Let $T(f(t)) = (f(0), f(1), f(2), f(3),\cdots)$ from $P$ to $V$, where $P$ denotes the space of all polynomials. Is $T$ linear and if so, is $T$ an isomorphism?
I feel like a counterexample is feasible, but I'm not sure if I should arbitrarily pick some polynomial of any degree. Suggestions?
EDIT:
$V$ denotes the space of infinite sequences of real numbers. Apologies for not clarifying earlier.
It's clear that T is linear :
$$T(f+\lambda g) = ((f+\lambda g)(0),(f+\lambda g)(1),(f+\lambda g)(2),\cdots) $$
$$= (f(0)+\lambda g(0),f(1)+\lambda g(1),f(2)+\lambda g(2),\cdots)$$
$$= (f(0),f(1),f(2),\cdots)+\lambda (g(0),g(1),g(2),\cdots)=T(f)+\lambda T(g)$$
Now is T bijective?
Take the sequence $(e^0,e^1,e^2,\cdots)$
If this was the image by T of a polynomial P, you would have $$\lim_{n\to +\infty} \frac{P(n)}{e^n} \neq 0 $$
But this limit is 0 for all polynomial. Hence T is not bijective.
Edit : notice that T is injective, because if $T(f-g)=0$, $f-g$ has an infinite number of roots, so it's null