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Problem:find all continuous functions $f:[0,+infty)to [0,+infty)$ such that $$f(y^2f(x)+x^2f(y))=xy(f(x)+f(y)),;forall x,yin [0,+infty)$$

Problem:find all continuous functions $f:[0,+infty)to [0,+infty)$ such that $$f(y^2f(x)+x^2f(y))=xy(f(x)+f(y)),;forall x,yin [0,+infty)$$

Solution:

We see that $ f(x)=0 $ is one solution.

We now proceed to find other solutions that are not identically zero. For these solutions we can find points $ a $ for which $ f(a) neq 0 $.

Also note that $ f(0)=0 $ :

$f(0)=f(0f(0)+0f(0))=0(f(0)+f(0))=0 $.

$ f(x) $ must be surjective:

Choose $ a neq 0 $ for which $ f(a) neq 0 $. Now : $f(y^2f(a)+a^2f(y))=ay(f(a)+f(y)) ge yaf(a) ,;forall yin [0,+infty)$ . The right part of this inequality can be made arbitrarily large. So values of $ f(x) $ range from $ 0 $ to $ infty $. Because of the continuity requirement all values in between must be reached somewhere $ implies f(x) $ is surjective.

Also the argument of $ f() $ on the left of the inequality above: $ y^2f(a)+a^2f(y) $ goes to $ 0 $ when $ y $ goes to $ 0 $ , and it goes to $ infty $ when $ y $ goes to $ infty $.


Because of the surjectivity, continuity and the fact that $ f(0)=0 $ the function: $ lambda(x) = 2x^2f(x) $ is also continuous, surjective and ranges from $ 0 $ to $ infty $.

In other words: every $ a in mathbb{R_{ge 0}} $ can be written as $ 2b^2f(b) $ for some $ b ge 0 $ .

Therefore we can change the coordinates of $ f() $ from $ x $ to $ lambda(x) = 2x^2f(x) $ without loss of generality.

$f(lambda)=f(2x^2f(x)) =f(x^2f(x)+x^2f(x))=x^2(f(x)+f(x))=2x^2f(x)=lambdaimplies $

The only continuous solutions are : $ f(x)=0 $ and $ f(x)=x $.