Let $(X, \| \cdot \|_X)$ $(Y, \| \cdot \|_Y)$ be normed space. A function $f \colon X \to Y$ shall be called an $n$-th degree (single variable) polynomial ($n \in \mathbb{N}\cup \{ 0\})$ if there exist:
- $(n+1)$-tuple of tensors $(T_i)_{i=0}^n$ such that for all $i \in \{ 0, \ldots,n\}$ $T_i$ is a continuous $i$-linear map (, so $T_0$ is a constant (here, possibly non-zero,) function).
- $x \in X$
such that for all $y \in X$ we have: $$f(y) = \sum_{i=0}^n T_i( y-x, \ldots, y-x).$$
Now, for a normed space $(V, \| \cdot \|_V)$, let:
\begin{cases} B_0^V &= \{ b \in C([0,1]; V) \mid \text{ b is constant}\},\\ B_n^V &= \left\{ [0,1] \ni t \mapsto (1-t)b_1(t) + tb_2(t)\mid b_1, b_2 \in B_{n-1}^V \right\}, \text{ for } n \in \mathbb{N}. \end{cases} Then, a function $f \colon X \to Y$ shall be called $k$-th order segmental if $$ f \circ B_1^X = \{ f \circ b \mid b \in B_1^X \} \subseteq B_{k}^Y. $$
It is not hard to see that every $n$-th degree polynomial $f \colon X \to Y$ is a continuous (I think, even smooth) $n$-th order segmental function.
Question
I'm more interested if there is a similar implication in the other direction, i.e., if one were to impose sufficient regularity condition on an $n$-th order segmental map, then it is a $n$-th degree polynomial.
Also, I'd be interested if there is some other, somewhat similar characterization of polynomials.
Note
The implication seems to hold for lipshitz (or just conitnuous) $1$-st order segmental maps (in fact, they are just affine maps, so they can be characterized as a linear map composed with a translation, i.e., a $1$-st degree polynomial).
