Subspace of polynomials vanishing to order $n$
Part b: Let $W = \Bbb R_m[x]$ denote the set of polynomials of degree at most $m$, where $m$ has been chosen so that $V \subset W$. Note that $\dim(W_n) = m+1-n$, and $W_{n+1} \subset W_n$. Let $S_n$ denote a subspace of $W_n$ such that $W_{n} = S_n \oplus W_{n+1}$; note that $\dim(S_n) = 1$.
Now, we note that $V_n = W_n \cap V$, and that $V_{n+1} \subset V_n$. With that, we have $\dim(V_{n+1}) \leq \dim(V_n)$ and $$ \begin{align} \dim(V_{n+1}) &= \dim(W_{n+1} \cap V) \\ & \geq \dim([W_{n+1} \oplus S] \cap V) - \dim(S \cap V) \\ & = \dim(W_n\cap V) - \dim(S \cap V) = \dim(V_n) - \dim(S \cap V) \\ & \geq \dim(V_n) - \dim(S) = \dim(V_n) - 1. \end{align} $$ The conclusion follows.
Regarding part c, it is helpful to note that $V_0 = V$, note that there exists an $n$ such that $V_n = \{0\}$, and consider the sequence $\dim(V_0),\dim(V_1),\dim(V_2),\dots$.
For $(b)$, let $\varphi \colon V_n \rightarrow \mathbb{R}$ be the linear functional given by $\varphi(f) = f^{(n)}(a)$. Note that
$$ V_n = \{ f \in V \, | \, f(a) = f'(a) = \dots = f^{(n-1)}(a) = 0 \}, \\ \ker(\varphi) = \{ f \in V_n \, | \, f^{(n)}(a) = 0 \} = \{ f \in V \, | \, f(a) = \dots = f^{(n-1)}(a) = f^{(n)}(a) = 0 \} = V_{n+1} $$
and so by rank-nullity or directly, there are two options:
- Either $\varphi = 0$ and then $V_{n+1} = \ker(\varphi) = V_n$.
- Or $\varphi \neq 0$ and then $\dim V_{n+1} = \dim \ker(\varphi) = \dim V_n - 1$.
For part $(c)$, let $M \geq k - 1$ be the maximal degree of an element in $V$ (this exists because $V$ is finite dimensional) and consider the sequence of descending subspaces $$ V := V_0 \supseteq V_1 \supseteq V_2 \supseteq V_3 \supseteq \dots \supseteq V_{M+1}. $$ Prove that $V_{M+1}= \{ 0 \}$. Taking dimensions, we get $$ k = \dim V \geq \dim V_1 \geq \dim V_2 \geq \dim V_3 \geq \dots \geq 0. $$ By part $(b)$, whenever we have a strict inequality the dimension decreases by $1$ and so there are precisely $k$ non-negative distinct integers $0 \leq n_1,\dots,n_k \leq M$ such that $\dim V_{n_i + 1} + 1 = \dim V_{n_i}$. Finally, note that $0\neq f \in V$ vanishes to order $n$ iff $f \in V_n \setminus V_{n+1}$ which implies that $\dim V_n > \dim V_{n+1}$ which happens iff $n = n_i$.