Let $(E_1, \| \cdot \|_{E_1}),(E_2, \|\cdot\|), \cdots, (E_n,\|\cdot\|_{E_n})$ be Banach spaces. Let $\displaystyle E = \mathop{\oplus}_{i=1}^{n} E_i.$ Define a norm $\|\cdot\|_{\infty}$ on $E$ by $$\|\underline {x}\|_{\infty} = \max\limits_{1 \leq i \leq n} \|x_i\|_{E_i},\ \underline {x} = (x_1,x_2,\cdots,x_n) \in E.$$ The pair $(E, \|\cdot\|_{\infty})$ is denoted by $\mathop{\oplus}\limits_{\infty} E_i.$
Proposition (Derivative of a multi-linear map). Let $T \in \mathcal L(E_1,E_2,\cdots,E_n;F)$ (linear space of bounded multi-linear maps from $\mathop{\oplus}\limits_{i=1}^{\infty} E_i$ to $F$) then $T$ viewed as a map from $\mathop{\oplus}\limits_{\infty} E_i$ to $F$ is Frechet differentiable with it's derivative given by $$DT(\underline {x}) (\underline {h}) = \sum\limits_{i=1}^{n} T(x_1,x_2,\cdots,x_{i-1},h_i,x_{i+1},\cdots,x_n)$$ where $\underline {x} = (x_1,x_2,\cdots,x_n)$ and $\underline {h} = (h_1,h_2,\cdots,h_n) \in E.$
Proof $:$ For every subset $J \subseteq \{1,2,\cdots,n\}$ let $v_J \in E$ be defined by $$ (v_J)_i = \left\{ \begin{array}{ll} x_i & \quad \text {if}\ i \in J \\ h_i & \quad \text {if}\ i \not\in J \end{array} \right. $$ Then $$f(\underline {x} + \underline {h}) = \sum\limits_{J \subseteq \{1,2,\cdots,n \}} T(v_J).$$ Therefore using $\displaystyle {\prod\limits_{i=1}^{n} \|(v_J)_i\|_{E_i} \leq {\|\underline {h}\|_{\infty}}^{n - |J|} {\|\underline {x}\|_{\infty}}^{|J|}}$ we get $$\begin{align*} \left \|T(\underline {x} + \underline {h}) - T(\underline {x}) - \sum\limits_{J\ |\ |J| = n-1} T(v_J) \right \|_F & = \left \| \sum\limits_{J\ |\ |J| \leq n-2} T(v_J) \right \|_F \\ & \leq \|T\| \sum\limits_{J\ |\ |J| \leq n-2} \prod\limits_{i=1}^{n} \|(v_J)_i\|_{E_i} \\ & \color{red} {\leq \|T\| {\|\underline {h}\|_{\infty}}^2 {\|\underline {x}\|_{\infty}}^{n-2} \sum\limits_{J\ |\ |J| \leq n-2} 1} \\ & = o (\|\underline {h}\|_{\infty}) \end{align*}$$ This completes the proof.
In the above proof I have understood everything except the inequality in red. Can anybody please help me in this regard?
Thanks for your time.