On Lp norms and the spectral radius of operators in Hilbert spaces

Mira Anisiu
(original), paper

Abstract

We prove that \(lim_{p\rightarrow \infty}k\) \(\left \Vert f\right \Vert _{+p}^{+p}\diagup \left \Vert f\right \Vert _{p}^{p}=\left \Vert f\right \Vert _{\infty}\) for \(f\neq0\) in the Bochner space \(L_{E}^{\infty}(TCIMACRO{\U{3bc} }BeginExpansion\mu EndExpansion)\), where \((E,|\bullet|)\) is a Banach space and \((X,A,TCIMACRO{\U{3bc} }BeginExpansion\mu EndExpansion\) a finite measure space. We discuss also the existence of \(lim_{n\rightarrow \infty}\left \Vert T^{n+1}\right \Vert \diagup \left \Vert T^{n}\right \Vert \) for continuous linear operators \(T\) in Hilbert spaces.

Authors

Mira-Cristiana Anisiu
Tiberiu Popoviciu Institute of Numerical Analysis Cluj-Napoca, Romanian Academy, Romania

Valeriu Anisiu
Babes-Bolyai University Cluj-Napoca, Romania

Keywords

\(L_{p}\) norms, linear operators, spectral radius.

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M.-C. Anisiu, V. Anisiu, On Lp norms and the spectral radius of operators in Hilbert spaces, Annals of the Tiberiu Popoviciu Seminar of Functional Equations, Approximation and Convexity 5 (2007), 9-16 (pdf file here)

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https://ictp.acad.ro/anisiu/papers/2007-Anisiu-Anisiu-OnLp.PDF

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2007-Anisiu-Anisiu-OnLp.PDF
Annals of the Tiberiu Popoviciu Seminar of Functional Equations, Approximation and Convexity ISSN 1584-4536, vol 5, 2007, pp. 3-10.

On L p L p L^(p)L^{p}Lp Norms and the Spectral Radius of Operators in Hilbert Spaces

Mira-Cristiana Anisiu Valeriu Anisiu(Cluj-Napoca) (Cluj-Napoca)

Abstract

We prove that lim p → ∞ ‖ f ‖ p + 1 p + 1 / ‖ f ‖ p p = ‖ f ‖ ∞ lim p → ∞   ‖ f ‖ p + 1 p + 1 / ‖ f ‖ p p = ‖ f ‖ ∞ lim_(p rarr oo)||f||_(p+1)^(p+1)//||f||_(p)^(p)=||f||_(oo)\lim _{p \rightarrow \infty}\|f\|_{p+1}^{p+1} /\|f\|_{p}^{p}=\|f\|_{\infty}limp→∞‖f‖p+1p+1/‖f‖pp=‖f‖∞ for f ≠ 0 f ≠ 0 f!=0f \neq 0f≠0 in the Bochner space L E ∞ ( μ ) L E ∞ ( μ ) L_(E)^(oo)(mu)L_{E}^{\infty}(\mu)LE∞(μ), where ( E , | ⋅ | ) ( E , | ⋅ | ) (E,|*|)(E,|\cdot|)(E,|⋅|) is a Banach space and ( X , A , μ X , A , μ X,A,muX, \mathcal{A}, \muX,A,μ ) a finite measure space. We discuss also the existence of lim n → ∞ ‖ T n + 1 ‖ / ‖ T n ‖ lim n → ∞   T n + 1 / T n lim_(n rarr oo)||T^(n+1)||//||T^(n)||\lim _{n \rightarrow \infty}\left\|T^{n+1}\right\| /\left\|T^{n}\right\|limn→∞‖Tn+1‖/‖Tn‖ for continuous linear operators T T TTT in Hilbert spaces.

Key Words: L p L p L^(p)L^{p}Lp norms, linear operators, spectral radius.
MSC 2000: 46E30, 47A75

