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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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    $\begingroup$ I have not had time to go through all of your question in detail, but my impression is that you are asking the following question: "suppose that a doubly-infinite matrix $A=(A_{ij})$ represents a bounded hermitian operator on $\ell^2$; is it trute that the matrix $(|A_{i,j}| )$ also represents a bounded operator on $\ell^2$?" $\endgroup$
    – Yemon Choi
    Commented Oct 2, 2016 at 14:07
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    $\begingroup$ If that is your question, then IIRC the answer is no, and I think there should be counterexamples which are Toeplitz matrices. Boundedness of a Toeplitz matrix corresponds to its symbol being a bounded function on the unit circle; if the answer to your question were positive, then for every real-valued bounded function on the circle, replacing each Fourier coefficient by its absolute value would also give a bounded function. But this is known to fail by old examples/calculations from the theory of Fourier series $\endgroup$
    – Yemon Choi
    Commented Oct 2, 2016 at 14:10
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    $\begingroup$ Dear Yemon Choi, your interpretation of the question is absolutely correct. I don't know Toeplitz matrices, but I will study them and I will try to elaborate your hint. Thank you very very much for your suggestion! $\endgroup$ Commented Oct 2, 2016 at 14:26
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    $\begingroup$ @Yemon Choi: I have now understood the wonderful relation you quoted among boundedness of a Toeplitz matrix and essential boundedness of its symbol: this classical result of Toeplitz (1911) is stated in Toeplitz Matrix and proved e.g. in Böttcher, Albrecht; Grudsky, Sergei M. (2012), Toeplitz Matrices, Asymptotic Linear Algebra, and Functional Analysis. Anyway, I could not find the counterexamples you quoted about the Fourier series of bounded real functions. Could you give me some reference, please? Thank you very much for your invaluable help! $\endgroup$ Commented Oct 2, 2016 at 16:34
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    $\begingroup$ I think the following example works. The function $f(z)=\log(1/(1-z)) - \log(1/(1-\bar{z}))$ is essentially bounded on the circle since it is $4\pi i$ of the argument of $1/(1-z)$. If you take absolute values of all the Fourier coefficients, you get $g(z)=\log(1/(1-z))+\log(1/(1-\bar{z}))$, which is not bounded near $z=1$. $\endgroup$
    – T. Le
    Commented Oct 2, 2016 at 17:44

2 Answers 2

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As requested, I've moved my comments to an answer.

The question is equivalent to the following:

Suppose that a doubly-infinite matrix $A=(A_{ij})_{i,j\in {\bf Z}}$ represents a bounded hermitian operator on $\ell^2$. Does the matrix $(|A_{ij}|)$ also represent a bounded operator on $\ell^2$ ?

(The original question just considers matrices indexed by ${\bf N}\times {\bf N}$ but the version I've stated is equivalent.)

To see that the answer is negative, we consider doubly-infinite Toeplitz matrices: that is, there is a sequence $(a_n)\subset {\bf C}$ such that $A_{ij}=a_{i-j}$ for all $i,j\in {\bf Z}$. (To get a Hermitian operator we need $a_{-n}=\overline{a_n}$ for all $n\in\bf Z$.)

Classical operator theory tells us that $A$ is bounded if and only if $(a_n)$ is the Fourier series of some $f\in L^\infty({\bf T})$, which is necessarily real-valued if $A$ is Hermitian. It is known that we can find such $f$ with the following property: the Fourier series $$ g(t) \sim \sum_{n\in{\bf Z}} \vert a_n \vert e^{int} $$ does not represent an essentially bounded function. Details should be in Katznelson's book, for instance, and no doubt also in Zygmund somewhere.

(An original version of this answer tried to give an explict example, but contained several stupid errors. See Trieu Le's comment to the main question for an explicit example.)

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  • $\begingroup$ Dear Yemon Choi, I am revising your example and developing it so to give a detailed and complete answer to the original question. I will publish it in the next days as a separate answer. Thank you very very much again! $\endgroup$ Commented Oct 3, 2016 at 20:52
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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

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