Let $A=(a\_{ij})\_{i,j=1}^{\infty}$ be an infinite matrix of complex numbers. For every positive integer $n$, we shall denote with $A\_n$ the $n \times n$ matrix $A\_n=(a\_{i,j})\_{i,j=1}^{n}$, and if $x \in \mathbb{C}^{n}$, we shall write $||x|| = \sqrt{\sum\_{i=1}^{n} |x\_i|^2}$. In the following, each $x \in \mathbb{C}^{n}$ will be considered as usual a column vector, and the transpose of a matrix $B$ will be denoted by $B^{\top}$. Finally, we shall denote with $|A|$ the infinite matrix $|A|=(|a\_{i,j}|)\_{i,j=1}^{\infty}$. 


Assume $A$ is hermitian, that is $a\_{i,j}=\bar{a}\_{j,i}$ for all $i,j=1,2,\dots$.
We shall say that $A$ satisfies the (BO) condition if there exists $M > 0$ such that for each positive integer $n$, by denoting with $x$ a vector in $\mathbb{C}^{n}$, we have 
\begin{equation}
\sup\_{||x|| \leq 1} \left| x^{\top} A\_n \bar{x} \right| \leq M.
\end{equation}
My questions are the following ones: can we find a hermitian infinite matrix $A$ such that $A$ satisfies the (BO) condition, but $|A|$ does not? Can we choose $A$ to be real symmetric?


Thank you very much in advance for your help.


P.S. I will explain the relevance of this question in the following two remarks.


Remark 1. First of all, let us note that for each fixed positive integer $n$, $A\_n$ is an $n \times n$ hermitian matrix. So by the Spectral Theorem, there exist a unitary $n \times n$ matrix $U$ such that $\bar{U}^{\top} A\_n U = \Lambda$ is a diagonal matrix, having as diagonal elements the eigenvalues $\lambda\_1,\dots, \lambda\_n$ of $A\_n$. Let $L = \max \{ |\lambda\_1|,\dots,|\lambda\_n| \}$. If we put $ \xi = U^{\top} x$, we get
\begin{equation}
\sup\_{||x|| \leq 1} \left| x^{\top} A\_n \bar{x} \right| = \sup\_{||\xi|| \leq 1} \left| \xi^{\top} \Lambda \bar{\xi} \right| = L.
\end{equation}
Let us also note that if $y \in \mathbb{C}^{n}$ and we put $\upsilon = U^{\top} y$, we have
\begin{equation}
\sup\_{||x|| \leq 1, ||y|| \leq 1} \left| x^{\top} A\_n \bar{y} \right| = \sup\_{||\xi|| \leq 1, ||\upsilon|| \leq 1} \left| \xi^{\top} \Lambda \bar{\upsilon} \right| = L.
\end{equation}
From these observations, we derive two conditions equivalent to the (BO) condition. First, $A$ satisfies the (BO) condition if and only if there exists $M > 0$ such that for each positive integer $n$, by denoting with $x, y$ vectors in $\mathbb{C}^{n}$, we have 
\begin{equation}
\sup\_{||x|| \leq 1, ||y|| \leq 1} \left| x^{\top} A\_n \bar{y} \right| \leq M.
\end{equation}
Second, $A$ satisfies the (BO) condition if and only if there exists $M > 0$ such that for each positive integer $n$ and each eigenvalue $\lambda$ of $A\_n$, we have $|\lambda | \leq M$.


