Proving ƒ(x) is the Identity Function

[Derek Hart]

Homework Statement

I have been going through a textbook trying to solve some of of these with somewhat formal proofs. This is a former Putnam exam question. (Seemingly the easiest one I have attempted, which worries me).

Consider a polynomial function ƒ with real coefficients having the property ƒ(g(x)) = g(ƒ(x)) for all polynomial functions g with real coefficients. Determine and prove the nature of ƒ.

Homework Equations

No equations, just the tricky fact that constant functions are polynomials.

The Attempt at a Solution

A proof by contradiction that ƒ(x) is the identity function:

We can consider the function h(x) = 1 and its derivative, which I will denote as g, namely to state the fact that ƒ((g+h)(x)) = ƒ(1) = g(ƒ(x)) + h(ƒ(x)), which then = ƒ(0) + ƒ(1); therefore ƒ(0) = 0.
Now, suppose that ƒ is not the identity function, or, specifically, that ƒ(c) ≠ c. We can consider the polynomial function k(x) which has a zero at c but not at ƒ(c). Then ƒ(k(c)) = ƒ(0) = 0 ≠ k(ƒ(c)); a contradiction occurs. This proves that ƒ must be the identity function.

 

[PeroK]

The Attempt at a Solution

A proof by contradiction that ƒ(x) is the identity function:

We can consider the function h(x) = 1 and its derivative, which I will denote as g, namely to state the fact that ƒ((g+h)(x)) = ƒ(1) = g(ƒ(x)) + h(ƒ(x)), which then = ƒ(0) + ƒ(1); therefore ƒ(0) = 0.
Now, suppose that ƒ is not the identity function, or, specifically, that ƒ(c) ≠ c. We can consider the polynomial function k(x) which has a zero at c but not at ƒ(c). Then ƒ(k(c)) = ƒ(0) = 0 ≠ k(ƒ(c)); a contradiction occurs. This proves that ƒ must be the identity function.

I can't see what you are doing in the first part. There is a very simple way to show that $f(0) = 0$.

The second part is sound. Although, this could be simplified by doing a straight proof, as opposed to using contradiction.