1 A limit involving L p L p L^(p)L^{p}Lp and L ∞ L ∞ L^(oo)L^{\infty}L∞ norms

Let ( X , A , μ X , A , μ X,A,muX, \mathcal{A}, \muX,A,μ ) be a measure space. If μ μ mu\muμ is finite and f ∈ L ∞ ( μ ) f ∈ L ∞ ( μ ) f inL^(oo)(mu)f \in L^{\infty}(\mu)f∈L∞(μ), the L ∞ L ∞ L^(oo)L^{\infty}L∞ norm of the real function f f fff can be obtained as the limit
(1) ‖ f ‖ ∞ = lim p → ∞ ‖ f ‖ p (1) ‖ f ‖ ∞ = lim p → ∞   ‖ f ‖ p {:(1)||f||_(oo)=lim_(p rarr oo)||f||_(p):}\begin{equation*} \|f\|_{\infty}=\lim _{p \rightarrow \infty}\|f\|_{p} \tag{1} \end{equation*}(1)‖f‖∞=limp→∞‖f‖p
qquad\qquad
qquad\qquad
This result can be found in [9], p. 34.
It is known that for a sequence of real numbers a p > 0 a p > 0 a_(p) > 0a_{p}>0ap>0, the equality
lim p → ∞ ( a p ) 1 / p = lim p → ∞ a p + 1 a p lim p → ∞   a p 1 / p = lim p → ∞   a p + 1 a p lim_(p rarr oo)(a_(p))^(1//p)=lim_(p rarr oo)(a_(p+1))/(a_(p))\lim _{p \rightarrow \infty}\left(a_{p}\right)^{1 / p}=\lim _{p \rightarrow \infty} \frac{a_{p+1}}{a_{p}}limp→∞(ap)1/p=limp→∞ap+1ap
holds, provided that the second limit exists (Stolz-Cesàro) [1, p. 150].
The problem we are going to solve is:
For a p = ‖ f ‖ p p a p = ‖ f ‖ p p a_(p)=||f||_(p)^(p)a_{p}=\|f\|_{p}^{p}ap=‖f‖pp, does the limit lim p → ∞ a p + 1 a p lim p → ∞   a p + 1 a p lim_(p rarr oo)(a_(p+1))/(a_(p))\lim _{p \rightarrow \infty} \frac{a_{p+1}}{a_{p}}limp→∞ap+1ap exist?
Remark 1.1 There are known several conditions on the sequences a n , b n a n , b n a_(n),b_(n)a_{n}, b_{n}an,bn insuring that a n b n → L ⟹ a n + 1 − a n b n + 1 − b n → L a n b n → L ⟹ a n + 1 − a n b n + 1 − b n → L (a_(n))/(b_(n))rarr L Longrightarrow(a_(n+1)-a_(n))/(b_(n+1)-b_(n))rarr L\frac{a_{n}}{b_{n}} \rightarrow L \Longrightarrow \frac{a_{n+1}-a_{n}}{b_{n+1}-b_{n}} \rightarrow Lanbn→L⟹an+1−anbn+1−bn→L. They apply for example for Traian Lalescu's sequence [5]: ( n + 1 ) ! n + 1 − n ! n → 1 / e ( n + 1 ) ! n + 1 − n ! n → 1 / e root(n+1)((n+1)!)-root(n)(n!)rarr1//e\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!} \rightarrow 1 / e(n+1)!n+1−n!n→1/e. Similar sequences were studied by T. Popoviciu [7] and recently by many other mathematicians.
As a special case, for a n := ln ⁡ a n a n := ln ⁡ a n a_(n):=ln a_(n)a_{n}:=\ln a_{n}an:=ln⁡an and b n := n b n := n b_(n):=nb_{n}:=nbn:=n, it follows that a n n → L ⟹ a n + 1 a n → L a n n → L ⟹ a n + 1 a n → L root(n)(a_(n))rarr L Longrightarrow(a_(n+1))/(a_(n))rarr L\sqrt[n]{a_{n}} \rightarrow L \Longrightarrow \frac{a_{n+1}}{a_{n}} \rightarrow Lann→L⟹an+1an→L.
Unfortunately, these conditions do not apply for the problem to be studied. We prove directly the following result (in Bochner spaces).
Theorem 1.1 Let ( E , | ⋅ | E , | ⋅ | E,|*|E,|\cdot|E,|⋅| ) be a Banach space, ( X , A , μ X , A , μ X,A,muX, \mathcal{A}, \muX,A,μ ) a finite measure space ( μ ( X ) < ∞ ) ( μ ( X ) < ∞ ) (mu(X) < oo)(\mu(X)<\infty)(μ(X)<∞) and f ∈ L E ∞ ( μ ) ∖ { 0 } f ∈ L E ∞ ( μ ) ∖ { 0 } f inL_(E)^(oo)(mu)\\{0}f \in L_{E}^{\infty}(\mu) \backslash\{0\}f∈LE∞(μ)∖{0}. Then
lim p → ∞ ∫ | f | p + 1 d μ ∫ | f | p d μ = ‖ f ‖ ∞ lim p → ∞   ∫ | f | p + 1 d μ ∫ | f | p d μ = ‖ f ‖ ∞ lim_(p rarr oo)(int|f|^(p+1)(d)mu)/(int|f|^(p)(d)mu)=||f||_(oo)\lim _{p \rightarrow \infty} \frac{\int|f|^{p+1} \mathrm{~d} \mu}{\int|f|^{p} \mathrm{~d} \mu}=\|f\|_{\infty}limp→∞∫|f|p+1 dμ∫|f|p dμ=‖f‖∞
Proof. Replacing f f fff by | f | / ‖ f ‖ ∞ | f | / ‖ f ‖ ∞ |f|//||f||_(oo)|f| /\|f\|_{\infty}|f|/‖f‖∞, one may suppose that 0 ≤ f ≤ 1 0 ≤ f ≤ 1 0 <= f <= 10 \leq f \leq 10≤f≤1 and ‖ f ‖ ∞ = 1 ‖ f ‖ ∞ = 1 ||f||_(oo)=1\|f\|_{\infty}=1‖f‖∞=1. Let us denote
r p = ∫ | f | p + 1 d μ ∫ | f | p d μ . r p = ∫ | f | p + 1 d μ ∫ | f | p d μ . r_(p)=(int|f|^(p+1)(d)mu)/(int|f|^(p)(d)mu).r_{p}=\frac{\int|f|^{p+1} \mathrm{~d} \mu}{\int|f|^{p} \mathrm{~d} \mu} .rp=∫|f|p+1 dμ∫|f|p dμ.
Then 0 ≤ r p ≤ 1 0 ≤ r p ≤ 1 0 <= r_(p) <= 10 \leq r_{p} \leq 10≤rp≤1, so,
(2) lim sup p → ∞ r p ≤ 1 (2) lim sup p → ∞   r p ≤ 1 {:(2)l i m   s u p_(p rarr oo)r_(p) <= 1:}\begin{equation*} \limsup _{p \rightarrow \infty} r_{p} \leq 1 \tag{2} \end{equation*}(2)lim supp→∞rp≤1
For 0 < a < 1 0 < a < 1 0 < a < 10<a<10<a<1 we denote A a = { x ∈ X : f ( x ) ≥ a } , B a = X ∖ A A a = { x ∈ X : f ( x ) ≥ a } , B a = X ∖ A A_(a)={x in X:f(x) >= a},B_(a)=X\\AA_{a}=\{x \in X: f(x) \geq a\}, B_{a}=X \backslash AAa={x∈X:f(x)≥a},Ba=X∖A. We have μ ( A a ) > 0 μ A a > 0 mu(A_(a)) > 0\mu\left(A_{a}\right)>0μ(Aa)>0 because ‖ f ‖ ∞ = 1 ‖ f ‖ ∞ = 1 ||f||_(oo)=1\|f\|_{\infty}=1‖f‖∞=1. We show that
