I know how to prove that it is one to one, but i don't know how to prove that it's not onto, can some1 help me ?
ps: P(R) is a polynomial
The endomorphism is:
$\displaystyle T:P(\mathbb{R})\rightarrow P(\mathbb{R})\,,\quad T[p(x)]=\int_0^x p(t)\;dt$
Suppose there exists $\displaystyle p(x)\in P(\mathbb{R})$ such that $\displaystyle T[p(x)]=1$ , use the Fundamental Theorem of Calculus and you will find a contradiction.
Fernando Revilla
I meant the whole expression "p(x)=0" is not a pol. but an equation, of course.
And in this case it is the zero polynomial, not the zero function. Had he written "p(x):=0" things would
have been clearer (for me, at least).
Tonio
Ps. By the way, I think my first message shows easily and clearly why the given transf. isn't surjective, but
there hasn't been any discussion on it...