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I have a midterm tomorrow and have been able to cover all other topics except this. I don't even have an idea how to start these questions. If someone could give me some tips I would very much appreciate it. The questions I'm having trouble with are:

Find all functions $ f : \mathbb R \to \mathbb R $, which satisfy the equation $$ f \left( x ^ 2 - y ^2 \right) = x f ( x ) + y f ( y ) $$

Find all functions $ f : \mathbb Z \to \mathbb Z $, which satisfy $$ f \big( f ( x ) \big) = x + 1 $$

The lecture slides give us some tips, but I'm not sure how to use the tips to solve the question.

Some tips:

  • Substituting values with variables. For example, plug in $ y $ for $ x $ above or $ f ( x ) $ instead of $ y $.
  • Using Mathematical Induction.
  • Is $ f $ one-to-one or onto (injective or surjective)?
  • Finding fixed points or zeros of function.
  • Write $ f ( x ) = g ( x ) + h ( x ) $ where $ g ( x ) $ is an even function and $ f ( x ) $ is an odd function (Reminder: This is always possible!)
  • If given polynomials, checking their degrees might help.
  • Don't forget to check that all the functions you found are actually solutions to the problem!

Thank you very much for your help.

Mohsen Shahriari
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The_Questioner
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1 Answers1

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For the first one : If $x=y=0$, then $f(0)=0$, if $x=y$, then $f(0)=2xf(x)$, so $\forall x, 2xf(x)=0$, so $\forall x, f(x)=0$. So $f=0$.

For the second one : $f(f(f(x)))=f(x+1)$ and $f(f(f(x)))=f(x)+1$. So $f(x+1)=f(x)+1$. So $\forall x,f(x)=f(0)+x$. Then note that $f(f(0))=0+1=1$ with the property in the question and $f(f(0))=f(f(0))+1$ with the previous equality, so $1=f(f(0))=f(f(0))+1=1+1=2$. It is absurd, so there is no such $f$.

Bérénice
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  • Thanks. How did you get f(f(f(x))) = f(x) + 1? Also, how did you conclude that f(x) = f(0) + x. Sorry, I'm new at this – The_Questioner May 30 '16 at 22:37
  • I appplied $f \circ f$ to $f(x)$ thus I could apply the identity and obtain $f(x)+1$. And after you obtain that $f(x+1)=f(x)+1$ for all $x$, so you also have $f(x)=f(x-1+1)=f(x-1)+1$, so if you know the image of one integer by $f$, you obtain all the other images by induction. – Bérénice May 30 '16 at 22:42