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The Schur norm of a matrix $A$ is defined to be $\|A\|_S=\max\{\|A\circ X\|: \|X\|\leq 1\}$, where $\|\cdot \|$ is the operator norm of a matrix, i.e., the largest singular value.

Let $a_1,\ldots, a_m, b_1,\ldots, b_n$ be positive reals.Let $A$ be an $m\times n$ matrix defined to be $A_{i,j}=(a_i-b_j)/(a_i+b_j)$.

My question is how to compute $\|A\|_S$. Is it upper bounded by an absolute constant independent of $m, n$?

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  • $\begingroup$ Is it upper bounded by an absolute constant independent of $m,n$? You cannot get that much unless you have some extra restrictions on $a,b$. Let $m=n$ and let $a_j=b_j$ be a fast increasing sequence. Then your multiplier is essentially $-1$ if $i<j$, $0$ if $i=j$ and $1$ if $i>j$, which has norm about $\log n$ $\endgroup$
    – fedja
    Jun 9 '19 at 18:47
  • $\begingroup$ Isn't it just 0 matrix if $a_i=b_j$? $\endgroup$ Jun 9 '19 at 21:17
  • $\begingroup$ Not $a_i=b_j$ but $a_j=b_j$ (not a_i=b_j but a_j=b_j). Sorry for the small font in comments. $\endgroup$
    – fedja
    Jun 9 '19 at 22:54

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