About real functions with $f(f(x))+2f(x)=3x, \forall x \in A$

Question

Let $f:A \rightarrow A$ where $A \subset \mathbb{R}$ such that $f(f(x))+2f(x)=3x, \forall x \in A$. Prove the following:

1. $f$ is injective
2. If $A$ finite then $f(x)=x$
3. Does 2. remain true if $A$ infinite?

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1. is very easy to prove using injectivity definition. I cannot respond to 2. and 3.

Answer 1

Hint for 2: If $A$ is finite, it has a smallest element $y$. Then $y$ is the smallest element of the range of $f$, and $f(f(y))+2f(y)=3y$, hence $f(f(y))=y$ and $2f(y)=2y$. So $f(y)=y$. You can use this together with induction to the size of $A$ to prove it.

Answer 2

For the third question, consider $A=\{(-3)^n \colon n \in \mathbb Z \}$ and $f(x)=-3x$.