Can we assume $ f ( y ) = 0 $ in solving the functional equation $ f \big( x - f ( y ) \big) = f \big( f ( y ) \big) + x f ( y ) + f ( x ) - 1 $?

Question

Here's a problem from one of my books. I believe it's not really a compete solution.

Problem. Determine all functions $ f : \mathbb R \to \mathbb R $ such that $$ f \big( x - f ( y ) \big) = f \big( f ( y ) \big) + x f ( y ) + f ( x ) - 1 \tag i \label i $$ for all $ x , y \in \mathbb R $.

Solution. Put $ x = f ( y ) = 0 $ in Eq. \eqref{i}. Then $$ f ( 0 ) = f ( 0 ) + 0 + f ( 0 ) - 1 \text ; $$ $$ \therefore \ f ( 0 ) = 1 \text . $$ Again, put $ x = f ( y ) = \lambda $ in Eq. \eqref{i}. Then $$ f ( 0 ) = f ( \lambda ) + \lambda ^ 2 + f ( \lambda ) - 1 \text ; $$ $$ \implies 1 = 2 f ( \lambda ) + \lambda ^ 2 - 1 \text ; $$ $$ \therefore \ f ( \lambda ) = \frac { 2 - \lambda ^ 2 } 2 = 1 - \frac { \lambda ^ 2 } 2 \text . $$ Hence, $ f ( x ) = 1 - \frac { x ^ 2 } 2 $ is the unique function.

In the opening line of the solution, it assumes that $ f ( y ) = 0 $ exists. But we don't really know if it's a surjective function. So, even if we assume such and prove, then, as the question demands "all possible functions that satisfy …", I believe there's another part of the problem assuming that no such $ y $ exists for which $ f ( y ) = 0 $ which may be proved or disproved, but not left unattempted.

Right?

Answer 1

You're right. In fact, there is another error in the solution as well: it only shows that $f(\lambda)=1-\frac{\lambda^2}{2}$ under the assumption that $\lambda=f(y)$ for some $y$. So this solution really is only valid if you know $f$ is surjective (which is in fact not possible, since the function $f(x)=1-\frac{x^2}{2}$ is not surjective!).

A correct solution requires a much more careful analysis to show the range of $f$ must be large enough that you can in fact conclude $f(x)=1-\frac{x^2}{2}$ for all $x$. As Rohan commented, you can find a correct solution here (Problem 16).

Answer 2

I also don't see why we can assume there exists $y$ such that $f(y)=0 .$

Let $x=2f(y).$ Then $$f(f(y))=f(2f(y)-f(y))=f(x-f(y))=f(f(y))+xf(y)+f(x)-1$$ which simplifies to $$0=xf(y)+f(x)-1$$ which, with $f(y)=x/2,$ gives $$0=x^2/2+f(x)-1.$$ So $f(x)=1-x^2/2$ whenever $x\in \{2f(y):y\in \Bbb R\}.$

It's very late for me. Perhaps someone else is willing to finish this: What if $x\not \in \{2f(y): y\in \Bbb R\}$?