^2, y) = F(x, y), \quad x \ne 0, \quad |y| < 1, \quad y \ne 0,$$ where $F(x, y)$ is continuous.

I tried using the standard technique, i.e., assuming $F$ and $G$ both satisfy the above equation, but it did not lead anywhere. I am not an expert in functional equations, so any help would be appreciated.

Edit: It appears that in the degenerate case $y = 0$, the functional equation above does not hold and $F(1, 0) = 1$.

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  • $\begingroup$ Where is the domain of $F$ ? Is it continous ? $\endgroup$ – Gabriel Romon Jun 1 '14 at 8:03
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    $\begingroup$ Putting y=0 gives F(x,0)=0 wich seems to be a contradiction. $\endgroup$ – poolpt Jun 1 '14 at 8:11
  • $\begingroup$ @G.T.R Yes, $F$ is continuous and defined for all $x \ne 0$ and $|y| < 1$. $\endgroup$ – glebovg Jun 1 '14 at 11:09
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    $\begingroup$ What happens if you put x=1 and y=0? $\endgroup$ – poolpt Jun 1 '14 at 12:33
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    $\begingroup$ Before we can seriously tackle this question the point made by poolpt has to be clarified. $\endgroup$ – Christian Blatter Jun 1 '14 at 12:35
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In the following I assume $x\ne0$ and $0<y<1$. Put $$x:=e^u,\quad y:=e^v,\quad F:=e^G\ .\tag{1}$$ Then the given functional equation goes over into $$u+v+G(e^{u+2v},e^v)=G(e^u,e^v)\ .$$ We now introduce the new unknown function $$H(u,v):=G(e^u,e^v)\tag{2}$$ that has to satisfy $$H(u+2v)=H(u)-(u+v)\ .$$ It follows that the function $$K(u,v):=H(u,v)+{u^2\over 4v}$$ satisfies $$\eqalign{K(u+2v,v)&=H(u+2v,v)+{(u+2v)^2\over 4v}\cr &=H(u,v)-(u+v)+{u^2+4uv+4v^2\over 4v}\cr &=K(u,v)\ ,\cr}$$ which implies that $K$ is periodic with period $2v$ in the first variable.

Therefore an infinity of solutions of the original problem can be obtained as follows: Choose an arbitrary continuous $1$-periodic function $\psi:\ {\mathbb R}\to{\mathbb R}$ and put $$H(u,v):=\psi\left({u\over 2v}\right)-{u^2\over 4v}\qquad(v<0)\ ;$$ then use $(2)$ and $(1)$ to obtain $F$ in terms of $x$ and $y$.

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