let $(a)_{ij}$ be a $M\times N$ Matrix with real entries ,is that possible to prove that:
for any $x \in [-1,1]^n, y \in [-1,1]^m$ we have:
$$|\underset{i,j}{\sum}a_{ij}x_iy_j| \leq \underset{u,v \in \{-1,1\}^n }{sup} |\underset{i,j}{\sum} a_{ij}u_iv_j|$$
where $1\leq i \leq m, 1\leq j \leq n$
Let $f:[-1,1]^M\times[-1,1]^N\to\mathbb R$ be given by $f(x,y)=|\sum a_{ij}x_iy_j|$. By compactness and continuity, $f$ attains its supremum $S=f(r,s)$ at at least one point $(r,s)\in[-1,1]^M\times[-1,1]^N$. Now consider the function $g:[-1,1]^M\to\mathbb R$ given by $g(x)=f(x,s)$. It is convex, and hence attains its supremum at an extreme point of $[-1,1]^M$, that is, at a vector $u$ all of whose coordinates are $\pm1$. (Clearly $g(u)=S$.) Now look at $h(y)=f(u,y)$. It is also convex, so it attains its supremum at an extreme point of $[-1,1]^N$, call it $v$, all of whose coordinates are $\pm1$. Obviously $h(v)=S$. But $h(v)=f(u,v)$ so $f(u,v)=S$ as well.