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.
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 $f$.