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?
 A: 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.
A: Consider $X=[0,1]$ with Lebesgue measure. Define the sequence of intervals $I_n=(\frac{1}{n+1},\frac{1}{n})$, $n\geq 1$.
The sequence of functions $$\chi_n=\sqrt{n(n+1)} \; \; {\bf 1}_{I_n}, \; n\geq 1$$ is orthonormal in $L^2([0,1])$. Similarly for $u_n(x) = \sqrt{2} \sin (2\pi nx)$, $n\geq 1$. The kernel:
 $$ k(x,y) = \sum_{n\geq 1} u_n(x) \chi_n(y) $$
then defines an operator $K$ of norm 1 on $L^2$. On the other hand
  $$ |k|(x,y) = \sum_{n\geq 1} \sqrt{2}\; |\sin(2\pi n x)|  \; \chi_n(y) $$
defines an unbounded operator on $L^2$. "Acting" on the constant function 1 we obtain:
  $$ \phi(x) =  \sum_{n\geq 1} \sqrt{\frac{2}{n(n+1)}}\; |\sin(2\pi n x)| $$
which is not in $L^2$ (use e.g. that $\int_0^1 |u_n(x)u_m(x)| dx \geq c>0$ for some $c$ and uniformly in $n,m$).
