Sequences of bounds for the spectral radius of a positive operator

TL;DR

This paper extends sequences of bounds for the spectral radius from nonnegative matrices to positive operators on L^2-spaces, providing a broader theoretical framework for spectral analysis.

Contribution

It generalizes existing bounds for the spectral radius from matrices to positive operators on L^2-spaces, enhancing the theoretical understanding.

Findings

Sequences of bounds are extended to positive operators on L^2-spaces.

Theoretical framework for spectral radius bounds is broadened.

Results unify matrix and operator spectral analysis.

Abstract

In 1992, Szyld provided a sequence of lower bounds for the spectral radius of a nonnegative matrix $A$, based on the geometric symmetrization of powers of $A$. In 1998, Ta\c{s}\c{c}i and Kirkland proved a companion result by giving a sequence of upper bounds for the spectral radius of $A$, based on the arithmetic symmetrization of powers of $A$. In this note, we extend both results to positive operators on $L^2$-spaces.

Equations

$$A=\left(\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ 1& 0& 0\end{array}\right),$$

$$w(A):=\sup \{|⟨Af,f⟩|:f\in L^2(X,\mu ),\Vert f{\Vert }_2=1\}.$$

$$w(A)=\sup \{⟨Af,f⟩:f\in L^2(X,\mu ),f\ge 0,\Vert f{\Vert }_2=1\}.$$

$$|⟨Af,f⟩|\le {\int }_X|Af||f|d\mu \le ⟨A|f|,|f|⟩$$

$$r(A)\le w(A)\le \Vert A\Vert$$

$$w(M(A))=w(A).$$

$$r(S(K))\le r(K).$$

$$S(K^2)\ge S(K{)}^2.$$

$$\sqrt{\left({\int }_{X}k(x,z)k(z,y)d\mu (z)\right)\left({\int }_{X}k(y,z)k(z,x)d\mu (z)\right)}.$$

$${\int }_X\sqrt{k(x,z)k(z,x)}\sqrt{k(z,y)k(y,z)}d\mu (z).$$

$${\rho }_0\le {\rho }_1\le \dots \le {\rho }_n\le r(K).$$

$$r(S(K^2))\ge r(S(K){)}^2.$$

$${\sigma }_0\ge {\sigma }_1\ge \dots \ge {\sigma }_n\ge r(A).$$

$$r(M(A^{2^n}))=w(M(A^{2^n}))=w(A^{2^n})\ge r(A^{2^n})=r(A{)}^{2^n},$$

$$⟨A^{\ast }Af,f⟩+⟨A{A}^{\ast }f,f⟩\ge 2⟨A^{2}f,f⟩$$

$$⟨(A+A^{\ast }{)}^{2}f,f⟩\ge 4⟨A^{2}f,f⟩,$$

$$r(M(A^{2^n}))=w(M(A^{2^n}))=w(A^{2^n})\le \Vert A^{2^n}\Vert ,$$

$$r(A)\le {\sigma }_n\le \Vert A^{2^n}{\Vert }^{2^{-n}}.$$

$$r(K)\ge \frac{4}{{\pi }^2}.$$

Full text

Key words: numerical radius, spectral radius, positive operators, kernel operators

Math. Subj. Classification (2010): 47A12, 47A10, 47B34, 47B60

1. Introduction

Let $A=(a_{ij})_{i,j=1}^{n}$ be a nonnegative matrix, i.e., $a_{i,j}\geq 0$ for all $i$ and $j$. In the literature, much attention has been paid to provide upper and lower bounds for the spectral radius $r(A)$ of $A$. Here $r(A)$ is defined as $\max\{|\lambda_{1}|,\ldots,|\lambda_{n}|\}$, where $\{\lambda_{i}\}_{i=1}^{n}$ are the eigenvalues of $A$. This number is an eigenvalue of $A$, and it is called the Perron root of $A$. Szyld [5] gave an increasing sequence of lower bounds for $r(A)$ that are based on the geometric symmetrization of powers of $A$. More precisely, let $S(A)$ be the matrix whose $(i,j)$ entry is equal to $\sqrt{a_{ij}a_{ji}}$, and define $\rho_{k}=r(S(A^{2^{k}}))^{2^{-k}}$. It is shown in [5] that $\rho_{0}\leq\rho_{1}\leq\ldots\leq\rho_{k}\leq r(A)$ for all $k$. Easy examples show that the sequence $\{\rho_{k}\}_{k\in\mathbb{N}}$ does not converge to $r(A)$ in general; if we take e.g.

[TABLE]

then $\rho_{0}=\rho_{1}=\ldots=0$, while $r(A)=1$. On the other hand, Taşçi and Kirkland [6] provided a decreasing sequence of upper bounds for $r(A)$ that are based on the arithmetic symmetrization of powers of $A$. Specifically, let $M(A)=(A+A^{T})/2$ and define $\sigma_{k}=r(M(A^{2^{k}}))^{2^{-k}}$. It is proved in [6] that $\sigma_{0}\geq\sigma_{1}\geq\ldots\geq\sigma_{k}\geq r(A)$ for all $k$ and that the sequence $\{\sigma_{k}\}_{k\in\mathbb{N}}$ converges to $r(A)$. In this paper we extend both results to positive operators on $L^{2}$-spaces. We should mention that in [1] the inequality $\sigma_{0}=r(M(A))\geq r(A)$ was already extended to this setting.

