Find all functions $f:\mathbb{R} \rightarrow \mathbb{R}$ such that $f[f(x)^2+f(y)]=xf(x)+y$ for all real numbers $x$ and $y$.

The answer to this has already been posted, but it doesn't explain why this function is injective and surjective. I would really appreciate it if someone did.

Link to that question: Functions satisfying $f\left( f(x)^2+f(y) \right)=xf(x)+y$


Surjective because $f[f(0)^2+f(y)] = y$ for any $y$. In particular, there is an $x_0$ such that $f(x_0)=0$, and using $x=x_0$ in the identity gives $f(f(y))=y$, which shows injectivity.

  • 1
    $\begingroup$ Why does $f(f(y))=y$ show injectivity? sorry for my naivety. And also why does $f[f(0)^2+f(y)]=y$ show surjectivity? I apologize for my ignorance. $\endgroup$ – Dominic Stone Apr 12 '13 at 6:58
  • 3
    $\begingroup$ @DominicStone go with defition $f(x)=f(y) \Rightarrow f(f(x))=f(f(y))\Rightarrow x=y$. For the surjective part since $f[f(f(0)^2+f(y)]=y$ and $y$ can take any value in $\mathrm{R}$ and by the previous equation $y$ is also in the image of $f$ since it is the image of $f[f(0)^2+f(y)]$ we conclude $\endgroup$ – clark Apr 12 '13 at 7:17
  • 1
    $\begingroup$ Thanks you SO MUCH. I now completely understand. I was being VERY blind. Thanks again! $\endgroup$ – Dominic Stone Apr 12 '13 at 7:23

Injectivity and surjectivity follow from relatively abstract properties of the form of the equation.

Injectivity follows because for any $x$, the left side (LHS) is a function of $f(y)$ while the right side (RHS) is an injective function of $y$.

Surjectivity because the RHS is a surjective function of $y$, for any $x$, while the LHS is $f(...)$.

Hence the arguments written down in comments under the answer that led to this question:

(to prove surjectivity) adjust $y$ so as to hit any desired value.

The proof of injectivity ... is to compare the functional equation written using $(x,y)$ and $(x,z)$ as the variables. If $f(y)=f(z)$ for fixed $x$ then the left sides of the equations are equal, which compels $y=z$ by looking at the right hand sides.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.