[Math] Solve the following functional equation $f(xf(y))+f(yf(x))=2xy$

functional-equations

Find all function $f:\mathbb{R}\rightarrow \mathbb{R}$ so that $f(xf(y))+f(yf(x)=2xy$.

By putting $x=y=0$ we get $f(0)=0$ and by putting $x=y=1$ we get $f(f(1))=1$.
Let $y=f(1)\Rightarrow f(x)+f(f(x)f(1))=2x$, which tells us that $f$ is an injective function.
The only solutions I came up with so far are $f(x)=x$ and $f(x)=-x$.

Best Answer

Putting $x=1$ in $f(x)+f(f(x)f(1))=2xf(1)$ we get that $f(1)=\pm 1$

Let $f(1)=1$.

$x\rightarrow xf(x), y\rightarrow \frac{1}{x} \Rightarrow f(xf(x)f(\frac{1}{x}))+f(x)=2f(x) \rightarrow f(x)f(\frac{1}{x})=1$.

Putting now $x\rightarrow \frac{1}{x}, y \rightarrow 1$ we get $\frac{1}{f(x)}+ \frac{1}{f(f(x))}=\frac{2}{x}$. Using $f(f(x))=2x-f(x)$ we get $(f(x)-x)^2=0$, and there we get one of our two solutions, $f(x)=x$. We get $f(x)=-x$ as the other solution by putting $f(1)=-1$.

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