adityaguharoy wrote: Does there exist any function $f: \mathbb{R}^+ \to \mathbb{R}$ such that for every real number $x,y$ the following is true : $$f(xf(x)+yf(y)) = xy$$ No. Just choose $x=y=0$ (domain of functional equation is "$\forall x,y\in\mathbb R$") and LHS becomes undefined (since $0\notin\mathbb R^+$, domain of function)

pco wrote: adityaguharoy wrote: Does there exist any function $f: \mathbb{R}^+ \to \mathbb{R}$ such that for every real number $x,y$ the following is true : $$f(xf(x)+yf(y)) = xy$$ No. Just choose $x=y=0$ (domain of functional equation is "$\forall x,y\in\mathbb R$") and LHS becomes undefined (since $0\notin\mathbb R^+$, domain of function) That was a typo : I wanted to say : for all positive real numbers $x,y$.

adityaguharoy wrote: Does there exist any function $f: \mathbb{R}^+ \to \mathbb{R}$ such that for every positive real number $x,y$ the following is true : $$f(xf(x)+yf(y)) = xy$$ Let $P(x,y)$ be the assertion $f(xf(x)+yf(y))=xy$ $P(x,x)$ $\implies$ $f(2xf(x))=x^2$ and so (in order LHS be defined) $f(x)>0$ $\forall x>0$ $P(1,\frac 12)$ $\implies$ $\exists u>0$ such that $f(u)=\frac 12$ Then $P(u,u)$ $\implies$ $u=\frac 1{\sqrt 2}$ $P(x,\frac 1{2x})$ $\implies$ $xf(x)+\frac 1{2x}f(\frac 1{2x})=\frac 1{\sqrt 2}$ So $xf(x)<\frac 1{\sqrt 2}$ $\forall x>0$ As a consequence $xy(xf(x)+yf(y))<\frac 1{\sqrt 2}$ $\forall x,y>0$, which is clearly impossible since, choosing $x$ great enough, we can get $xy(xf(x)+yf(y))>x(y^2f(y))$ as great as we want, and so $>\frac 1{\sqrt 2}$ And so $\boxed{\text{No such function}}$

After proving $xf(x)+yf(y)<\frac{1}{\sqrt 2 .xy}$ $0<xf(x)<xf(x)+yf(y)<\frac{1}{\sqrt 2. xy}$ Now sending , $y\rightarrow\infty$ We get $xf(x)=0$ which is not possible. Hence no such functions

TomMarvoloRiddle wrote: After proving $xf(x)+yf(y)<\frac{1}{\sqrt 2 .xy}$ $0<xf(x)<xf(x)+yf(y)<\frac{1}{\sqrt 2. xy}$ Now sending , $y\rightarrow\infty$ We get $xf(x)=0$ which is not possible. Hence no such functions No we don't get $xf(x)=0$. We get $xf(x) \to 0$ and these are very different. Just as an example : the function $f(x)=1/x$ for non zero $x$ obeys $f(x) > 0 \forall x >0$ but $f(x) \to 0$ as $x \to \infty$.