Derivatives of a polynomials on a Banach space as a multilinear maps

Question

I'm reading "Differential Equations Driven by Rough Paths" by Lyon, Caruana and Lévy and I can't wrap my head around the following part (beginning of Section 1.4.2).

Let $V$ be a finite-dimensional Banach space. For an integer $K \geq 0$ let $P: V \to \mathbb{R}$ be a polynomial of degree $k$. Let $X : [0,T] \to V$ be a Lipschitz continuous path. Then, if $P^1:V \to L(V,\mathbb{R})$ denotes the derivative of $P$, then Taylor's theorem asserts that, for all $t \in [0,T]$, $$ P(X_t) = P(X_0) + \int_0^t P^1(X_s)dX_s. $$
Let now $P^2 : V \to L(V \otimes V, R)$ denote the second derivative of $P$. Recall that $L(V \otimes V, R)$ is the space of bilinear forms on $V$ . Then, by substitution into the last expression, we get $$ P(X_t) = P(X_0) + P^1(X_0) \int_{0<u_1<t}dX_{u_1} + \iint_{0<u_1<u_2<t} P^2(X_u) dX_{u_1} \otimes dX_{u_2}. $$ Integrals are understood in Young's sense.

I have trouble understanding the following concepts:

1. I'm only familiar with polynomials over rings, not Banach spaces. Where does the multiplication come from?
2. Why is the derivative of a polynomial a map from $V$ to the space of linear maps from $V$ to $\mathbb{R}$? I'm completely lost here.
3. How do we obtain the first Taylor representation?
4. I guess that the formulation of the second derivative as a bilinear map follows from the same reasoning as for the first derivative, but nonetheless. What is the intuition?

Any help is appreciated!

Answer 1

An homogenous polynomial of degree $r$ in a vector space is the composition of a $r$-linear function $N$ (with values in some vector space) and the $r$-diagonal map $v \mapsto v^{(r)} = (v, \ldots, v) \in V^r.$ Thus, a homogeneous polynomial of degree $r$ is $f_r(v) = N(v^{(r)}).$ A polynomial of degree at most $n \in \mathbf{N},$ is the sum up to $r \leq n$ of homogenous polynomials of degree $r.$ Polynomials in vector spaces in general do not have product defined. However, there is the following construction, suppose $W_1, W_2$ are two vector spaces and $P:W_1 \times W_2 \to X$ a "product" (i.e., bilinear map, $X$ being another vector space). Suppose $f_s$ and $g_t$ are homogenous polynomials of degree $s$ and $t$ with values in $W_1$ and $W_2,$ respectively. The product of $f_s$ and $g_t$ relative to $P$ is $P \circ (f_s, g_t).$ This is a homogeneous polynomial of degree $s+t.$ To see this, consider $s$-linear and $t$-linear functions $M$ and $N$ such that $f_s(v) = M(v^{(s)})$ and $g_t(v) = N(v^{(t)}).$ Then $(P \circ (f_s, g_t))(v) = P\circ(M(v^{(s)}), N(v^{(t)}))$ and it is easy to see that $(v_1, \ldots, v_s, v_{s+1}, \ldots, v_{s+t}) \mapsto P(M(v_1, \ldots, v_s), N(v_{s+1},\ldots,v_{s+t}))$ is $(s+t)$-linear. If now $f = \sum_{s \leq m} f_s$ and $g = \sum_{t \leq n}$ are two polynomials (not necessarily homogenous), then their product is $P \circ (f,g),$ and by bilinearity of $P,$ we can write this as $\sum_{s \leq m,t \leq n} P \circ (f_s, g_t),$ which is a sum of homogenous polynomials of degree $s+t,$ and therefore the product of $f$ and $g$ relative to $P$ is a polynomial of degree at most $m+n.$ The authors are using $P:\mathbf{R} \times \mathbf{R} \to \mathbf{R}$ the usual product of numbers $P(a,b) = ab.$

I hope the previous answered your first question; all other questions should receive their own thread.