1
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.