Matrix representation of operators on $\ell^2$. Let $A=(a_{nk})_{n,k=1}^{\infty}$ be an infinite matrix that defines a bounded operator on $\ell^2(\mathbb{N})$ by matrix multiplication. Suppose that $B=(b_{nk})_{n,k=1}^{\infty}$ satisfies $|b_{nk}|\leq |a_{nk}|$ for all $n,k\in \mathbb{N}$.
Question: Does $B$ act boundedly on $\ell^2(\mathbb{N})$ by matrix multiplication?
Remark: It's easy to construct examples where $\|B\|_{\ell^2\to\ell^2}>\|A\|_{\ell^2\to\ell^2}$ but I haven't been able to construct a pair of matrices such that the operator induced by $B$ is unbounded.
 A: For fixed $n$ let $A_n$ be the matrix with entries $$a_{kl}=e^{2\pi ikl/n},\qquad 1\le k,l\le n$$  and $a_{kl}=0$ otherwise. Then $A^*A=nP_n,$ where $P_n$ denotes the projection onto the first $n$ coordinates in the space $\ell^2.$
Therefore $\|A\|=\sqrt{n}.$ On the other hand the matrix $B_n$ with entries $b_{kl}=|a_{kl}|$ satisfies $B_n^2=nB_n.$ Therefore $\|B_n\|=n.$  By applying direct sum with coefficients $n^{-1/2}$ we get a desired example.
It is possible to construct a similar example, but with real entries. Consider the finite group $G_n=\mathbb{Z}_2^{*n}, $ where $\mathbb{Z}_2=\{1,-1\}$ is the two element group with multiplication. The dual group $\Gamma_n$ is isomorphic to $G_n.$ By the classical result the elements $\gamma\in \Gamma_n$ (i.e. the characters of $G_n$) are orthogonal:
$$\sum_{g\in G_n} \gamma(g)\gamma'(g)=2^n\delta_{\gamma,\gamma'}$$ Let $A_n$ be the $2^n\times 2^n$ matrix defined by $$a_{ij}=\gamma_j(g_i),$$ where $g_i$ and $\gamma_j$ are arbitrary enumerations of elements in $G_n$ and $\Gamma_n,$ respectively. Then by the orthogonality we get $$A_n^tA_n=2^nI,$$ hence $\|A_n\|=2^{n/2}$ On the other hand the matrix $B_n$ of the same size with all entries equal $1=|a_{ij}|$ satisfies $B_n^2=2^nB_n,$ thus $\|B_n\|=2^n.$
Remark If you do not like the group theoretical background you can treat $\gamma_j,$ for $1\le j\,\le 2^n,$ as the finite Rademacher system.