Throughout the note, let $\mu$ be a $\sigma$-finite positive measure on a set $X$. We consider bounded (linear) operators on the complex Hilbert space $L^{2}(X,\mu)$. The norm in $L^{2}(X,\mu)$ is denoted by $\|\cdot\|_{2}$. An operator $A$ on $L^{2}(X,\mu)$ is said to be positive if it maps nonnegative functions to nonnegative ones. Given operators $A$ and $B$ on $L^{2}(X,\mu)$, we write $A\geq B$ if the operator $A-B$ is positive. The operator norm and the spectral radius of an operator are denoted by $\|\cdot\|$ and $r(\cdot)$, respectively. The numerical radius of an operator $A$ on $L^{2}(X,\mu)$ is defined by

[TABLE]

If, in addition, $A$ is positive, then we have

[TABLE]

Indeed, this follows from the estimate

[TABLE]

that holds for any $f\in L^{2}(X,\mu)$. It is well-known [4] that

[TABLE]

for all bounded operators $A$ on $L^{2}(X,\mu)$. If, in particular, $A$ is selfadjoint, then we have $r(A)=w(A)=\|A\|$.

Let $A$ be a positive operator on $L^{2}(X,\mu)$. The arithmetic symmetrization $M(A)$ of $A$ is the positive selfadjoint operator on $L^{2}(X,\mu)$ defined by $M(A)=(A+A^{*})/2$. Since $\langle M(A)f,f\rangle=\langle Af,f\rangle$ for any nonnegative function $f\in L^{2}(X,\mu)$, we have

[TABLE]

Let $K$ be a positive kernel operator on $L^{2}(X,\mu)$ with a kernel $k$, that is, $k:X\times X\rightarrow[0,\infty)$ is a measurable function such that $(Kf)(x)=\int_{X}k(x,y)f(y)d\mu(y)$ for all $f\in L^{2}(X,\mu)$ and for almost all $x\in X$. The geometric symmetrization $S(K)$ of $K$ is the positive selfadjoint kernel operator on $L^{2}(X,\mu)$ with the kernel equal to $\sqrt{k(x,y)k(y,x)}$ at a point $(x,y)\in X\times X$. Note that $S(K)$ is well-defined on the whole $L^{2}(X,\mu)$, because the kernel of $S(K)$ is smaller than or equal to the kernel of $M(K)$ by the inequality of arithmetic and geometric means. It was proved in [2, Proposition 2.7] that

[TABLE]

Note that our results do not generalize beyond $L^{2}$-spaces.

2. Results

We begin with an observation that seems to be new also in the finite-dimensional case.

Lemma 2.1**.**

If $K$ is a positive kernel operator on $L^{2}(X,\mu)$, then

[TABLE]

Proof.

Let us compare the kernels of both operators. If $k$ is the kernel of $K$, then the kernel of $K^{2}$ at a point $(x,y)\in X\times X$ is equal to $\int_{X}k(x,z)k(z,y)d\mu(z)$, and so the kernel of $S(K^{2})$ at a point $(x,y)$ equals

[TABLE]

On the other hand, the kernel of $S(K)^{2}$ at a point $(x,y)$ is equal to

[TABLE]

Now the desired inequality is proved by an application of the Cauchy-Schwarz inequality. ∎

We now extend the finite-dimensional result due to Szyld [5, Theorem 2.2].

Theorem 2.2**.**

Let $K$ be a positive kernel operator on $L^{2}(X,\mu)$, and let $\rho_{n}=r(S(K^{2^{n}}))^{2^{-n}}$, $n\in\mathbb{N}\cup\{0\}$. Then, for each $n$,

[TABLE]

Proof.

By (2), we have $r(S(K^{2^{n}}))\leq r(K^{2^{n}})=r(K)^{2^{n}}$, which implies that $\rho_{n}\leq r(K)$ for all $n\in\mathbb{N}\cup\{0\}$. To finish the proof, it is enough to show that

[TABLE]

By Lemma 2.1, we have $S(K^{2}))\geq S(K)^{2}$ which implies easily that $r(S(K^{2}))\geq r(S(K)^{2})=r(S(K))^{2}$ as desired.

∎

The following theorem is an infinite-dimensional generalization of [6, Theorems 1 and 2].

Theorem 2.3**.**

Let $A$ be a positive operator on $L^{2}(X,\mu)$, and let $\sigma_{n}=r(M(A^{2^{n}}))^{2^{-n}}$, $n\in\mathbb{N}\cup\{0\}$. Then, for each $n$,

[TABLE]

Furthermore, the sequence $\{\sigma_{n}\}_{n\in\mathbb{N}}$ converges to $r(A)$.

Proof.

By (1), we have

[TABLE]

and so $\sigma_{n}\geq r(A)$ for all $n$.

