Electronic Journal of Differential Equations, Vol. 2022 (2022), No. 29, pp. 1–7.

ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu

BOUNDS FOR THE SPECTRAL RADIUS OF POSITIVE

OPERATORS

MOHAMMED SAID EL KHANNOUSSI, ABDERRAHIM ZERTITI

Abstract. We use partially ordered spaces to obtain bounds for the spectral

radius of positive linear operators, and to improve the corresponding fixed
point theorems.

1. Introduction

In [9], methods of partially ordered spaces are used for obtaining bounds for the
spectral radius of positive operators. In Theorem 2.2 below we present a result
which generalizes those in [9], by weakening the hypothesis that the operator under
consideration is bounded by some u0 with “u0 is a quasi-interior element of the
cone”. In Theorem 2.5 we improve a variant of the well known contraction mapping
principle given by Krasnosel’skii and Zabreiko in [9] by requiring only estimates for
the difference of the values of the operator on comparable elements. Also, we use
the previous results to prove that the operator given in Theorem 2.5 satisfies all
the conditions of the converse to the Banach contraction theorem.

2. Main results

Let (E, ‖ · ‖E) be a real Banach space and P be a nonempty closed convex set
in E. P is called a cone if it satisfies the following two conditions:

(i) if x ∈ P and λ ≥ 0, then λx ∈ P ;
(ii) if x ∈ P and −x ∈ P , then x = θ, where θ is the zero element in E.

A cone P is said to be reproducing if E = P − P , i.e., every element x ∈ E can
be represented in the form x = u− v where u, v ∈ P . The cone P defines a linear
ordering in E by

x ≤ y if and only if y − x ∈ P.
The cone P is said to be normal if there exists a constant K > 0 such that

θ ≤ x ≤ y =⇒ ‖x‖ ≤ K‖y‖, ∀x, y ∈ P.

For L : E → E, a bounded linear operator, we define its spectral radius by

σ(L) = lim
n→+∞

‖Ln‖1/n.

2020 Mathematics Subject Classification. 47H10, 55M20, 54H25.
Key words and phrases. Contraction mapping principle; fixed point theorems; spectral radius;

successive approximations, unique solutions.
©2022. This work is licensed under a CC BY 4.0 license.

Submitted March 4, 2021. Published April 8, 2022.

1



2 M. S. KHANNOUSSI, A. ZERTITI EJDE-2022/29

The linear operator L is called positive if it transforms the cone P into itself. It
is not hard to see that it follows from x ≤ y for arbitrary elements x, y ∈ E that
Lx ≤ Ly.

Let u0 be a non-zero element of P . Following Krasnosel’skii [8], we say that the
positive operator L is u0-bounded above if there exists a natural number m such
that for every non-zero element x ∈ P there exists a positive number β(x) such
that

Lmx ≤ β(x)u0.

Arguing as in [2, Lemma 1.4.2] we have the following result.

Theorem 2.1. Let P be a normal reproducing cone. If the positive linear operator
L is u0-bounded above then there exist constants τ > 0 and m > 0 such that for
any x ∈ E there exist an element y ∈ P and a constant β(x) > 0 such that

x ≤ y, Lmy ≤ β(x)u0, ‖y‖ ≤ τ‖x‖, ‖β(x)u0‖ ≤ τ‖x‖.
Proof. Since P is a reproducing cone, then for x ∈ E there exist y, z ∈ P such that
x = y − z. From the u0-boundedness above of L there exists an integer m > 0
such that Lmy ≤ β(x)u0 for some constant β(x) > 0. From which it follows that
E = ∪∞n=1En, where

En =
{
x ∈ E : there are y ∈ P and β(x) > 0 such that x ≤ y,
Lmy ≤ β(x)u0, ‖y‖ ≤ n‖x‖, ‖β(x)u0‖ ≤ n‖x‖

}
,

for n = 1, 2, 3, . . . . By the Baire-Hausdorff’s Theorem (that says nonempty com-
plete metric spaces are second Baire sets) there exist positive integer n1, x0 ∈ E
and R > r > 0 satisfying

B0 = {x ∈ E : r < ‖x− x0‖ < R} ⊂ En1
.

Let −x0 = y0− z0 where y0, z0 ∈ P . Take positive constant β0 and positive integer
n2 satisfying Lmy0 ≤ β0u0, ‖y0‖ ≤ n2‖x0‖, and ‖β0u0‖ ≤ n2‖x0‖.

Let B = {x ∈ E : r < ‖x‖ < R}, and choose some integer n3 satisfying

n3 > n1 +
1

r
(n1 + n2)‖x0‖.

