question -
What are the continuous functions on $\mathbb{R}$ which are solutions of the equation $$ x f(y)+y f(x)=(x+y) f(x) f(y) ? $$
my try -
by putting $y=x$ i get $f(x)=0$ or $1$ for all $x$ not equal to $0$... now my answer is same as mention in book but i think it is incorrect because they are not valid for all $x$ ???
can someone tell how to fix this hole with the help of continuity ... i know this is simple question but i want to clear my doubt...
thankyou
Wit $x=y$, we arrive at the necessary condition $$2xf(x)=2xf(x)^2 $$ and hence $$\tag1f(x)\in\{0,1\}\quad\text{for }x\ne0. $$ The only continuous functions with this property are the constant functions $$f(x)=0$$ and $$f(x)=1. $$ Both are directly verified to indeed solve the original functional equation.
We still have $(1)$, but note that $f$ may jump discontinuously between $0$ and $1$. Suppose $f(x_0)=0$ for at least one $x_0\ne 0$. Then with $y=x_0$, we get $x_0f(x)=0 $ and hence $$ f(x)=0\quad\text{for all } x$$ as one solution (again).
So assume $f(x)=1$ for all $x\ne 0$. Then nothing can be said about $f(0)$, i.e., for any $c$, $$f(x)=\begin{cases}c&x=0\\1&x\ne 0\end{cases} $$ is a solution, as one readily verifies.
Suppose $f(x) =0$ for all $x \neq 0.$ Since $f$ is continuous, so $f(0)=\lim_{x \to 0} f(x)=0$. Hence $f(x)=0$ for all $x \in \mathbb{R}.$
Similarly, the other case follows.