Remark 2. Let $C=(c\_{i,j})\_{i,j=1}^{\infty}$ be an infinite matrix of complex numbers (hermitian or not). Moreover let $\mathcal{H}$ be a separable Hilbert space and fix an orthonormal basis $(e\_j)\_{j=1}^{\infty}$ of $\mathcal{H}$. Then a classical result (see Akhiezer and Glazman, Theory of Linear Operators in Hilbert Space, Volume I, Section 26) asserts that $C$ represents a bounded operator defined on all $\mathcal{H}$ with respect to the basis $(e\_j)\_{j=1}^{\infty}$ if and only if there exists $M > 0$ such that for each positive integer $n$, by denoting with $x, y$ vectors in $\mathbb{C}^{n}$, we have 
\begin{equation}
\sup\_{||x|| \leq 1, ||y|| \leq 1} \left| x^{\top} C\_n \bar{y} \right| \leq M.
\end{equation}
Note that if $C$ is hermitian, in view of the previous remark, $C$ represents a bounded operator defined on all $\mathcal{H}$ with respect to the basis $(e\_j)\_{j=1}^{\infty}$ if and only if $C$ satisfies the (BO) condition (from this fact we derived the name "BO", which stands for Bounded Operator).
We have the following simple result.


Theorem
Let $C=(c\_{i,j})\_{i,j=1}^{\infty}$ be an infinite matrix of complex numbers. If $|C|$ represents a bounded operator defined on all $\mathcal{H}$ with respect to the basis $(e\_j)\_{j=1}^{\infty}$, then also $C$ does.


Proof. Let $M > 0$ be such that for each positive integer $n$, by denoting with $x, y$ vectors in $\mathbb{C}^{n}$, we have 
\begin{equation}
\sup\_{||x|| \leq 1, ||y|| \leq 1} \left| x^{\top} |C|\_n \bar{y} \right| \leq M.
\end{equation}
If $x=(x\_1,\dots,x\_n)$ and $y=(y\_1,\dots,y\_n)$, then we have 
\begin{equation}
\left| x^{\top} |C|\_n \bar{y} \right| = \left| \sum\_{i,j=1}^{n} |c\_{i,j}| x\_{i} \bar{y}\_j \right| \leq 
\sum\_{i,j=1}^{n} |c\_{i,j}| |x\_{i}| |\bar{y}\_j|.
\end{equation}
On the other hand, if $x\_1,\dots, x\_n$ and $y\_1,\dots,y\_n$ are real and nonnegative (we write succinctly $x \geq 0$, and $y \geq 0$), we have
\begin{equation}
\left| \sum\_{i,j=1}^{n} |c\_{i,j}| x\_{i} \bar{y}\_j \right| = 
\sum\_{i,j=1}^{n} |c\_{i,j}| |x\_{i}| |\bar{y}\_j|.
\end{equation}
Since we clearly have
\begin{equation}
\sup\_{||x|| \leq 1, ||y|| \leq 1} \sum\_{i,j=1}^{n} |c\_{i,j}| |x\_{i}| |\bar{y}\_j| = \sup\_{\substack{||x|| \leq 1, ||y|| \leq 1 \\ x \geq 0, y \geq 0}} \sum\_{i,j=1}^{n} |c\_{i,j}| |x\_{i}| |\bar{y}\_j|,
\end{equation}
we get
\begin{equation}
\sup\_{||x|| \leq 1, ||y|| \leq 1} \sum\_{i,j=1}^{n} |c\_{i,j}| |x\_{i}| |\bar{y}\_j| = \sup\_{||x|| \leq 1, ||y|| \leq 1} \left| x^{\top} |C |\_n \bar{y} \right| \leq M.
\end{equation}
From this we conclude that
\begin{equation}
\sup\_{||x|| \leq 1, ||y|| \leq 1} \left| x^{\top} C\_n \bar{y} \right| \leq \sup\_{||x|| \leq 1, ||y|| \leq 1} \sum\_{i,j=1}^{n} |c\_{i,j}| |x\_{i}| |\bar{y}\_j| \leq M.
\end{equation}
QED


I think that the converse of this result does not hold, even if we make the additional assumption that $C$ is hermitian or real symmetric. Anyway any effort I made to find a counterexample has failed up to now.



Matrices Representing Bounded Operators and Absolute Values
By following Yemon Choi's hint, I give here a complete answer to my question.


