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} 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 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} for f!=0f \neq 0 in the Bochner space L_(E)^(oo)(mu)L_{E}^{\infty}(\mu), where (E,|*|)(E,|\cdot|) is a Banach space and ( X,A,muX, \mathcal{A}, \mu ) a finite measure space. We discuss also the existence of lim_(n rarr oo)||T^(n+1)||//||T^(n)||\lim _{n \rightarrow \infty}\left\|T^{n+1}\right\| /\left\|T^{n}\right\| for continuous linear operators TT in Hilbert spaces.
Key Words: L^(p)L^{p} norms, linear operators, spectral radius.
MSC 2000: 46E30, 47A75
1 A limit involving L^(p)L^{p} and L^(oo)L^{\infty} norms
Let ( X,A,muX, \mathcal{A}, \mu ) be a measure space. If mu\mu is finite and f inL^(oo)(mu)f \in L^{\infty}(\mu), the L^(oo)L^{\infty} norm of the real function ff can be obtained as the limit
{:(1)||f||_(oo)=lim_(p rarr oo)||f||_(p):}\begin{equation*}
\|f\|_{\infty}=\lim _{p \rightarrow \infty}\|f\|_{p} \tag{1}
\end{equation*}
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) > 0a_{p}>0, the equality
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}}
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}, does the limit lim_(p rarr oo)(a_(p+1))/(a_(p))\lim _{p \rightarrow \infty} \frac{a_{p+1}}{a_{p}} exist?
Remark 1.1 There are known several conditions on the sequences a_(n),b_(n)a_{n}, b_{n} insuring that (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 L. They apply for example for Traian Lalescu's sequence [5]: root(n+1)((n+1)!)-root(n)(n!)rarr1//e\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!} \rightarrow 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} and b_(n):=nb_{n}:=n, it follows that 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 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,|\cdot| ) be a Banach space, ( X,A,muX, \mathcal{A}, \mu ) a finite measure space (mu(X) < oo)(\mu(X)<\infty) and f inL_(E)^(oo)(mu)\\{0}f \in L_{E}^{\infty}(\mu) \backslash\{0\}. Then
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}
Proof. Replacing ff by |f|//||f||_(oo)|f| /\|f\|_{\infty}, one may suppose that 0 <= f <= 10 \leq f \leq 1 and ||f||_(oo)=1\|f\|_{\infty}=1. Let us denote
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} .
Then 0 <= r_(p) <= 10 \leq r_{p} \leq 1, so,
{:(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*}
For 0 < a < 10<a<1 we denote A_(a)={x in X:f(x) >= a},B_(a)=X\\AA_{a}=\{x \in X: f(x) \geq a\}, B_{a}=X \backslash A. We have mu(A_(a)) > 0\mu\left(A_{a}\right)>0 because ||f||_(oo)=1\|f\|_{\infty}=1. We show that
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 .
Let us choose bb so that a < b < 1a<b<1. Then A_(b)subeA_(a)A_{b} \subseteq A_{a} and
{:[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}
We obtain for l i m i n f_(p rarr oo)r_(p)\liminf _{p \rightarrow \infty} r_{p} the following estimation
{:[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}
But a in(0,1)a \in(0,1) is arbitrary, so
{:(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*}
From (2) and (3) it follows lim_(p rarr oo)r_(p)=1\lim _{p \rightarrow \infty} r_{p}=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)!=0r(T) \neq 0
Let EE be a Hilbert space and TT a linear continuous operator. Then the spectral radius r(T)r(T) of the operator TT is given by
r(T)=lim_(n rarr oo)||T^(n)||^(1//n)r(T)=\lim _{n \rightarrow \infty}\left\|T^{n}\right\|^{1 / n}
The result in section 1 suggest the following problem: If r(T)!=0r(T) \neq 0, is it true that lim_(n rarr oo)||T^(n+1)||//||T^(n)||\lim _{n \rightarrow \infty}\left\|T^{n+1}\right\| /\left\|T^{n}\right\| does exist?
We mention the following interesting related result due to Kellogg [3],[8,p.240][3],[8, \mathrm{p} .240], which provides an algorithm for finding an eigenvalue for a compact self-adjoint operator.
Theorem 2.1 Let EE be a Hilbert space, TT a compact self-adjoint operator, x_(0)in Ex_{0} \in E such that Tx_(0)!=0T x_{0} \neq 0. Then, for x_(n)=T^(n)x_(0)x_{n}=T^{n} x_{0}, one has that x_(n)!=0x_{n} \neq 0, the sequence ||x_(n+1)||//||x_(n)||\left\|x_{n+1}\right\| /\left\|x_{n}\right\| is increasing and convergent to r > 0r>0 such that either rr or -r-r is an eigenvalue for TT.
In [2, p. 222], the definition of operators of class K\mathcal{K} was given.
Definition 2.1 The linear continuous operator TT is of class K\mathcal{K} if for each x in E,m inN,m >= 2x \in E, m \in \mathbb{N}, m \geq 2 and k in{1,2,dots,m-1}k \in\{1,2, \ldots, m-1\}
{:(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*}
where C_(m,k)C_{m, k} are constants.
