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Let's define for every pair of vectors $u,v\in\mathbb{R}^n$, a quantity as follows: $$f(u,v) = \sum_{1\leq i,j\leq n}|u_iu_j-v_iv_j|.$$

I want to find: $$M(n)= \max \{f(u,v): u,v\in \mathbb{R}^n, |u|=|v|=1, u\perp v\}.$$

An easy estimate using the triangle inequality gives $M(n) <2n$, but it seems there should be better upper bounds. (The vectors $u = \frac{1}{\sqrt{n}}(1,1,\dots,1)$ and $v = \frac{1}{\sqrt{2}}(1,-1,0,0,\dots,0)$ gives also $n<M(n)$ for $n>6$.)

Let's define for every pair of vectors $u,v\in\mathbb{R}^n$, a quantity as follows: $$f(u,v) = \sum_{1\leq i,j\leq n}|u_iu_j-v_iv_j|.$$

I want to find: $$M(n)= \max \{f(u,v): u,v\in \mathbb{R}^n, |u|=|v|=1, u\perp v\}.$$

An easy estimate using the triangle inequality gives $M(n) <2n$, but it seems there should be better upper bounds.

Remark. 1) Let $a = \cos(\pi/8), b = \sin(\pi/8)$, then if for even numbers $n$ we define the vectors $u = \sqrt{\frac{2}{n}}(a,b,a,b,\dots)$ and $ v = \sqrt{\frac{2}{n}}(b,-a,b,-a,\dots)$, we have $f(u,v)=\sqrt{2}n\leq M(n)$. 2)A similar construction shows that for every positive integers $n,k$, $\frac{M(n)}{n}\leq \frac{M(nk)}{nk}.$

Let's define for every pair of vectors $u,v\in\mathbb{R}^n$, a quantity as follows: $$f(u,v) = \sum_{1\leq i,j\leq n}|u_iu_j-v_iv_j|.$$

I want to find: $$M(n)= \max \{f(u,v): u,v\in \mathbb{R}^n, |u|=|v|=1, u\perp v\}.$$

An easy estimate using the triangle inequality gives $M(n) <2n$, but it seems there should be better upper bounds. (The vectors $u = \frac{1}{\sqrt{n}}(1,1,\dots,1)$ and $v = \frac{1}{\sqrt{n}}(1,-1,0,0,\dots,0)$ gives also $n<M(n)$ for $n>6$.)

Let's define for every pair of vectors $u,v\in\mathbb{R}^n$, a quantity as follows: $$f(u,v) = \sum_{1\leq i,j\leq n}|u_iu_j-v_iv_j|.$$

I want to find: $$M(n)= \max \{f(u,v): u,v\in \mathbb{R}^n, |u|=|v|=1, u\perp v\}.$$

An easy estimate using the triangle inequality gives $M(n) <2n$, but it seems there should be better upper bounds. (The vectors $u = \frac{1}{\sqrt{n}}(1,1,\dots,1)$ and $v = \frac{1}{\sqrt{2}}(1,-1,0,0,\dots,0)$ gives also $n<M(n)$ for $n>6$.)