We denote by $\mathbb{Z}$ the set of all integer numbers.
Let $f:[-\pi, \pi) \rightarrow \mathbb{R}$ be defined by $f(x)=x$. The Fourier coefficients of $f$ are:
\begin{equation}
c\_{0}=\frac{1}{2\pi} \int\_{-\pi}^{\pi} x dx = 0,
\end{equation}
\begin{equation}
c\_{n}= \frac{1}{2 \pi} \int\_{-\pi}^{\pi} x e^{-inx} dx = - \frac{xe^{-inx}}{2 \pi i n} \bigg|\_{-\pi}^{\pi} + \int\_{-\pi}^{\pi} \frac{e^{-inx}}{2 \pi i n} dx = \frac{i (-1)^{|n|}}{n}, \qquad (n= \pm 1, \pm 2, \dots).
\end{equation}
So $(c\_n)\_{n=-\infty}^{\infty}$ are the Fourier coefficients of a real bounded function. We shall prove that $(|c\_n|)\_{n=-\infty}^{\infty}$ are not the Fourier coefficients of an essentially bounded function. The partial sums of the Fourier series corresponding to the coefficients $(|c\_n|)\_{n=-\infty}^{\infty}$ are
\begin{equation}
s\_N(x) = \sum\_{n=-N}^{N} |c\_n| e^{inx} = \sum\_{n=1}^{N} \frac{e^{inx} }{n} + \sum\_{n=1}^{N} \frac{e^{-inx} }{n} \quad (x \in [-\pi,\pi) \quad \textrm{and} \quad N=1,2,\dots).
\end{equation}
Now note that for every complex number $z$, with $|z| < 1$, we have the power expansion
\begin{equation}
\sum\_{n=1}^{\infty} \frac{z^n}{n} = - \sum\_{n=1}^{\infty} (-1)^{n+1} \frac{(-z)^n}{n} = -\log(1-z).
\end{equation}
Moreover by [R], Theorem (3.44) the series
\begin{equation}
\sum\_{n=1}^{\infty} \frac{z^n}{n}
\end{equation}
converges for each complex number $z$ such that $|z|=1$, with $z \neq 1$. Fix $z$ such that $|z|=1$, with $z \neq 1$. Then for any real $t \in (-1,1)$, an application of Abel's Theorem to the series
\begin{equation}
\sum\_{n=1}^{\infty} \frac{z^n}{n} t^n =-\log(1-tz),
\end{equation}
shows that
\begin{equation}
\sum\_{n=1}^{\infty} \frac{z^n}{n} =-\log(1-z).
\end{equation}
By replacing $z$ by $\bar{z}$ we also get that for any complex number $z$ such that $|z| \leq 1$, with $z \neq 1$, we have
\begin{equation}
\sum\_{n=1}^{\infty} \frac{\bar{z}^n}{n} =-\log(1-\bar{z}).
\end{equation}
In particular we get that for any $x \in [-\pi,\pi) \backslash \{0\}$ we have 
\begin{equation}
\lim\_{N \rightarrow \infty} s\_N(x) =-2 \mathfrak{R}(\log(1-e^{ix})).
\end{equation}


Now let $g$ be a function in the equivalent class $[g] \in L^{2}([-\pi,\pi))$ which has $(|c\_n|)\_{n=-\infty}^{\infty}$ has Fourier coefficients (note that for sure $(|c\_n|)\_{n=-\infty}^{\infty} \in \ell^{2}(\mathbb{Z})$ as we can immediately check and as it must be since the $(c\_n)\_{n=-\infty}^{\infty}$ are the Fourier coefficients of $[f] \in L^{2}([-\pi,\pi))$). Since $(s\_{N})\_{N=1}^{\infty}$ converges to $g$ in $L^{2}([-\pi,\pi))$, there is a subsequence $(s\_{N\_k})\_{k=1}^{\infty}$ which converges pointwise a.e. to $g$. We conclude that
\begin{equation}
g(x)=-2 \mathfrak{R}(\log(1-e^{ix})) \quad \textrm{a.e. on}\quad [-\pi,\pi),
\end{equation}
so that $g$ is not essentially bounded.


