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.

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