Infinite matrices, integral operators and absolute values. If $k: X \times X \to \mathbb{C}$ (or $\mathbb{R}$) is the kernel of a bounded integral operator $K: L^2 (X, \Omega, \mu) \to L^2 (X, \Omega, \mu)$, is $\lvert k \rvert$ also the kernel of a bounded integral operator?
($\lvert k \rvert (x, y) = \lvert k (x, y) \rvert$)
Depending on one's taste, this can also be formulated in the special case where $X = \mathbb{N}$ and $\mu$ is the counting measure:
If $A_{ij}$ is an infinite matrix that determines a bounded operator on $\ell^2$, does the matrix $\lvert A_{ij} \rvert$ also determine a bounded operator?

Here is a counterexample for the matrix version. Define a sequence of matrices $(H_n)$ inductively by $H_0=\begin{pmatrix}1 \end{pmatrix}$ and $H_{n+1} = \begin{pmatrix}H_n & H_n \\ H_n & -H_n \end{pmatrix}$ for $n\ge 0$. Then $H_n$ is a square matrix of size $2^{n}$ whose columns are mutually orthogonal vectors of length $2^{n/2}$; hence, $\|H_n\| = 2^{n/2}$.  (These are known as Hadamard matrices). Let $A$ be the infinite block-diagonal matrix whose diagonal blocks are orthogonal matrices $2^{-n/2}H_n$. Then $A$ is a unitary operator on $\ell^2$. 
On the other hand, taking the absolute values of the matrix entries we get a block-diagonal matrix with blocks $2^{-n/2}\mathbf 1_{2^n\times 2^n}$ where $\mathbf 1_{2^n\times 2^n}$ is a matrix of all ones. The operator norm of $\mathbf 1_{2^n\times 2^n}$ is $2^n$ (consider its action on the all-ones vector). Thus, the diagonal blocks have operator norms $2^{n/2}\to \infty$, and so the infinite block-diagonal matrix does not define a bounded operator.