Now take $A=(a\_{i,j})\_{i,j=0}^{\infty}$, with $a\_{i,j}=c\_{i-j}$ for all $i,j=0,1,2,\dots$. The result we have proved combined with Toeplitz's Theorem (see e.g. [BG] , Theorem (1.1)) shows that $A$ is a hermitian matrix with the required properties. 


To get an example with $A$ real symmetric, let us consider the following even real-valued function $f:[-\pi, \pi) \rightarrow \mathbb{R}$ defined by
\begin{equation}
f(x) = \begin{cases}
1 \quad \textrm{if} & x \in \left[ - \frac{\pi}{2}, \frac{\pi}{2} \right), \\
-1 \quad \textrm{otherwise}.
\end{cases}
\end{equation}
The Fourier coefficients of $f$ are:
\begin{equation}
c\_n = \begin{cases}
0 \quad \textrm{if} & n=2k \quad (k \in \mathbb{Z}), \\
4 \frac{(-1)^{|k|}}{n} \quad \textrm{if} & n=2k+1 \quad (k \in \mathbb{Z}).
\end{cases}
\end{equation}
We shall now prove that $(|c\_n|)\_{n=-\infty}^{\infty}$ are not the Fourier coefficients of an essentially bounded function. From we have proved above, we get that for any complex number $z$ such that $|z| \leq 1$, with $z \neq 1$, we have
\begin{equation}
\sum\_{k=0}^{\infty} \frac{z^{2k+1}}{2k+1} = \frac{1}{2} \left[ \log(1+z) - \log(1-z) \right].
\end{equation}
By replacing $z$ by $\bar{z}$, we also have that for any complex number $z$ such that $|z| \leq 1$, with $z \neq 1$
\begin{equation}
\sum\_{k=0}^{\infty} \frac{\bar{z}^{2k+1}}{2k+1} = \frac{1}{2} \left[ \log(1+\bar{z}) - \log(1-\bar{z}) \right].
\end{equation}
So if we consider the partial sums of the Fourier series corresponding the coefficients $(|c\_n|)\_{n=-\infty}^{\infty}$:
\begin{equation}
s\_N(x) = \sum\_{n=-N}^{N} |c\_n| e^{inx} \quad (x \in [-\pi,\pi) \quad \textrm{and} \quad N=1,2,\dots),
\end{equation}
we get that for any $x \in [-\pi,\pi) \backslash \{0\}$ we have 
\begin{multline}
\lim\_{N \rightarrow \infty} s\_N(x) = 2 \left[ \log(1+e^{ix}) - \log(1 - e^{ix}) \right] + 2 \left[ \log(1+e^{-ix}) - \log(1 - e^{-ix}) \right] = \\ = 2 \left[ \mathfrak{R}(\log(1+e^{ix})) - \mathfrak{R}(\log(1 - e^{ix})) \right].
\end{multline}
Since the function
\begin{equation}
g(x) = 2 \left[ \mathfrak{R}(\log(1+e^{ix})) - \mathfrak{R}(\log(1 - e^{ix})) \right] \quad (x \in [-\pi,\pi))
\end{equation}
is not essentially bounded, by proceeding as in the previous example we conlcude that $(|c\_n|)\_{n=-\infty}^{\infty}$ are not the Fourier coefficients of an essentially bounded function. Finally, take $A=(a\_{i,j})\_{i,j=0}^{\infty}$, with $a\_{i,j}=c\_{i-j}$ for all $i,j=0,1,2,\dots$. The result we have proved combined with Toeplitz's Theorem (see e.g. [BG] , Theorem (1.1)) shows that $A$ is a real symmetric matrix with the required properties.


References


[R] Rudin, Principles of Mathematical Analysis, Third Edition


[BG] Böttcher, Albrecht; Grudsky, Sergei M. (2012), Toeplitz Matrices, Asymptotic Linear Algebra, and Functional Analysis