lim p → ∞ ∫ B a f p d μ ∫ A a f p d μ = 0 . lim p → ∞   ∫ B a   f p d μ ∫ A a   f p d μ = 0 . lim_(p rarr oo)(int_(B_(a))f^(p)(d)mu)/(int_(A_(a))f^(p)(d)mu)=0.\lim _{p \rightarrow \infty} \frac{\int_{B_{a}} f^{p} \mathrm{~d} \mu}{\int_{A_{a}} f^{p} \mathrm{~d} \mu}=0 .limp→∞∫Bafp dμ∫Aafp dμ=0.
Let us choose b b bbb so that a < b < 1 a < b < 1 a < b < 1a<b<1a<b<1. Then A b ⊆ A a A b ⊆ A a A_(b)subeA_(a)A_{b} \subseteq A_{a}Ab⊆Aa and
0 ≤ ∫ B a f p d μ ∫ A a f p d μ ≤ a p μ ( B a ) ∫ A b f p d μ ≤ a p μ ( B a ) b p μ ( A b ) = ( a b ) p μ ( B a ) μ ( A b ) → 0 ( p → ∞ ) . 0 ≤ ∫ B a   f p d μ ∫ A a   f p d μ ≤ a p μ B a ∫ A b   f p d μ ≤ a p μ B a b p μ A b = a b p μ B a μ A b → 0 ( p → ∞ ) . {:[0 <= (int_(B_(a))f^(p)(d)mu)/(int_(A_(a))f^(p)(d)mu) <= (a^(p)mu(B_(a)))/(int_(A_(b))f^(p)(d)mu) <= (a^(p)mu(B_(a)))/(b^(p)mu(A_(b)))=],[((a)/(b))^(p)(mu(B_(a)))/(mu(A_(b)))rarr0(p rarr oo).]:}\begin{aligned} & 0 \leq \frac{\int_{B_{a}} f^{p} \mathrm{~d} \mu}{\int_{A_{a}} f^{p} \mathrm{~d} \mu} \leq \frac{a^{p} \mu\left(B_{a}\right)}{\int_{A_{b}} f^{p} \mathrm{~d} \mu} \leq \frac{a^{p} \mu\left(B_{a}\right)}{b^{p} \mu\left(A_{b}\right)}= \\ & \left(\frac{a}{b}\right)^{p} \frac{\mu\left(B_{a}\right)}{\mu\left(A_{b}\right)} \rightarrow 0(p \rightarrow \infty) . \end{aligned}0≤∫Bafp dμ∫Aafp dμ≤apμ(Ba)∫Abfp dμ≤apμ(Ba)bpμ(Ab)=(ab)pμ(Ba)μ(Ab)→0(p→∞).
We obtain for lim inf p → ∞ r p lim inf p → ∞   r p l i m   i n f_(p rarr oo)r_(p)\liminf _{p \rightarrow \infty} r_{p}lim infp→∞rp the following estimation
lim inf p → ∞ r p = lim inf p → ∞ ∫ A a f p + 1 d μ + ∫ B a f p + 1 d μ ∫ A a f p d μ + ∫ B a f p d μ = lim inf p → ∞ ∫ A a f p + 1 d μ ∫ A a f p d μ ⋅ 1 + ∫ B a f p + 1 d μ ∫ A a f p + 1 d μ 1 + ∫ B a f p d μ ∫ A a f p d μ = lim inf p → ∞ ∫ A a f p + 1 d μ ∫ A a f p d μ ⋅ 1 ≥ lim inf p → ∞ ∫ A a f p a d μ ∫ A a f p d μ = a lim inf p → ∞   r p = lim inf p → ∞   ∫ A a   f p + 1 d μ + ∫ B a   f p + 1 d μ ∫ A a   f p d μ + ∫ B a   f p d μ = lim inf p → ∞   ∫ A a   f p + 1 d μ ∫ A a   f p d μ ⋅ 1 + ∫ B a   f p + 1 d μ ∫ A a   f p + 1 d μ 1 + ∫ B a   f p d μ ∫ A a   f p d μ = lim inf p → ∞   ∫ A a   f p + 1 d μ ∫ A a   f p d μ ⋅ 1 ≥ lim inf p → ∞   ∫ A a   f p a d μ ∫ A a   f p d μ = a {:[l i m   i n f_(p rarr oo)r_(p)=l i m   i n f_(p rarr oo)(int_(A_(a))f^(p+1)(d)mu+int_(B_(a))f^(p+1)(d)mu)/(int_(A_(a))f^(p)(d)mu+int_(B_(a))f^(p)(d)mu)=],[l i m   i n f_(p rarr oo)(int_(A_(a))f^(p+1)(d)mu)/(int_(A_(a))f^(p)(d)mu)*(1+(int_(B_(a))f^(p+1)(d)mu)/(int_(A_(a))f^(p+1)(d)mu))/(1+(int_(B_(a))f^(p)(d)mu)/(int_(A_(a))f^(p)(d)mu))=],[l i m   i n f_(p rarr oo)(int_(A_(a))f^(p+1)(d)mu)/(int_(A_(a))f^(p)(d)mu)*1 >= l i m   i n f_(p rarr oo)(int_(A_(a))f^(p)a(d)mu)/(int_(A_(a))f^(p)(d)mu)=a]:}\begin{aligned} & \liminf _{p \rightarrow \infty} r_{p}=\liminf _{p \rightarrow \infty} \frac{\int_{A_{a}} f^{p+1} \mathrm{~d} \mu+\int_{B_{a}} f^{p+1} \mathrm{~d} \mu}{\int_{A_{a}} f^{p} \mathrm{~d} \mu+\int_{B_{a}} f^{p} \mathrm{~d} \mu}= \\ & \liminf _{p \rightarrow \infty} \frac{\int_{A_{a}} f^{p+1} \mathrm{~d} \mu}{\int_{A_{a}} f^{p} \mathrm{~d} \mu} \cdot \frac{1+\frac{\int_{B_{a}} f^{p+1} \mathrm{~d} \mu}{\int_{A_{a}} f^{p+1} \mathrm{~d} \mu}}{1+\frac{\int_{B_{a}} f^{p} \mathrm{~d} \mu}{\int_{A_{a}} f^{p} \mathrm{~d} \mu}}= \\ & \liminf _{p \rightarrow \infty} \frac{\int_{A_{a}} f^{p+1} \mathrm{~d} \mu}{\int_{A_{a}} f^{p} \mathrm{~d} \mu} \cdot 1 \geq \liminf _{p \rightarrow \infty} \frac{\int_{A_{a}} f^{p} a \mathrm{~d} \mu}{\int_{A_{a}} f^{p} \mathrm{~d} \mu}=a \end{aligned}lim infp→∞rp=lim infp→∞∫Aafp+1 dμ+∫Bafp+1 dμ∫Aafp dμ+∫Bafp dμ=lim infp→∞∫Aafp+1 dμ∫Aafp dμ⋅1+∫Bafp+1 dμ∫Aafp+1 dμ1+∫Bafp dμ∫Aafp dμ=lim infp→∞∫Aafp+1 dμ∫Aafp dμ⋅1≥lim infp→∞∫Aafpa dμ∫Aafp dμ=a
But a ∈ ( 0 , 1 ) a ∈ ( 0 , 1 ) a in(0,1)a \in(0,1)a∈(0,1) is arbitrary, so
(3) lim inf p → ∞ r p ≥ 1 . (3) lim inf p → ∞   r p ≥ 1 . {:(3)l i m   i n f_(p rarr oo)r_(p) >= 1.:}\begin{equation*} \liminf _{p \rightarrow \infty} r_{p} \geq 1 . \tag{3} \end{equation*}(3)lim infp→∞rp≥1.
From (2) and (3) it follows lim p → ∞ r p = 1 lim p → ∞   r p = 1 lim_(p rarr oo)r_(p)=1\lim _{p \rightarrow \infty} r_{p}=1limp→∞rp=1.
Equality (1) can be obtained as a consequence of Theorem 1.1, using the Stolz-Cesàro result.
qquad\qquad
qquad\qquad

