Why are multilinear maps continuous?

Question

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

$$\|A(x_1 , \ldots , \ldots x_N) \|_Y \le C \prod_{k=1}^{N}\|x_i\|_{X_i}$$

Answer 1

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 $$ \norm{A(x_1, \ldots, x_N)}_Y \le \underbrace{\max_{j_1, \ldots, j_N}\norm{A(e_{1,j_1}, \ldots, e_{N,j_N})}_Y \cdot \prod_{i=1}^N C_i}_{L:=} \cdot \prod_{i=1}^N\norm{x_i}_{X_i} $$

Answer 2

I'll first demonstrate how to show continuity for $N=2$. To this end, note that \begin{align} A(x_1,y_1)-A(x_2,y_2)&=A(x_1,y_1)-A(x_1,y_2)+A(x_1,y_2)-A(x_2,y_2)\\ &= A(x_1,y_1-y_2)-A(x_1-x_2,y_2). \end{align} So we see that if we choose $x_1-x_2$ and $y_1-y_2$ small, then we can invoke the result that was already proven, namely that $\exists C>0$ such that $$ \Vert A(x,y)\Vert\leq C\Vert x\Vert\cdot\Vert y\Vert. $$ For $N=3$ we would have \begin{align} A(x_1,y_1,z_1)-A(x_2,y_2,z_2)&=A(x_1-x_2,y_1,z_1)+A(x_2,y_1,z_1)-A(x_2,y_2,z_2)\\ &= A(x_1-x_2,y_1,z_1)+A(x_2,y_1-y_2,z_1)+A(x_2,y_2,z_1)-A(x_2,y_2,z_2)\\ &= A(x_1-x_2,y_1,z_1)+A(x_2,y_1-y_2,z_1)+A(x_2,y_2,z_1-z_2). \end{align} We can now guess (if we didn't already) how the expression looks like for arbitrary $N$: $$ A(x_1,\dots,x_N)-A(x_1',\dots,x_N')=\sum_{i=1}^N A(x_1',\dots,x_{i-1}',x_i-x_i',x_{i+1},\dots,x_N). $$ This can be shown induction. So assume that for $N$ it's true. We have \begin{multline} A(x_1,\dots,x_{N+1})-A(x_1',\dots,x_{N+1}')=A(x_1,\dots,x_{N+1})-A(x_1',\dots,x_N',x_{N+1})+A(x_1',\dots,x_N',x_{N+1})-A(x_1',\dots,x_{N+1}'). \end{multline} Note that the map $A(-,\dots,-,x_{N+1})$ ($x_{N+1}$ is fixed) is an $N$-linear map, so we may apply our induction hypothesis: $$ A(x_1,\dots,x_{N+1})-A(x_1',\dots,x_N',x_{N+1})=\sum_{i=1}^N A(x_1',\dots,x_{i-1}',x_i-x_i',\dots,x_{N+1}). $$ Combining this with $$ A(x_1',\dots,x_N',x_{N+1})-A(x_1',\dots,x_{N+1}')=A(x_1',\dots,x_n',x_{N+1}-x_{N+1}') $$ yields the result.