Remark 2.1 1. If TT is invertible, the minimal constants C_(m,k)C_{m, k} in (4) must satisfy (see [2, p. 223])
{:(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*}
if TT is normal, C_(m,k)=1C_{m, k}=1 for each mm and kk.
2. For T_(i),i=1,dots,4T_{i}, i=1, \ldots, 4 linear continuous operators on EE, the following two inequalities regarding the spectral radius
{:[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}
have been proved in [4].
We state the following
Conjecture 2.1 If TT is of class K\mathcal{K} and r(T)!=0r(T) \neq 0, then lim_(n rarr oo)||T^(n+1)||//||T^(n)||\lim _{n \rightarrow \infty}\left\|T^{n+1}\right\| /\left\|T^{n}\right\| 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\| \Leftrightarrow\left\|T^{n}\right\|=\|T\|^{n}, for all n inNn \in \mathbb{N}.
Proof. The spectral mapping theorem implies that r(T^(n))=r(T)^(n)r\left(T^{n}\right)=r(T)^{n}, so if 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\left(T^{n}\right) \leq\left\|T^{n}\right\| \leq\|T\|^{n}, hence ||T^(n)||=||T||^(n)\left\|T^{n}\right\|=\|T\|^{n}.
Conversely, if ||T^(n)||=||T||^(n)\left\|T^{n}\right\|=\|T\|^{n}, for all n inNn \in \mathbb{N} then 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\|.
If r(T)=||T||r(T)=\|T\|, obviously ||T^(n+1)||//||T^(n)||=||T||\left\|T^{n+1}\right\| /\left\|T^{n}\right\|=\|T\| and conjecture 2.1 holds. Note also that if TT is normal, then r(T)=||T||r(T)=\|T\|; but TT may not be normal and yet lim_(n rarr oo)||T^(n+1)||//||T^(n)||\lim _{n \rightarrow \infty}\left\|T^{n+1}\right\| /\left\|T^{n}\right\| exists (=r(T))(=r(T)); see ex 2.3.
Remark 2.2 Let TT be the Volterra operator in L^(2)([0,1])L^{2}([0,1]),
(Tx)(t)=int_(0)^(t)xdlambda(T x)(t)=\int_{0}^{t} x \mathrm{~d} \lambda
Then 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 . .. where alpha\alpha is the smallest positive root of the equation ( {:e^(a)+e^(-a))cos(a)=-2\left.e^{a}+e^{-a}\right) \cos (a)=-2, see [6, p. 259]. The norms ||T^(n)||\left\|T^{n}\right\| are more difficult to find for n > 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)=1r(T)=1 for which lim_(n rarr oo)||T^(n+1)||//||T^(n)||\lim _{n \rightarrow \infty}\left\|T^{n+1}\right\| /\left\|T^{n}\right\| does not exist.
Let T=[[1,1],[0,-1]].\operatorname{Let} T=\left[\begin{array}{cc}
1 & 1 \\
0 & -1
\end{array}\right] .
Then 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.. In this case, ||T^(n)||^(1//n)rarr r(T)\left\|T^{n}\right\|^{1 / n} \rightarrow r(T) but ||T^(n+1)||//||T^(n)||\left\|T^{n+1}\right\| /\left\|T^{n}\right\| diverges.
Actually, this behaviour is almost generic. We give below the Maple code computing r(T)r(T) and the sequence ||T^(n+1)||//||T^(n)||\left\|T^{n+1}\right\| /\left\|T^{n}\right\| for a linear operator in R^(d)(d=2)\mathbb{R}^{d}(d=2) with randomly selected entries from {-5,-4,dots,4,5}\{-5,-4, \ldots, 4,5\}. Note that for d > 4d>4 this can be done only approximately.
We display the values of the sequences ||T^(n+1)||//||T^(n)||\left\|T^{n+1}\right\| /\left\|T^{n}\right\| and ||T^(n)||^(1//n)\left\|T^{n}\right\|^{1 / n}.
Example 2.2 The operator T=[[4,5],[-4,4]]T=\left[\begin{array}{cc}4 & 5 \\ -4 & 4\end{array}\right] has r(T)=2r(T)=2 and ||T^(n+1)||//||T^(n)||\left\|T^{n+1}\right\| /\left\|T^{n}\right\| 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]]:}\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}
Example 2.3 However, for T:=[[-2,2],[5,-3]]T:=\left[\begin{array}{cc}-2 & 2 \\ 5 & -3\end{array}\right] one obtains: r(T)=(5+sqrt41)/(2)≃5.702r(T)= \frac{5+\sqrt{41}}{2} \simeq 5.702 and the sequence ||T^(n+1)||//||T^(n)||\left\|T^{n+1}\right\| /\left\|T^{n}\right\| converges (to r(A)r(A) ). Note that the numerical results show that this sequence coverges faster than ||T^(n)||^(1//n)\left\|T^{n}\right\|^{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]]:}\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}
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qquad\qquad
qquad\qquad
[3] O.D. Kellogg. On the existence of closure of sets of characteristic functions, Math. Ann., 86 (1922), 14-17.
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