2 On a limit concerning operators with spectral radius r ( T ) ≠ 0 r ( T ) ≠ 0 r(T)!=0r(T) \neq 0r(T)≠0

Let E E EEE be a Hilbert space and T T TTT a linear continuous operator. Then the spectral radius r ( T ) r ( T ) r(T)r(T)r(T) of the operator T T TTT is given by
r ( T ) = lim n → ∞ ‖ T n ‖ 1 / n r ( T ) = lim n → ∞   T n 1 / n r(T)=lim_(n rarr oo)||T^(n)||^(1//n)r(T)=\lim _{n \rightarrow \infty}\left\|T^{n}\right\|^{1 / n}r(T)=limn→∞‖Tn‖1/n
The result in section 1 suggest the following problem: If r ( T ) ≠ 0 r ( T ) ≠ 0 r(T)!=0r(T) \neq 0r(T)≠0, is it true that lim n → ∞ ‖ T n + 1 ‖ / ‖ T n ‖ lim n → ∞   T n + 1 / T n lim_(n rarr oo)||T^(n+1)||//||T^(n)||\lim _{n \rightarrow \infty}\left\|T^{n+1}\right\| /\left\|T^{n}\right\|limn→∞‖Tn+1‖/‖Tn‖ does exist?
We mention the following interesting related result due to Kellogg [ 3 ] , [ 8 , p .240 ] [ 3 ] , [ 8 , p .240 ] [3],[8,p.240][3],[8, \mathrm{p} .240][3],[8,p.240], which provides an algorithm for finding an eigenvalue for a compact self-adjoint operator.
Theorem 2.1 Let E E EEE be a Hilbert space, T T TTT a compact self-adjoint operator, x 0 ∈ E x 0 ∈ E x_(0)in Ex_{0} \in Ex0∈E such that T x 0 ≠ 0 T x 0 ≠ 0 Tx_(0)!=0T x_{0} \neq 0Tx0≠0. Then, for x n = T n x 0 x n = T n x 0 x_(n)=T^(n)x_(0)x_{n}=T^{n} x_{0}xn=Tnx0, one has that x n ≠ 0 x n ≠ 0 x_(n)!=0x_{n} \neq 0xn≠0, the sequence ‖ x n + 1 ‖ / ‖ x n ‖ x n + 1 / x n ||x_(n+1)||//||x_(n)||\left\|x_{n+1}\right\| /\left\|x_{n}\right\|‖xn+1‖/‖xn‖ is increasing and convergent to r > 0 r > 0 r > 0r>0r>0 such that either r r rrr or − r − r -r-r−r is an eigenvalue for T T TTT.
In [2, p. 222], the definition of operators of class K K K\mathcal{K}K was given.
Definition 2.1 The linear continuous operator T T TTT is of class K K K\mathcal{K}K if for each x ∈ E , m ∈ N , m ≥ 2 x ∈ E , m ∈ N , m ≥ 2 x in E,m inN,m >= 2x \in E, m \in \mathbb{N}, m \geq 2x∈E,m∈N,m≥2 and k ∈ { 1 , 2 , … , m − 1 } k ∈ { 1 , 2 , … , m − 1 } k in{1,2,dots,m-1}k \in\{1,2, \ldots, m-1\}k∈{1,2,…,m−1}
(4) ‖ T k x ‖ ≤ C m , k ‖ x ‖ 1 − k m ‖ T m x ‖ k m , (4) T k x ≤ C m , k ‖ x ‖ 1 − k m T m x k m , {:(4)||T^(k)x|| <= C_(m,k)||x||^(1-(k)/(m))||T^(m)x||^((k)/(m))",":}\begin{equation*} \left\|T^{k} x\right\| \leq C_{m, k}\|x\|^{1-\frac{k}{m}}\left\|T^{m} x\right\|^{\frac{k}{m}}, \tag{4} \end{equation*}(4)‖Tkx‖≤Cm,k‖x‖1−km‖Tmx‖km,
where C m , k C m , k C_(m,k)C_{m, k}Cm,k are constants.
Remark 2.1 1. If T T TTT is invertible, the minimal constants C m , k C m , k C_(m,k)C_{m, k}Cm,k in (4) must satisfy (see [2, p. 223])
(5) C m , k ≤ ‖ T k ‖ 1 − k m ‖ T − ( m − k ) ‖ k m ; (5) C m , k ≤ T k 1 − k m T − ( m − k ) k m ; {:(5)C_(m,k) <= ||T^(k)||^(1-(k)/(m))||T^(-(m-k))||^((k)/(m));:}\begin{equation*} C_{m, k} \leq\left\|T^{k}\right\|^{1-\frac{k}{m}}\left\|T^{-(m-k)}\right\|^{\frac{k}{m}} ; \tag{5} \end{equation*}(5)Cm,k≤‖Tk‖1−km‖T−(m−k)‖km;
if T T TTT is normal, C m , k = 1 C m , k = 1 C_(m,k)=1C_{m, k}=1Cm,k=1 for each m m mmm and k k kkk.
2. For T i , i = 1 , … , 4 T i , i = 1 , … , 4 T_(i),i=1,dots,4T_{i}, i=1, \ldots, 4Ti,i=1,…,4 linear continuous operators on E E EEE, the following two inequalities regarding the spectral radius
