Prove $S_{m}^{m}(\Delta)$={$s:s\in C^{m}[a,b]$ and $s$ is a polynomial of order $m$ in each $[x_{i},x_{i+1}]$}=$P_{m}$

Question

Basically, I don't understand clearly. The point is to prove that these two spaces are equal or that the polynomial $s \in S_{m}^{m}(\Delta)$ is unique/the same in each $[x_{i},x_{i+1}]$? where $(\Delta)$ is a segmentation of [a,b]

Answer 1

So here is my answer:

Let $s_{i-1}$ be the polynomial in the $[x_{i-1},x_{i}]$ region of $[a,b]$ and $s_{i}$ the polynomial in the $[x_{i},x_{i+1}]$. Then by the continuity of all the $m$ derivatives of $s$ in the internal nodes of $[a,b]$ we have that:

$s_{i-1}^{(m)}(x_{i}) = c_{i-1}$$\quad$ and$\quad$ $s_{i} ^{(m)}(x_{i}) = c_{i}$ $\quad$ $\rightarrow$ $\quad$$c_{i-1}= c_{i}$

$s_{i-1}^{(m-1)}(x_{i}) =a_{i-1}x_{i}+c_{i}$$\quad$ and$\quad$ $s_{i} ^{(m-1)}(x_{i}) = a_{i}x_{i}+c_{i}$$\quad$$\rightarrow$$\quad$ $a_{i-1}=a_{i}$

etc.

So we have that $s_{i-1}=s_{i}$ i.e. we have that $s$ is the same polynomial in each $[x_{i},x_{i+1}]$.

p.s These self answers in StackEchange are kinda creepy.