Why are multilinear maps continuous? Let $X_1,...,X_N$ be a finite sequence of finite-dimensional normed real vector spaces. Let $Y$ be another normed real vector space, and let $A:\prod_{i=1}^N X_i \to Y$ be multilinear.
How to prove that $A$ is continuous? That is how to show there exist $C>0$ such that
$$\|A(x_1 , \ldots , \ldots x_N) \|_Y \le C  \prod_{k=1}^{N}\|x_i\|_{X_i}$$ 
 A: We try to follow the proof of the fact for $N=1$, that is the fact that a linear map $A \colon X \to Y$ from a finite-dimensional normed space is always bounded, that is, we use bases: 
Let $A \colon \prod_{i=1}^N X_i \to Y$ be multilinear, let $(e_{i,j})_{j=1}^{\dim X_i}$ be a basis of $X_i$, with $\def\norm#1{\left\|#1\right\|} \norm{e_{i,j}}_{X_i} = 1$, all $i$, $j$. We have for any $x_i = \sum_{j=1}^{\dim X_i} x_{i,j}e_{i,j} \in X_i$, that:
\begin{align*} 
  \norm{A(x_1, \ldots, x_N)}_Y &= \norm{A\left(\sum_{j_1=1}^{\dim X_1} x_{1,j_1}e_{1,j_1}, \ldots, \sum_{j_N=1}^{\dim X_N} x_{N,j_N}e_{N,j_N}\right)}_Y\\
    &= \norm{\sum_{j_1=1}^{\dim X_1} x_{1,j_1}\cdots \sum_{j_N=1}^{\dim X_N} x_{N,j_N}A(e_{i,j_1}, \ldots, e_{N,j_N})}_Y\\
    &\le \sum_{j_1=1}^{\dim X_1} \def\abs#1{\left|#1\right|}\abs{x_{1,j_1}}\cdots \sum_{j_N=1}^{\dim X_N} \abs{x_{N,j_N}} \max_{j_1, \ldots, j_N}\norm{A(e_{1,j_1}, \ldots, e_{N,j_N})}_Y\\
    &= \norm{(x_{1,j_1})_{j_1}}_{\ell^1(\dim X_1)} \cdots \norm{(x_{N,j_N})_{j_N}}_{\ell^1(\dim X_N)} \cdot \max_{j_1, \ldots, j_N}\norm{A(e_{1,j_1}, \ldots, e_{N,j_N})}_Y\\
\end{align*}
Now, as $X_i$ is finite-dimensional, all norms on $X_i$ are equivalent, hence there is a $C_i$ such that 
$$ \norm{(x_{i,j_i})}_{\ell^1(\dim X_i)} \le C_i\norm{x_i}_{X_i}, \qquad \forall x_i = \sum_{j_i=1}^{\dim X_i} x_{i,j_i} e_{i,j_i}  $$
This gives 
$$ \norm{A(x_1, \ldots, x_N)}_Y \le  \underbrace{\max_{j_1, \ldots, j_N}\norm{A(e_{1,j_1}, \ldots, e_{N,j_N})}_Y \cdot \prod_{i=1}^N C_i}_{L:=}  \cdot \prod_{i=1}^N\norm{x_i}_{X_i} $$