r ( [ T 1 T 2 T 3 T 4 ] ) ≤ r ( [ ‖ T 1 ‖ ‖ T 2 ‖ ‖ T 3 ‖ ‖ T 4 ‖ ] ) r ( T 1 T 2 + T 3 T 4 ) ≤ 1 2 ( ‖ T 2 T 1 ‖ + ‖ T 4 T 3 ‖ ) + ( ‖ T 2 T 1 ‖ − ‖ T 4 T 3 ‖ ) 2 + 4 ‖ T 2 T 3 ‖ ⋅ ‖ T 4 T 1 ‖ r T 1 T 2 T 3 T 4 ≤ r T 1 T 2 T 3 T 4 r T 1 T 2 + T 3 T 4 ≤ 1 2 T 2 T 1 + T 4 T 3 + T 2 T 1 − T 4 T 3 2 + 4 T 2 T 3 ⋅ T 4 T 1 {:[quad r([[T_(1),T_(2)],[T_(3),T_(4)]]) <= r([[||T_(1)||,||T_(2)||],[||T_(3)||,||T_(4)||]])],[r(T_(1)T_(2)+T_(3)T_(4)) <= (1)/(2)(||T_(2)T_(1)||+||T_(4)T_(3)||)],[+sqrt((||T_(2)T_(1)||-||T_(4)T_(3)||)^(2)+4||T_(2)T_(3)||*||T_(4)T_(1)||)]:}\begin{aligned} & \quad r\left(\left[\begin{array}{ll} T_{1} & T_{2} \\ T_{3} & T_{4} \end{array}\right]\right) \leq r\left(\left[\begin{array}{ll} \left\|T_{1}\right\| & \left\|T_{2}\right\| \\ \left\|T_{3}\right\| & \left\|T_{4}\right\| \end{array}\right]\right) \\ & r\left(T_{1} T_{2}+T_{3} T_{4}\right) \leq \frac{1}{2}\left(\left\|T_{2} T_{1}\right\|+\left\|T_{4} T_{3}\right\|\right) \\ & +\sqrt{\left(\left\|T_{2} T_{1}\right\|-\left\|T_{4} T_{3}\right\|\right)^{2}+4\left\|T_{2} T_{3}\right\| \cdot\left\|T_{4} T_{1}\right\|} \end{aligned}r([T1T2T3T4])≤r([‖T1‖‖T2‖‖T3‖‖T4‖])r(T1T2+T3T4)≤12(‖T2T1‖+‖T4T3‖)+(‖T2T1‖−‖T4T3‖)2+4‖T2T3‖⋅‖T4T1‖
have been proved in [4].
We state the following
Conjecture 2.1 If T T TTT is of class K K K\mathcal{K}K and r ( T ) ≠ 0 r ( T ) ≠ 0 r(T)!=0r(T) \neq 0r(T)≠0, then lim n → ∞ ‖ T n + 1 ‖ / ‖ T n ‖ lim n → ∞   T n + 1 / T n lim_(n rarr oo)||T^(n+1)||//||T^(n)||\lim _{n \rightarrow \infty}\left\|T^{n+1}\right\| /\left\|T^{n}\right\|limn→∞‖Tn+1‖/‖Tn‖ do exist.
We prove the next result mentioned in [2, p. 216].
Proposition 2.1 r ( T ) = ‖ T ‖ ⇔ ‖ T n ‖ = ‖ T ‖ n 2.1 r ( T ) = ‖ T ‖ ⇔ T n = ‖ T ‖ n 2.1 r(T)=||T||<=>||T^(n)||=||T||^(n)2.1 r(T)=\|T\| \Leftrightarrow\left\|T^{n}\right\|=\|T\|^{n}2.1r(T)=‖T‖⇔‖Tn‖=‖T‖n, for all n ∈ N n ∈ N n inNn \in \mathbb{N}n∈N.
Proof. The spectral mapping theorem implies that r ( T n ) = r ( T ) n r T n = r ( T ) n r(T^(n))=r(T)^(n)r\left(T^{n}\right)=r(T)^{n}r(Tn)=r(T)n, so if r ( T ) = ‖ T ‖ r ( T ) = ‖ T ‖ r(T)=||T||r(T)=\|T\|r(T)=‖T‖ then ‖ T ‖ n = r ( T ) n = r ( T n ) ≤ ‖ T n ‖ ≤ ‖ T ‖ n ‖ T ‖ n = r ( T ) n = r T n ≤ T n ≤ ‖ T ‖ n ||T||^(n)=r(T)^(n)=r(T^(n)) <= ||T^(n)|| <= ||T||^(n)\|T\|^{n}=r(T)^{n}=r\left(T^{n}\right) \leq\left\|T^{n}\right\| \leq\|T\|^{n}‖T‖n=r(T)n=r(Tn)≤‖Tn‖≤‖T‖n, hence ‖ T n ‖ = ‖ T ‖ n T n = ‖ T ‖ n ||T^(n)||=||T||^(n)\left\|T^{n}\right\|=\|T\|^{n}‖Tn‖=‖T‖n.
Conversely, if ‖ T n ‖ = ‖ T ‖ n T n = ‖ T ‖ n ||T^(n)||=||T||^(n)\left\|T^{n}\right\|=\|T\|^{n}‖Tn‖=‖T‖n, for all n ∈ N n ∈ N n inNn \in \mathbb{N}n∈N then r ( T ) = lim n → ∞ ‖ T n ‖ 1 / n = lim n → ∞ ‖ T ‖ n ⋅ 1 / n = ‖ T ‖ r ( T ) = lim n → ∞   T n 1 / n = lim n → ∞   ‖ T ‖ n ⋅ 1 / n = ‖ T ‖ r(T)=lim_(n rarr oo)||T^(n)||^(1//n)=lim_(n rarr oo)||T||^(n*1//n)=||T||r(T)=\lim _{n \rightarrow \infty}\left\|T^{n}\right\|^{1 / n}=\lim _{n \rightarrow \infty}\|T\|^{n \cdot 1 / n}=\|T\|r(T)=limn→∞‖Tn‖1/n=limn→∞‖T‖n⋅1/n=‖T‖.
If r ( T ) = ‖ T ‖ r ( T ) = ‖ T ‖ r(T)=||T||r(T)=\|T\|r(T)=‖T‖, obviously ‖ T n + 1 ‖ / ‖ T n ‖ = ‖ T ‖ T n + 1 / T n = ‖ T ‖ ||T^(n+1)||//||T^(n)||=||T||\left\|T^{n+1}\right\| /\left\|T^{n}\right\|=\|T\|‖Tn+1‖/‖Tn‖=‖T‖ and conjecture 2.1 holds. Note also that if T T TTT is normal, then r ( T ) = ‖ T ‖ r ( T ) = ‖ T ‖ r(T)=||T||r(T)=\|T\|r(T)=‖T‖; but T T TTT may not be normal and yet lim n → ∞ ‖ T n + 1 ‖ / ‖ T n ‖ lim n → ∞   T n + 1 / T n lim_(n rarr oo)||T^(n+1)||//||T^(n)||\lim _{n \rightarrow \infty}\left\|T^{n+1}\right\| /\left\|T^{n}\right\|limn→∞‖Tn+1‖/‖Tn‖ exists ( = r ( T ) ) ( = r ( T ) ) (=r(T))(=r(T))(=r(T)); see ex 2.3.
