# $|\underset{i,j}{\sum}a_{ij}x_iy_j| \leq \underset{u,v \in \{-1,1\}^n }{sup} |\underset{i,j}{\sum} a_{ij}u_iv_j|$?

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 $$1\leq i \leq m, 1\leq j \leq n$$

## 1 Answer

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.