Olympiad functional equations proof verification: $f\bigl(2f(x)+f(y)\bigr)=2x+y$ on $\mathbb Q$ and $f\bigl(xf(x)+f(y)\bigr)=f(x)^2+y$ on $\mathbb R$

contest-mathfunctional-equationsproof-explanationsolution-verification

First, let me be clear: I have absolutely no experience with olympiad functional equations, but I will start with a book in the very near future.

OK, but often, when I'm bored, I read solutions to functional equations on AOPS and/or YouTube, and often I find them interesting. This one time, I found a couple of problems (I don't exactly remember where, but they were from some early 2000's national olympiads), and I ran into the same issue with both these problems.

  1. Find all functions $ f : \mathbb Q \to \mathbb Q $ such that
    $$ f \bigl ( 2 f ( x ) + f ( y ) \bigr ) = 2 x + y \text . $$
  2. Find all functions $ f : \mathbb R \to \mathbb R $ such that
    $$ f \bigl ( x f ( x ) + f ( y ) \bigr ) = f ( x ) ^ 2 + y \text . $$

Here are my solutions:

  1. Let $ P ( x , y ) $ denote the assertion that $ f \bigl ( 2 f ( x ) + f ( y ) \bigr ) = 2 x + y $.
    $$ P ( x , x ) \implies f \bigl ( 3 f ( x ) \bigr ) = 3 x \label i \tag i \text . $$
    $$ P \bigl ( 3 f ( x ) , 3 f ( y ) \bigr ) , \eqref{i} \implies f \bigl ( 6 f ( x ) + 3 f ( y ) \bigr ) = 6 f ( x ) + 3 f ( y ) \text . $$
    Now letting $ a = 6 f ( x ) + 3 f ( y ) $, we have $ f ( a ) = a $.
  2. Let $ Q ( x , y ) $ denote the assertion that $ f \bigl ( x f ( x ) + f ( y ) \bigr ) = f ( x ) ^ 2 + y $.
    $$ Q ( 0 , 0 ) \implies f ^ 2 ( 0 ) = f ( 0 ) ^ 2 \text , $$
    where $ f ^ 2 ( t ) $ is $ f \bigl ( f ( t ) \bigr ) $.
    $$ Q ( 0 , y ) \implies f ^ 2 ( y ) = f ( 0 ) ^ 2 + y \text . $$
    $$ Q \bigl ( 0 , f ( y ) \bigr ) \implies f \left ( f ( 0 ) ^ 2 + f ( y ) \right ) = f ( 0 ) ^ 2 + f ( y ) \text , $$
    using the previous two relations.
    Now letting $ t = f ( 0 ) ^ 2 + f ( y ) $, we have $ f ( t ) = t $.

For both of the questions $ f ( x ) = x $ was indeed a valid solution, but by just messing around I saw $ f ( x ) = – x $ is also a valid solution for both.

I'm not sure if my solution is wrong or correct but I think one possible reason it may be wrong is that letting $ a = \dots $ and $ t = \dots $ step because not all numbers in the domain can be written as that, that is my guess.

Please let me know where my solution is wrong, and also I would highly appreciate it if you could tell me some resource a novice like myself can start from.

Thanks!

Best Answer

  1. Just after (i), $P(3f(x),3f(y))=P(6x+3y)$, not $P(6f(x)+3f(y))$

  2. $f(f(0)^2+f(y))$ should be $f(f(0)^2+y)$

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