Remark 2.2 Let T T TTT be the Volterra operator in L 2 ( [ 0 , 1 ] ) L 2 ( [ 0 , 1 ] ) L^(2)([0,1])L^{2}([0,1])L2([0,1]),
( T x ) ( t ) = ∫ 0 t x d λ ( T x ) ( t ) = ∫ 0 t   x d λ (Tx)(t)=int_(0)^(t)xdlambda(T x)(t)=\int_{0}^{t} x \mathrm{~d} \lambda(Tx)(t)=∫0tx dλ
Then r ( T ) = 0 , ‖ T ‖ = 2 / π = 0.6366197722 … , ‖ T 2 ‖ = 1 / α 2 = .2844128717 . . r ( T ) = 0 , ‖ T ‖ = 2 / π = 0.6366197722 … , T 2 = 1 / α 2 = .2844128717 . . r(T)=0,||T||=2//pi=0.6366197722 dots,||T^(2)||=1//alpha^(2)=.2844128717..r(T)=0,\|T\|=2 / \pi=0.6366197722 \ldots,\left\|T^{2}\right\|=1 / \alpha^{2}= .2844128717 . .r(T)=0,‖T‖=2/π=0.6366197722…,‖T2‖=1/α2=.2844128717... where α α alpha\alphaα is the smallest positive root of the equation ( e a + e − a ) cos ⁡ ( a ) = − 2 e a + e − a cos ⁡ ( a ) = − 2 {:e^(a)+e^(-a))cos(a)=-2\left.e^{a}+e^{-a}\right) \cos (a)=-2ea+e−a)cos⁡(a)=−2, see [6, p. 259]. The norms ‖ T n ‖ T n ||T^(n)||\left\|T^{n}\right\|‖Tn‖ are more difficult to find for n > 2 n > 2 n > 2n>2n>2.
qquad\qquad
qquad\qquad
The next example shows that conjecture 2.1 does not hold for all linear continuous operators.
Example 2.1 An operator with r ( T ) = 1 r ( T ) = 1 r(T)=1r(T)=1r(T)=1 for which lim n → ∞ ‖ T n + 1 ‖ / ‖ T n ‖ lim n → ∞   T n + 1 / T n lim_(n rarr oo)||T^(n+1)||//||T^(n)||\lim _{n \rightarrow \infty}\left\|T^{n+1}\right\| /\left\|T^{n}\right\|limn→∞‖Tn+1‖/‖Tn‖ does not exist.
Let T = [ 1 1 0 − 1 ] . Let T = 1 1 0 − 1 . Let T=[[1,1],[0,-1]].\operatorname{Let} T=\left[\begin{array}{cc} 1 & 1 \\ 0 & -1 \end{array}\right] .LetT=[110−1].
Then r ( T ) = 1 , ‖ T n ‖ = { 1 , for n even ( 5 + 1 ) / 2 , for n odd r ( T ) = 1 , T n = 1 ,       for  n  even  ( 5 + 1 ) / 2 ,       for  n  odd  r(T)=1,||T^(n)||={[1","," for "n" even "],[(sqrt5+1)//2","," for "n" odd "]:}r(T)=1,\left\|T^{n}\right\|=\left\{\begin{array}{ll}1, & \text { for } n \text { even } \\ (\sqrt{5}+1) / 2, & \text { for } n \text { odd }\end{array}\right.r(T)=1,‖Tn‖={1, for n even (5+1)/2, for n odd . In this case, ‖ T n ‖ 1 / n → r ( T ) T n 1 / n → r ( T ) ||T^(n)||^(1//n)rarr r(T)\left\|T^{n}\right\|^{1 / n} \rightarrow r(T)‖Tn‖1/n→r(T) but ‖ T n + 1 ‖ / ‖ T n ‖ T n + 1 / T n ||T^(n+1)||//||T^(n)||\left\|T^{n+1}\right\| /\left\|T^{n}\right\|‖Tn+1‖/‖Tn‖ diverges.
Actually, this behaviour is almost generic. We give below the Maple code computing r ( T ) r ( T ) r(T)r(T)r(T) and the sequence ‖ T n + 1 ‖ / ‖ T n ‖ T n + 1 / T n ||T^(n+1)||//||T^(n)||\left\|T^{n+1}\right\| /\left\|T^{n}\right\|‖Tn+1‖/‖Tn‖ for a linear operator in R d ( d = 2 ) R d ( d = 2 ) R^(d)(d=2)\mathbb{R}^{d}(d=2)Rd(d=2) with randomly selected entries from { − 5 , − 4 , … , 4 , 5 } { − 5 , − 4 , … , 4 , 5 } {-5,-4,dots,4,5}\{-5,-4, \ldots, 4,5\}{−5,−4,…,4,5}. Note that for d > 4 d > 4 d > 4d>4d>4 this can be done only approximately.
We display the values of the sequences ‖ T n + 1 ‖ / ‖ T n ‖ T n + 1 / T n ||T^(n+1)||//||T^(n)||\left\|T^{n+1}\right\| /\left\|T^{n}\right\|‖Tn+1‖/‖Tn‖ and ‖ T n ‖ 1 / n T n 1 / n ||T^(n)||^(1//n)\left\|T^{n}\right\|^{1 / n}‖Tn‖1/n.
Example 2.2 The operator T = [ 4 5 − 4 4 ] T = 4 5 − 4 4 T=[[4,5],[-4,4]]T=\left[\begin{array}{cc}4 & 5 \\ -4 & 4\end{array}\right]T=[45−44] has r ( T ) = 2 r ( T ) = 2 r(T)=2r(T)=2r(T)=2 and ‖ T n + 1 ‖ / ‖ T n ‖ T n + 1 / T n ||T^(n+1)||//||T^(n)||\left\|T^{n+1}\right\| /\left\|T^{n}\right\|‖Tn+1‖/‖Tn‖ diverges.