To show that the sequence $\{\sigma_{n}\}_{n\in\mathbb{N}}$ is decreasing, it is enough to see that $r(M(A^{2}))\leq r(M(A))^{2}=r(M(A)^{2})$, or equivalently $w(M(A^{2}))\leq w(M(A)^{2})$. It follows from the inequality $\|Af-A^{*}f\|^{2}_{2}\geq 0$ that we have

[TABLE]

for all nonnegative functions $f\in L^{2}(X,\mu)$. This implies that

[TABLE]

and so we obtain the desired inequality $w(M(A)^{2})\geq w(A^{2})=w(M(A^{2}))$, where we have also used (1).

For each $n\in\mathbb{N}\cup\{0\}$ we have

[TABLE]

and so

[TABLE]

Since $\|A^{2^{n}}\|^{2^{-n}}\rightarrow r(A)$ as $n\rightarrow\infty$, the sequence $\{\sigma_{n}\}_{n\in\mathbb{N}}$ converges to $r(A)$. This completes the proof. ∎

Examples in [6] explain Theorem 2.3 in the finite-dimensional case. The following example further illustrates it in the infinite-dimensional setting.

Example 2.4**.**

Let $A$ be a weighted unilateral shift on $l^{2}$ with weights $\{1,4,1,4,1,4,\ldots\}$, that is, the operator defined by $A(x_{1},x_{2},\ldots)=(0,x_{1},4x_{2},x_{3},4x_{4},\ldots)$. Then $M(A)x=\frac{1}{2}(x_{2},x_{1}+4x_{3},4x_{2}+x_{4},x_{3}+4x_{5},4x_{4}+x_{6},\ldots)$. It is not difficult to verify that $r(M(A))=\|M(A)\|=5/2$, and so $\sigma_{0}=5/2$. Furthermore, $A^{2}x=(0,0,4x_{1},4x_{2},4x_{3},\ldots)=4S^{2}x$, where $S$ is the unilateral shift on $l^{2}$. Therefore, $r(A)^{2}=r(A^{2})=4r(S)^{2}=4$, and so $r(A)=2$. Since $M(A^{2})=2\,(S^{2}+(S^{*})^{2})$, we have $r(M(A^{2}))=\|M(A^{2})\|=4$, so that $\sigma_{1}=2$. Similarly, we obtain that $\sigma_{n}=2$ for all $n=2,3,4,\ldots$.

More generally, let $p$ be a positive integer, and let $A_{p}$ be a weighted unilateral shift on $l^{2}$ with weights $\{\underbrace{1,1,\ldots,1}_{2^{p}-1},2^{2^{p}},\underbrace{1,1,\ldots,1}_{2^{p}-1},2^{2^{p}},\ldots\}$, so that $A_{1}=A$. Then $A_{p}^{2^{p}}=2^{2^{p}}S^{2^{p}}$, and $r(A_{p})^{2^{p}}=r(A_{p}^{2^{p}})=2^{2^{p}}r(S)^{2^{p}}=2^{2^{p}}$, so that $r(A_{p})=2$. Since $M(A_{p}^{2^{p}})=2^{2^{p}-1}\,(S^{2^{p}}+(S^{*})^{2^{p}})$, we have $r(M(A_{p}^{2^{p}}))=\|M(A_{p}^{2^{p}})\|=2^{2^{p}}$, so that $\sigma_{p}=2$. Then $\sigma_{n}=2$ for all $n\geq p$ by Theorem 2.3. One can also show that $\sigma_{0}=r(M(A_{p}))=\|M(A_{p})\|\geq 2^{2^{p}-1}$ and that $\sigma_{n}>2$ for all $n<p$.

We complete this note with an application of the inequality (2). Several authors have studied the spectrum and the spectral radius of selfadjoint kernel operators. If $K$ is a positive kernel operator on $L^{2}(X,\mu)$, then $S(K)$ is a selfadjoint kernel operator. Therefore, if we can compute $r(S(K))$, then the inequality (2) provides the lower bound for $r(K)$. Let us illustrate this with the following proposition.

Proposition 2.5**.**

Let $g:[0,1]\times[0,1]\rightarrow[0,\infty)$ be a measurable function such that, for some $M\geq 1$, $\frac{1}{M}\leq g(x,y)\leq M$ and $g(x,y)g(y,x)=1$ for all $x$ and $y$. Then the function $k(x,y)=\min\{x,y\}\cdot g(x,y)$ is the kernel of the positive kernel operator $K$ on $L^{2}([0,1])$, and we have

[TABLE]

Proof.

Since $k(x,y)$ is a bounded nonnegative function, it defines the positive kernel operator $K$ on $L^{2}([0,1])$. The kernel of $S(K)$ at a point $(x,y)\in[0,1]\times[0,1]$ is equal to $\min\{x,y\}$. It follows from [3, Exercise 6.5.3, p.103 and p.271] that $r(S(K))=4/\pi^{2}$. Now, the proof is finished with an application of the inequality (2). ∎

Acknowledgment. The author acknowledges the financial support from the Slovenian Research Agency (research core funding No. P1-0222).