Good evening, I'm doing an exercise about the derivative of multilinear maps:
Let $E_1, \ldots, E_m$ be normed vector space and $F$ a Banach space. Prove that $\mathcal{L}\left(E_{1}, \ldots, E_{m} ; F\right)$ is a vector subspace of $\mathcal C^{1}\left(E_{1} \times \cdots \times E_{m}, F\right)$. Moreover, for $\varphi \in \mathcal{L}\left(E_{1}, \ldots, E_{m} ; F\right)$ and $\left(x_{1}, \ldots, x_{m}\right) \in E_{1} \times \cdots \times E_{m}$, we have $$\partial \varphi\left(x_{1}, \ldots, x_{m}\right)\left(h_{1}, \ldots, h_{m}\right)=\sum_{j=1}^{m} \varphi\left(x_{1}, \ldots, x_{j-1}, h_{j}, x_{j+1}, \ldots, x_{m}\right)$$
Could you please verify if my proof looks fine or contains logical gaps/errors? Thank you so much for your help!
My attempt:
Because $E_{1} \times \cdots \times E_{m}$ is finite dimensional, all norms on this space are equivalent. As such, we could endow $E_{1} \times \cdots \times E_{m}$ with the norm $$\|x\| = \sqrt{\sum_{i=1}^m x_i^2} \quad, \quad x=\left(x_{1}, \ldots, x_{m}\right) \in E_{1} \times \cdots \times E_{m}$$
We define the map $$\begin {array}{l|rcl} A_{x} & E_{1} \times \cdots \times E_{m} & \longrightarrow & F \\ & \left(h_{1}, \ldots, h_{m}\right) & \longmapsto & \sum_{j=1}^{m} \varphi\left(x_{1}, \ldots, x_{j-1}, h_{j}, x_{j+1}, \ldots, x_{m}\right) \end{array}$$
For $h,h' \in E_{1} \times \cdots \times E_{m}$ and $\lambda \in \mathbb K$, we have
$$\begin{aligned} A_x(h+\lambda h') &= \sum_{j=1}^{m} \varphi\left(x_{1}, \ldots, x_{j-1}, h_{j} +\lambda h_j', x_{j+1}, \ldots, x_{m}\right)\\ &= \sum_{j=1}^{m} \varphi\left(x_{1}, \ldots, x_{j-1}, h_{j} , x_{j+1}, \ldots, x_{m}\right)\\ & \quad + \lambda \sum_{j=1}^{m} \varphi\left(x_{1}, \ldots, x_{j-1}, h_j', x_{j+1}, \ldots, x_{m}\right)\\ &= A_x(h) + \lambda A_x(h') \end{aligned}$$
Hence $A_x \in L(E_{1} \times \cdots \times E_{m}, F)$. Moreover, $A_x$ is the sum of continuous maps, so it is a continuous one. As such, $A_x \in \mathcal L(E_{1} \times \cdots \times E_{m}, F)$. Let $y=x+h$. It follows from the multilinearity of $\varphi$ that $$\varphi\left(y_1, \ldots, y_m\right) = \varphi\left(x_{1}, \ldots, x_{m}\right)+\sum_{k=1}^{m} \varphi\left(x_{1}, \ldots, x_{k-1}, h_{k}, y_{k+1}, \ldots, y_{m}\right)$$
and \begin{align} \begin{split}\label{eq:1} &\varphi\left(x_{1}, \ldots, x_{k-1}, h_{k}, y_{k+1}, \ldots, y_{m}\right) =\varphi\left(x_{1}, \ldots, x_{k-1}, h_{k}, x_{k+1}, \ldots, x_{m}\right) \\ &+\sum_{j=k+1}^{m} \varphi\left(x_{1}, \ldots, x_{k-1}, h_{k}, x_{k+1}, \ldots, x_{j-1}, h_{j}, y_{j+1}, \ldots, y_{m}\right), \quad k=\overline{1,m} \end{split} \end{align}
Hence \begin{align} \begin{split}\label{eq:3} \varphi\left(y_{1}, \ldots, y_m\right) ={}& \varphi\left(x_{1}, \ldots, x_{m}\right) + \sum_{k=1}^{m} \Big [ \varphi\left(x_{1}, \ldots, x_{k-1}, h_{k}, x_{k+1}, \ldots, x_{m}\right)\\ & + \sum_{j=k+1}^{m} \varphi\left(x_{1}, \ldots, x_{k-1}, h_{k}, x_{k+1}, \ldots, x_{j-1}, h_{j}, y_{j+1}, \ldots, y_{m}\right) \Big] \end{split}\\ \begin{split}\label{eq:7} ={}& \varphi\left(x_{1}, \ldots, x_{m}\right) + \sum_{k=1}^{m} \varphi\left(x_{1}, \ldots, x_{k-1}, h_{k}, x_{k+1}, \ldots, x_{m}\right)\\ &+ \sum_{k=1}^{m} \sum_{j=k+1}^{m} \varphi\left(x_{1}, \ldots, x_{k-1}, h_{k}, x_{k+1}, \ldots, x_{j-1}, h_{j}, y_{j+1}, \ldots, y_{m}\right) \end{split}\\ \begin{split}\label{eq:2} ={}& \varphi\left(x_{1}, \ldots, x_{m}\right) + A_x (h)\\ &+ \sum_{k=1}^{m} \sum_{j=k+1}^{m} \varphi\left(x_{1}, \ldots, x_{k-1}, h_{k}, x_{k+1}, \ldots, x_{j-1}, h_{j}, y_{j+1}, \ldots, y_{m}\right) \end{split} \end{align}