> T:=randmatrix(2,2,entries=rand(-5..5));
    T:=[ch]}[\begin{array}{cc}{4}&{5}\\{-4}&{4}\end{array}
> Digits:=15:
> interface(displayprecision=3):
> m:=30:
> max(op(map(abs,[eigenvalues(T)]))); #r(T)
        2
> nT:=norm(T,2);
    nT:=9/2 + 1/2 * 65 1/2
> S:=evalf(evalm(T/nT));
    p:=evalf(seq( norm(S&^n,2),n=1..m)):
    S:=[cc}0.46890.5861][\begin{array}{cc}{-0.4689}&{-0.4689}\end{array}
> evalf([seq(nT*p[n+1]/p[n],n=1..m-1)]);
[ .4689 , 8.531 , .4689 , 8.531 , .4689 , 8.531 , .4689 , 8.531 , .4689 , 8.531 , .4689 , 8.531 , .4689 , 8.531 , .4689 , 8.531 , .4689 , 8.531 , .4689 , 8.531 , .4689 , 8.531 , .4689 , 8.531 , .4689 , 8.531 , .4689 , 8.531 , .4689 ] > evalf ( [ seq ( nT ∗ p [ n ] ∧ ( 1 / n ) , n = 1 . . m ) ] ) ; [ 8.531 , 2.000 , 3.244 , 2.000 , 2.673 , 2.000 , 2.461 , 2.000 , 2.350 , 2.000 , 2.282 , 2.000 , 2.236 , 2.000 , 2.203 , 2.000 , 2.178 , 2.000 , 2.159 , 2.000 , 2.143 , 2.000 , 2.130 , 2.000 , 2.119 , 2.000 , 2.110 , 2.000 , 2.103 , 2.000 ] [ .4689 , 8.531 , .4689 , 8.531 , .4689 , 8.531 , .4689 , 8.531 , .4689 , 8.531 , .4689 , 8.531 , .4689 , 8.531 , .4689 , 8.531 , .4689 , 8.531 , .4689 , 8.531 , .4689 , 8.531 , .4689 , 8.531 , .4689 , 8.531 , .4689 , 8.531 , .4689 ] >  evalf  seq nT ∗ p [ n ] ∧ ( 1 / n ) , n = 1 . . m ; [ 8.531 , 2.000 , 3.244 , 2.000 , 2.673 , 2.000 , 2.461 , 2.000 , 2.350 , 2.000 , 2.282 , 2.000 , 2.236 , 2.000 , 2.203 , 2.000 , 2.178 , 2.000 , 2.159 , 2.000 , 2.143 , 2.000 , 2.130 , 2.000 , 2.119 , 2.000 , 2.110 , 2.000 , 2.103 , 2.000 ] {:[[.4689","8.531",".4689","8.531",".4689","8.531",".4689","8.531","],[.4689","8.531",".4689","8.531",".4689","8.531",".4689","8.531","],[.4689","8.531",".4689","8.531",".4689","8.531",".4689","8.531","],[.4689","8.531",".4689","8.531",".4689]],[ > " evalf "([seq(nT**p[n]^(1//n),n=1..m)]);],[[8.531","2.000","3.244","2.000","2.673","2.000","2.461","2.000","],[2.350","2.000","2.282","2.000","2.236","2.000","2.203","2.000","],[2.178","2.000","2.159","2.000","2.143","2.000","2.130","2.000","],[2.119","2.000","2.110","2.000","2.103","2.000]]:}\begin{aligned} & {[.4689,8.531, .4689,8.531, .4689,8.531, .4689,8.531,} \\ & .4689,8.531, .4689,8.531, .4689,8.531, .4689,8.531, \\ & .4689,8.531, .4689,8.531, .4689,8.531, .4689,8.531, \\ & .4689,8.531, .4689,8.531, .4689] \\ & >\text { evalf }\left(\left[\mathrm{seq}\left(\mathrm{nT} * \mathrm{p}[\mathrm{n}]^{\wedge}(1 / \mathrm{n}), \mathrm{n}=1 . . \mathrm{m}\right)\right]\right) ; \\ & {[8.531,2.000,3.244,2.000,2.673,2.000,2.461,2.000,} \\ & 2.350,2.000,2.282,2.000,2.236,2.000,2.203,2.000, \\ & 2.178,2.000,2.159,2.000,2.143,2.000,2.130,2.000, \\ & 2.119,2.000,2.110,2.000,2.103,2.000] \end{aligned}[.4689,8.531,.4689,8.531,.4689,8.531,.4689,8.531,.4689,8.531,.4689,8.531,.4689,8.531,.4689,8.531,.4689,8.531,.4689,8.531,.4689,8.531,.4689,8.531,.4689,8.531,.4689,8.531,.4689]> evalf ([seq(nT∗p[n]∧(1/n),n=1..m)]);[8.531,2.000,3.244,2.000,2.673,2.000,2.461,2.000,2.350,2.000,2.282,2.000,2.236,2.000,2.203,2.000,2.178,2.000,2.159,2.000,2.143,2.000,2.130,2.000,2.119,2.000,2.110,2.000,2.103,2.000]
