Theorem: If $f$ is a polynomial of degree $\leq n$, then all of the divided differences $f[x_0,x_1,\cdots, x_i]$ is zero for $i\geq n+1$.
This can be proved by using the result that $f[x_0,x_1,\cdots, x_i]=\frac{f^{(n)}(c)}{n!}$ for some $c\in (a,b)$, for $n$th continuously differentiable function $f$ on $[a,b]$.
Now I want to know the converse of above theorem . Can I say that if $f$ is $n$-times continuously differentiable on $[a,b]$ having all of the divided differences $f[x_0,x_1,\cdots, x_i]$ is zero for $i\geq n+1$ then $f$ must be polynomial of degree $\leq n$?
According me converse is true as $f[x_0,x_1,\cdots, x_i]=\frac{f^{(n)}(c)}{n!}=0$ for some $c$ will give $n$th derivative of $f$ as zero as $c$ varies according $x_i$ and hence $f$ must be a polynomial. Am I right ? Please comment. Thank you.
I think the answer is yes. Suppose $f \colon [a, b] \to \mathbb{R}$ and that $$f[x_0, x_1, \dots, x_{n + 1}] = 0$$ for any set of distinct points $\{x_0, x_1, \dots, x_{n + 1}\} \subseteq [a, b]$. Choose $n$ distinct points $x_0, x_1, \dots, x_n$ in $[a, b]$. Let $$p_n(x) = \sum_{i = 0}^{n}f[x_0, \dots, x_i](x - x_0) \dots (x - x_{i - 1})$$ be the interpolation polynomial for $f$ through $x_0, \dots, x_n$. Now fix an arbitrary $c \in [a, b]$. We can add $c$ to our list of interpolation points to get a polynomial $p_{n + 1}$ that matches $f$ at $x_0, x_1, \dots, c$. Thus $$f(c) = p_{n + 1}(c) = p_n(c) + f[x_0, x_1, \dots, x_n, c](x - x_0)(x - x_1) \dots (x - x_n) = p_n(c).$$ Thus $f(x) = p_n(x)$ for all $x \in [a, b]$.