For each $f \in P_{n-1}$ there exists $g \in P_n$ such that $f(x)=g(x+1)-g(x)$ Assume that $P_n$ & $P_{n-1}$ are defined as the vector spaces of the polynomials of degree $n$ & $n-1$ over field $\mathbb R$, respectively.
Prove that for each $f \in P_{n-1}$ there exists a polynomial $g \in P_n$ such that $f(x)=g(x+1)-g(x)$.  
My try :  
I defined a linear-map $T: P_{n} \to P_{n-1}$ such that $T(g)=g(x+1)-g(x)$. I need to prove that $T$ is surjective. That's what i don't know how to show ...  
Thanks in advance.
 A: Simply show that the kernel of $T$ are the constant polynomials of $P_n$ and use a dimensional argument.
A: Here is a direct construction. Define for $x\in\Bbb R$ and $k\in\Bbb N$ the symbol $\binom xk$ as $\frac{x(x-1)\ldots(x-k+1)}{k!}$ (when $x$ is a natural number, this is in fact a binomial coefficient), which is a polynomial in$~x$ of degree$~k$. The interest of these polynomials for this question is that they satisfy $\binom{x+1}k-\binom xk=\binom x{k-1}$, as is easily checked. (For $x\in\Bbb N$, this is equivalent to Pascal's recursion relation for binomial coefficients.)
Let $b_k$ designate the function $x\mapsto\binom xk$, for any $k\in\Bbb N$. Then the space $P_n$ of polynomial functions of degree at most$~n$ has a basis $[b_0,b_1,\ldots,b_n]$. So for any $f\in P_{n-1}$ there are coefficients $c_0,c_1,\ldots,c_{n-1}$ such that  $f=c_0b_0+c_1b_1+\cdots+c_{n-1}b_{n-1}$. Then the function $g=c_0b_1+c_1b_2+\cdots+c_{n-1}b_n$ satisfies $g\in P_n$ and $\forall x\in\Bbb R:g(x+1)-g(x)=f(x)$, as desired.
A: A constructive proof:
Let $f\in P_{n-1}$ be given. We are looking for some function $g\in P_{n}$ such
$$ f(x) = g(x+1) - g(x) = \int_0^1 g'(x+t) dt. $$
Since this property is linear in $f$, checking it for a basis is sufficient. 
Consider $f_1(x) = 1$. Then, such a function would be $g_1(x) = x$. For $f_2(x) = x$ we have
$$ f_2(x) = \int_0^x f_1(y) dy = \int_0^1 \Bigl (\underbrace{g_1(x+t) - \int_0^1 g_1(s)ds}_{=:g_2'(x+t)} \Bigr) dt. $$
Now, inductively we obtain a basis with the desired property. 