Example 2.3 However, for T := [ − 2 2 5 − 3 ] T := − 2 2 5 − 3 T:=[[-2,2],[5,-3]]T:=\left[\begin{array}{cc}-2 & 2 \\ 5 & -3\end{array}\right]T:=[−225−3] one obtains: r ( T ) = 5 + 41 2 ≃ 5.702 r ( T ) = 5 + 41 2 ≃ 5.702 r(T)=(5+sqrt41)/(2)≃5.702r(T)= \frac{5+\sqrt{41}}{2} \simeq 5.702r(T)=5+412≃5.702 and the sequence ‖ T n + 1 ‖ / ‖ T n ‖ T n + 1 / T n ||T^(n+1)||//||T^(n)||\left\|T^{n+1}\right\| /\left\|T^{n}\right\|‖Tn+1‖/‖Tn‖ converges (to r ( A ) r ( A ) r(A)r(A)r(A) ). Note that the numerical results show that this sequence coverges faster than ‖ T n ‖ 1 / n T n 1 / n ||T^(n)||^(1//n)\left\|T^{n}\right\|^{1 / n}‖Tn‖1/n.
> evalf ( [ seq ( nT ∗ p [ n + 1 ] / p [ n ] , n = 1 . . m − 1 ) ] ) ; [ 5.550 , 5.719 , 5.699 , 5.702 , 5.702 , 5.702 , 5.702 , 5.702 , 5.702 , 5.702 , 5.702 , 5.702 , 5.702 , 5.702 , 5.702 , 5.702 , 5.702 , 5.702 , 5.702 , 5.702 , 5.702 , 5.702 , 5.702 , 5.702 ] > evalf ( [ seq ( nT ∗ p [ n ] ∧ ( 1 / n ) , n = 1 . . m ) ] ) ; [ 6.451 , 5.983 , 5.894 , 5.845 , 5.816 , 5.797 , 5.783 , 5.773 , 5.765 , 5.758 , 5.753 , 5.749 , 5.745 , 5.742 , 5.739 , 5.737 , 5.735 , 5.733 , 5.731 , 5.730 , 5.729 , 5.727 , 5.726 , 5.725 , 5.724 ] > evalf ( [ seq ( nT ∗ p [ n + 1 ] / p [ n ] , n = 1 . . m − 1 ) ] ) ; [ 5.550 , 5.719 , 5.699 , 5.702 , 5.702 , 5.702 , 5.702 , 5.702 , 5.702 , 5.702 , 5.702 , 5.702 , 5.702 , 5.702 , 5.702 , 5.702 , 5.702 , 5.702 , 5.702 , 5.702 , 5.702 , 5.702 , 5.702 , 5.702 ] > evalf seq nT ∗ p [ n ] ∧ ( 1 / n ) , n = 1 . . m ; [ 6.451 , 5.983 , 5.894 , 5.845 , 5.816 , 5.797 , 5.783 , 5.773 , 5.765 , 5.758 , 5.753 , 5.749 , 5.745 , 5.742 , 5.739 , 5.737 , 5.735 , 5.733 , 5.731 , 5.730 , 5.729 , 5.727 , 5.726 , 5.725 , 5.724 ] {:[ > evalf([seq(nT**p[n+1]//p[n]","n=1..m-1)]);],[[5.550","5.719","5.699","5.702","5.702","5.702","5.702","5.702","],[5.702","5.702","5.702","5.702","5.702","5.702","5.702","5.702","],[5.702","5.702","5.702","5.702","5.702","5.702","5.702","5.702]],[ > evalf([seq(nT**p[n]^(1//n),n=1..m)]);],[[6.451","5.983","5.894","5.845","5.816","5.797","5.783","5.773","],[5.765","5.758","5.753","5.749","5.745","5.742","5.739","5.737","],[5.735","5.733","5.731","5.730","5.729","5.727","5.726","5.725","],[5.724]]:}\begin{aligned} & >\operatorname{evalf}([\mathrm{seq}(\mathrm{nT} * \mathrm{p}[\mathrm{n}+1] / \mathrm{p}[\mathrm{n}], \mathrm{n}=1 . . \mathrm{m}-1)]) ; \\ & {[5.550,5.719,5.699,5.702,5.702,5.702,5.702,5.702,} \\ & 5.702,5.702,5.702,5.702,5.702,5.702,5.702,5.702, \\ & 5.702,5.702,5.702,5.702,5.702,5.702,5.702,5.702] \\ & >\operatorname{evalf}\left(\left[\mathrm{seq}\left(\mathrm{nT} * \mathrm{p}[\mathrm{n}]^{\wedge}(1 / \mathrm{n}), \mathrm{n}=1 . . \mathrm{m}\right)\right]\right) ; \\ & {[6.451,5.983,5.894,5.845,5.816,5.797,5.783,5.773,} \\ & 5.765,5.758,5.753,5.749,5.745,5.742,5.739,5.737, \\ & 5.735,5.733,5.731,5.730,5.729,5.727,5.726,5.725, \\ & 5.724] \end{aligned}>evalf([seq(nT∗p[n+1]/p[n],n=1..m−1)]);[5.550,5.719,5.699,5.702,5.702,5.702,5.702,5.702,5.702,5.702,5.702,5.702,5.702,5.702,5.702,5.702,5.702,5.702,5.702,5.702,5.702,5.702,5.702,5.702]>evalf([seq(nT∗p[n]∧(1/n),n=1..m)]);[6.451,5.983,5.894,5.845,5.816,5.797,5.783,5.773,5.765,5.758,5.753,5.749,5.745,5.742,5.739,5.737,5.735,5.733,5.731,5.730,5.729,5.727,5.726,5.725,5.724]

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qquad\qquad
qquad\qquad
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  1. ⋄ ⋄ ^(diamond){ }^{\diamond}⋄ Mira-Cristiana Anisiu, Tiberiu Popoviciu Institute of Numerical Analysis, email: mira@math.ubbcluj.ro
    ⋄ ⋄ ^(diamond){ }^{\diamond}⋄ Valeriu Anisiu, Babeş-Bolyai University, email: anisiu@math.ubbcluj.ro
2007

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