I am reading up on the proof of the following statement:
Let $p:\mathbb{X} \rightarrow \mathbb{Y}$ be a polynomial. Then $p$ is Frechet differentiable at every $x \in \mathbb{X}$, with $$dp(x) v = \sum_{k=1}^K \sum_{j=1}^k A_k(x^{(j-1)}, v, x^{(k-j)})$$
Everything is OK until the final stage of the proof, where they assert:
$$\Big|\Big|p(x+v) - p(x) - \sum_{j=1}^kA_k(x^{(j-1)}, v, x^{(k-j)})\Big|\Big|_{\mathbb{Y}} \leq \sum_{d=2}^k {{k}\choose{d}}a_k ||x||^{k-d}_{\mathbb{X}}||v||^{d}_{\mathbb{X}} $$
where $a_k = ||A_k||_{L^k(\mathbb{X};\mathbb{Y})}$.
I am really not sure how they obtained the bound.
I have tried to expand inside the norm, and I think that $p(x+v) - p(x) = A_k (v, v, \cdots, v)$, but how to obtain the binomial coefficient and the powers on $||x||$ and $||v||$, I am not sure at all. I am not even sure why the sum starts at 2!
Definition: $p: \mathbb{X} \rightarrow \mathbb{Y}$ is a polynomial if $p(x) = \sum_{k=0}^K A_k(x, \cdots, x)$, where $A_k$ is a $k$-multilinear map.
A step of the proof that is omitted above is where the problem is reduced to the case $p(x) = A_k(x, \dots, x)$; this can be done because the sum of differentiable functions is differentiable, with the derivative being the sum of derivatives.
Consider the expression $p(x+v)-p(x)$ as a polynomial with respect to $v$. Its constant term is zero. Its linear term comes from expanding $A_k(x+v, \cdots, x+v)$ into $2^k$ terms and taking only those where $v$ is present once (this is what makes the term linear). So, the linear term of $p(x+v)-p(x)$ is $$ A_k(v, x, x, \dots, x) + A_k(x, v, x, \dots, x) + \cdots + A_k(x, x, x, \dots, v) $$ which is denoted $\sum_{j=1}^k A_k(x^{(j-1)}, v, x^{(k-j)})$ for brevity.
Once the linear term is subtracted off, we have only terms of degrees $d\ge 2$ to consider. The term of degree $d$ comes from expanding $A_k(x+v, \cdots, x+v)$ into $2^k$ terms and taking only those where $v$ appears exactly $d$ times. There are $\binom{k}{d}$ such terms, and each of them is estimated by $a_k\|x\|^{k-d}\|v\|^d$. The rest is the triangle inequality.
Lastly, the Fréchet differentiability follows from $$ \sum_{d=2}^k {{k}\choose{d}}a_k \|x\|^{k-d}_{\mathbb{X}}\|v\|^{d}_{\mathbb{X}} = O(\|v\|^2), \quad v\to 0 $$