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The vector space $\mathcal P_m(\Bbb R)$ is the set of polynomial functions with real coefficients of degree at most $m$. I know that it standard dual basis is the set

$$B:=\left\{\frac{p^{(j)}(0)}{j!}:j\in\{0,\ldots,m\}\right\}\tag{1}$$

where $p^{(j)}$ is the $j$-th derivative of $p$.

However it is clear that the functional defined by

$$p\mapsto\int_a^bp(x)\mathrm dx\quad\text{and}\quad p\mapsto p^{(j)}(a),\quad a,b\in\Bbb R,p\in\mathcal P_m(\Bbb R)\tag{2}$$

belong to the dual space of $\mathcal P_m(\Bbb R)$, hence it must exists a linear combination of elements of $B$ that equal these functional, but I cant see how this is possible.

Can someone show me how is possible to write functional as the described in $(2)$ as a linear combination of elements in $(1)$? Or there is something about Im wrong in my thoughts? Thank you in advance.

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    $\begingroup$ for example $p(a) = \sum_{i}a^i \cdot \dfrac{p^{(i)}(0)}{i!}$ $\endgroup$ – JJR May 17 '17 at 11:32
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First off, your dual basis element $B_j:p\mapsto\frac {p^{(j)}(0)}{j!}$ is just the operation of taking the coefficient (in $p$) of the monomial $X^j$. By linearity of functionals, it suffices to understand you integration functional on each of the monomials $X^k$. But clearly $\int_a^bx^k\mathrm dx=\frac{b^{k+1}-a^{k+1}}{k+1}$. Then your functional is $\sum_{k=0}^m\frac{b^{k+1}-a^{k+1}}{k+1}B_k$.

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