Let's take two diffeomorphisms $F,G: \mathbb{R}^{n} \mapsto \mathbb{R}^{n}$. Let $x \in \mathbb{R}^{n}$, and $x^{n} = F^{n}(x)$, where $n \in \mathbb{Z}$. Suppose that $F$ has a first integral, i.e. there exists a function $I: \mathbb{R}^{n} \mapsto \mathbb{R}$ such that $I(x) = I(F(x))$.
Suppose that I have a function $H(x)$ (that arises in the theory of invariant manifolds) that is defined to be
\begin{equation} H(x) = \sum_{n= -\infty}^{n= \infty} { \langle DF^{n}(x^{-n})G(x^{-n-1}), I(x) \rangle } \label{one} \end{equation}
where the angle brackets indicate inner product, and $D$ is the derivative. I would like to manipulate the above expression for $H(x)$ into the following form:
$$H(x)= \sum_{n= -\infty}^{n= \infty} { \langle G(x^{n-1}), \nabla I(x^{n}) \rangle }.$$
Now I can use that $I(x) = I(F^{n}(x^{-n}))$, and $\nabla I(x^{n}) = \nabla_{x^{n}} I(x^{n+1}) = \frac{\partial I(F(x^{n})}{\partial F}DF(x^{n})$, with $\langle Ax,y \rangle = \langle x, A^{T}y\rangle$ to make some progress, but I still can't get the desired form.
I would appreciate any step by step help.
Observe that $\nabla I(x) = (DF(x))^{T} \nabla I(x^{1}) = (DF(x^{n-1}...DF(x))^{T} \nabla I(x^{n}) = (DF^{n}(x))^{T} \nabla I(x^{n})$.
Hence we have
$$ \sum_{n= -\infty}^{n= \infty} {\langle DF^{n}(x^{-n}) G(x^{-n-1}), \nabla I(x) \rangle} = \sum_{n= -\infty}^{n= \infty} {\langle G(x^{-n-1}), (DF^{n}(x^{-n})^{T}\nabla I(x) \rangle} = \sum_{n= -\infty}^{n= \infty} {\langle G(x^{n-1}), (DF^{-n}(x^{n})^{T}\nabla I(x) \rangle} = \sum_{n= -\infty}^{n= \infty} {\langle G(x^{n-1}), (DF^{-n}(x^{n})^{T} (DF^{n}(x))^{T}\nabla I(x^{n}) \rangle} = \sum_{n= -\infty}^{n= \infty} {\langle G(x^{n-1}), (DF^{n}(x) DF^{-n}(x^{n}))^{T}\nabla I(x^{n}) \rangle} = \sum_{n = -\infty}^{n = \infty} {\langle G(x^{n-1}), \nabla I(x^{n}) \rangle }$$
as required.