I got an exercise in which we define the application $\displaystyle \Delta : \mathbb{R}_{2p+1} \rightarrow \mathbb{R}_{2p+1}$ defined as $$ \Delta(P)(X)=P(X+1)-P(X) $$ I also have a sequence of polynomials element given by $$ \forall x \in \mathbb{R}, \forall n \in \mathbb{N}^{*}, \ \begin{cases} \displaystyle B_0(x)=1\\ \displaystyle B'_n(x)=B_{n-1}(x)\\ \displaystyle \int_{0}^{1}B_n(t)\text{d}t=0 \end{cases} $$ A first question was to find Ker$(\Delta)$ and to show that $\text{Im}\left(\Delta\right)=\mathbb{R}_p\left[X\right]$.
What I have done is to consider the equation $\displaystyle \forall X , \ P\left(X+1\right)=P\left(X\right)$ so $P(0)=P(n)$ for $n \in \mathbb{Z}$. Hence the polynomial $X \mapsto P(X)-P(0)$ has an infinite number of roots so $P(X)=P(0)$. With the reciprocal $$ \text{Ker}\left(\Delta\right)=\mathbb{R}_0\left[X\right] $$ And using that $\text{Im}\left(\Delta\right) \subset \mathbb{R}_p\left[X\right]$ and the equality of dimension i've the result. Is this okay ? However I need to prove now that there exists a unique polynomial $Q$ such that $$ \Delta\left(Q\right)\left(X\right)=\frac{X^p}{p!} \text{ and }\int_{0}^{1}Q\left(t\right)\text{d}t=0 $$ I know this polynomial $Q$ exists because of $\displaystyle X \mapsto X^p/p! \in \mathbb{R}_p\left[X\right]$. But supposing $Q$ and $S$ to be two polynomials satisfying those conditions, I 'm stuck proving they are equal. How can I do that ? I just obtained $$ \int_{0}^{1}S\left(X+1\right)\text{d}X=\int_{0}^{1}Q\left(X+1\right)\text{d}X=\frac{1}{\left(p+1\right)!} $$
You should have observed that for any $p$, $\Delta$ maps $\mathbb R_p[X]$ to $\mathbb R_{p-1}[X]$. As you showed, $\ker \Delta = \mathbb R_0[X]$, the set of constant polynomials. By the rank-nullity theorem, $\dim \operatorname{im} \Delta = p$, thus $\Delta$ is surjective (when considered as a linear map between $\mathbb R_p[X]$ and $\mathbb R_{p-1}[X]$).
Therefore, there exists $Q\in \mathbb R_{p+1}[X]$ such that $\Delta Q = X^p/p!$. Choose $\alpha\in \mathbb R$ such that $\int_0^1 (Q+\alpha) =0$ and note that we still have $\Delta(Q+\alpha) = X^p/p!$.
If $R$ is also such that $\Delta R = X^p/p!$ and $\int_0^1 R = 0$, then by linearity $\Delta (Q+\alpha-R)=0$, thus $Q+\alpha-R$ is constant. But $\int_0^1 (Q+\alpha-R)=0$, hence $Q+\alpha-R = 0$. So $Q+\alpha$ is unique.