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Rozwiązać równanie funkcyjne

\(\displaystyle{ f (f(x-y)f(x+y)) = x^2 - yf(y)}\)

gdy \(\displaystyle{ x, y \in \RR.}\)

\(\displaystyle{ f\left[ f(x-y)f(x+y)\right] =x^2-yf(y)}\)

Podstawiamy:

\(\displaystyle{ y=0}\)

\(\displaystyle{ f\left[ f^2(x)\right] =x^2}\)

\(\displaystyle{ f(0)=a}\)

\(\displaystyle{ y=x}\)

\(\displaystyle{ f\left[ af(2x)\right] =x^2-xf(x)}\)

\(\displaystyle{ y=-x}\)

\(\displaystyle{ f\left[ af(x)\right] =x^2+xf(-x)}\)

z tego:

(*) \(\displaystyle{ f(-x)=-f(x) , x \neq 0}\)

\(\displaystyle{ f\left[ f(x-y)f(x+y)\right] =x^2-yf(y)}\)

teraz podstawmy:

\(\displaystyle{ y:=x, x:=y}\)

\(\displaystyle{ f\left[ f(y-x)f(x+y)\right] =y^2-xf(x)}\)

ale z (*):

\(\displaystyle{ f\left[ f(y-x)f(x+y)\right] =f\left[ -f(x-y)f(x+y)\right] =-f\left[ f(x-y)f(x+y)\right]}\)

z tego:

\(\displaystyle{ f\left[ f(x-y)f(x+y)\right]=-y^2+xf(x)}\)

więc:

\(\displaystyle{ -y^2+xf(x)=x^2-yf(y)}\)

Podstawienie:

\(\displaystyle{ y=x}\)

\(\displaystyle{ -x^2+xf(x)=x^2-xf(x)}\)

\(\displaystyle{ 2xf(x)=2x^2}\)

\(\displaystyle{ f(x)=x}\)

Po podstawieniu pasuje...