I came across the following question while studying functional equations.
Let $f: \mathbb{R} \to \mathbb{R} $ be a function satisfying $$f \big(xf (y)-f(x)\big)=2f (x)+xy $$ $\forall x, y \in \mathbb {R} $, prove that $f $ is a surjective function.
I really have no idea that how to prove the aforementioned as I am very new to functional equations. The author has provided a very short proof without mentioning much details, making it very confusing and difficult to understand.
A pedantic proof would be highly helpful and will really be appreciated.
Thank you
Best regards !
If we fix $x = 1$, then the equation becomes $$f(f(y) - f(1)) = 2f(1) + y,$$ for any $y$. Intuitively, the right hand side can be made to be any value we want. The left hand side is $f$ of some expression of $y$ (it doesn't really matter what). So, we can choose a value of $y$ to make the right hand side whatever we want, then we have $f$ of an expression of $y$ that produces this value. This is what makes $f$ surjective.
If we want to prove it formally, fix $b \in \Bbb{R}$. Let $x = 1$ and $y = b - 2f(1)$. Then our equation turns into $$f(f(b - f(1)) - f(1)) = 2f(1) + b - 2f(1) = b.$$ If we let $a = f(b - f(1)) - f(1)$, then we have constructed an $a \in \Bbb{R}$ such that $f(a) = b$, proving surjectivity.