Moreover, $$\begin{aligned} & \left \| \sum_{k=1}^{m} \sum_{j=k+1}^{m} \varphi \left(x_{1}, \ldots, x_{k-1}, h_{k}, x_{k+1}, \ldots, x_{j-1}, h_{j}, y_{j+1}, \ldots, y_{m}\right) \right \| \\ \le & \sum_{k=1}^{m} \sum_{j=k+1}^{m} \left \| \varphi \left(x_{1}, \ldots, x_{k-1}, h_{k}, x_{k+1}, \ldots, x_{j-1}, h_{j}, y_{j+1}, \ldots, y_{m}\right) \right\| \\ \le & \sum_{k=1}^{m} \sum_{j=k+1}^{m} \|\varphi \| \cdot \|x_{1}\| \cdots \|x_{k-1}\| \cdot \| h_{k}\| \cdot \|x_{k+1}\| \cdots \|x_{j-1}\| \cdot \|h_{j}\| \cdot \|y_{j+1}\| \cdots \| y_{m}\| \\ \le & \sum_{k=1}^{m} \sum_{j=k+1}^{m} \|\varphi \| \cdot \|x\| \cdot \|y\| \cdot \|h_k\| \cdot \|h_{j}\| \\ \le & \sum_{k=1}^{m} \sum_{j=k+1}^{m} \|\varphi \| \cdot \|x\| \cdot \|y\| \cdot \|h\|^2 \\ =& \|h\|^2\sum_{k=1}^{m} \sum_{j=k+1}^{m} \|\varphi \| \cdot \|x\| \cdot \|y\| \end{aligned}$$
Finally, $$\begin{aligned} &\lim_{h \to 0} \frac{\|\varphi (x+h) -\varphi(h) - A_x(h)\|}{\|h\|}\\ = &\lim_{h \to 0} \frac{\left \| \sum_{k=1}^{m} \sum_{j=k+1}^{m} \varphi\left(x_{1}, \ldots, x_{k-1}, h_{k}, x_{k+1}, \ldots, x_{j-1}, h_{j}, y_{j+1}, \ldots, y_{m}\right) \right \|}{\|h\|} \\ \le&{}\lim_{h \to 0} \frac{\|h\|^2\sum_{k=1}^{m} \sum_{j=k+1}^{m} \|\varphi \| \cdot \|x\| \cdot \|y\|}{\|h\|} \\ =& \lim_{h \to 0} \|h\|\sum_{k=1}^{m} \sum_{j=k+1}^{m} \|\varphi \| \cdot \|x\| \cdot \|y\| = 0\end{aligned}$$
As such, $\partial \varphi(x) = A_x$. Next we show that $$\partial \varphi: E_{1} \times \cdots \times E_{m} \to \mathcal L(E_{1} \times \cdots \times E_{m}, F), \quad x \mapsto A_{x}$$ is continuous. We have $$\begin{aligned} \|\partial \varphi(x) - \partial \varphi(y)\| &= \|A_x - A_y\| \\ & = \sup_{h \in E_{1} \times \cdots \times E_{m} \atop \|h\|=1} \|(A_x-A_y)(h)\| \\ &= \sup_{h \in E_{1} \times \cdots \times E_{m} \atop \|h\|=1} \|A_x(h) -A_y(h)\| \\&=\sup_{h \in E_{1} \times \cdots \times E_{m} \atop \|h\|=1} \Bigg \| \sum_{j=1}^{m} \varphi\left(x_{1}, \ldots, x_{j-1}, h_{j}, x_{j+1}, \ldots, x_{m}\right) \\ & \quad \quad \quad \quad \quad \quad \quad - \sum_{j=1}^{m} \varphi\left(y_{1}, \ldots, y_{j-1}, h_{j}, y_{j+1}, \ldots, y_{m}\right) \Bigg \| \\ &\le \sup_{h \in E_{1} \times \cdots \times E_{m} \atop \|h\|=1} \sum_{j=1}^{m} \| \varphi\left(x_{1}, \ldots, x_{j-1}, h_{j}, x_{j+1}, \ldots, x_{m}\right) \\ & \quad \quad \quad \quad \quad \quad \quad - \varphi\left(y_{1}, \ldots, y_{j-1}, h_{j}, y_{j+1}, \ldots, y_{m}\right) \| \end{aligned}$$
It follows from the continuity of $\varphi$ that $$\| \varphi\left(x_{1}, \ldots, x_{j-1}, h_{j}, x_{j+1}, \ldots, x_{m}\right) - \varphi\left(y_{1}, \ldots, y_{j-1}, h_{j}, y_{j+1}, \ldots, y_{m}\right) \| \to 0$$ as $x \to y$
Hence $\partial \varphi(x) - \partial \varphi(y) \to 0$ as $x \to y$. This completes the proof.