Now we prove that B ⊂ En3
. Indeed, for any x ∈ B, we have y = x0 + x ∈ B0,

then there exists a sequence {xi} ⊂ En1
such that xi → y as i → ∞. Clearly, we

can assume that xi ∈ B0 for i = 1, 2, 3, . . . . Take elements yi ∈ P and constants
βi > 0 such that xi ≤ yi, L

myi ≤ βiu0, ‖yi‖ ≤ n1‖xi‖, and ‖βiu0‖ ≤ n1‖xi‖, then
we obtain xi − x0 ≤ yi + y0 and

‖yi + y0‖ ≤ n1‖xi‖+ n2‖x0‖
≤ (n1 + n2)‖x0‖+ n1‖xi − x0‖

≤
[
(n1 + n2)

‖x0‖
r

+ n1

]
‖xi − x0‖

≤ n3‖xi − x0‖.
On the other hand we have Lm(yi + y0) = Lmyi + Lmy0 ≤ (βi + β0)u0 and

‖(βi + β0)u0‖ ≤ n1‖xi‖+ n2‖x0‖ ≤ n3‖xi − x0‖,
from which it follows that xi − x0 ∈ En3 for n = 1, 2, 3, . . . . From the fact that
xi − x0 → y − x0 as i→∞ we obtain x ∈ En3

. Therefore B ⊂ En3
. Clearly, from

x ∈ En3
, we can easily prove that tx ∈ En3

, for all t ≥ 0. Consequently, E = En3
.



EJDE-2022/29 BOUNDS FOR THE SPECTRAL RADIUS 3

Finally, we show that E = E3n3
. Taking x ∈ E such that x 6= θ, then there

exists x1 ∈ En3 satisfying

‖x− x1‖ <
1

2
‖x‖.

Since x1 ∈ En3 , there exist y1 ∈ P and β1 > 0 such that

x1 ≤ y1, Lmy1 ≤ β1u0, ‖y1‖ ≤ n3‖x1‖, ‖β1u0‖ ≤ n3‖x1‖.

Similarly, there exist x2 ∈ En3 , y2 ∈ P , and β2 > 0 such that

‖x− x1 − x2‖ <
1

22
‖x‖, x2 ≤ y2,

Lmy2 ≤ β2u0, ‖y2‖ ≤ n3‖x2‖, ‖β2u0‖ ≤ n3‖x2‖.

Inductively, we find sequences {xk} ⊂ En3 , {yk} ⊂ P , and {βk} > 0 satisfying

‖x− x1 − x2 − · · · − xk‖ <
1

2k
‖x‖, xk ≤ yk,

Lmyk ≤ βku0, ‖yk‖ ≤ n3‖xk‖, ‖βku0‖ ≤ n3‖xk‖

for k = 1, 2, 3 . . . . Clearly, x =
∑∞
k=1 xk and

‖xk‖ ≤ ‖x−
k−1∑
i=1

xi‖+ ‖x−
k∑
i=1

xi‖ <
3‖x‖
2k

, k = 1, 2, . . . .

From which it follows that
∞∑
k=1

‖yk‖ ≤ n3
∞∑
k=1

‖xk‖ ≤ 3n3‖x‖ <∞,

∞∑
k=1

‖βku0‖ ≤ n3
∞∑
k=1

‖xk‖ ≤ 3n3‖x‖ <∞.

Consequently the series
∑∞
k=1 yk and

∑∞
k=1 βk converge to and element y ∈ P and

a constant β > 0, respectively. Clearly

x =

∞∑
k=1

xk ≤
∞∑
k=1

yk = y, ‖y‖ ≤
∞∑
k=1

‖yk‖ ≤ 3n3‖x‖.

On the other hand by using that a linear operator which maps a reproducing cone
into a normal cone must be continuous (see [6]), Lm is continuous. Then

Lmy = Lm
( ∞∑
k=1

yk

)
=

∞∑
k=1

Lmyk ≤
∞∑
k=1

βku0 = βu0,

‖βu0‖ ≤
∞∑
k=1

‖βku0‖ ≤ 3n3‖x‖.

Therefore, x ∈ E3n3 , which implies that E = E3n3 . �

Our main result reads as follows.

Theorem 2.2. Let (E,P ) be an ordered Banach space with normal, reproducing
cone P . Let L : E → E be u0-bounded above positive linear operator. If there exist
a positive integer n and a λ ≥ 0 such that Lnu0 ≤ λu0, then

r(L) ≤ λ1/n



4 M. S. KHANNOUSSI, A. ZERTITI EJDE-2022/29

Proof. First, suppose that λ > 0. By Theorem 2.1 there exist positive constants τ
and m such that for every x ∈ E there exist y ∈ P, β(x) > 0 such that x ≤ y,
Lmy ≤ β(x)u0, ‖y‖ ≤ τ‖x‖, ‖β(x)u0‖ ≤ τ‖x‖.

Taking z = y − x ∈ P , then by Theorem 2.1 we can find an element ω ∈ P
and a constant β(z) > 0 such that z ≤ ω, Lmω ≤ β(z)u0, ‖ω‖ ≤ τ‖z‖, and
‖β(z)u0‖ ≤ τ‖z‖. If u = y + z, then −u ≤ x ≤ u and

Lmu = Lmy + Lmz ≤ β(x)u0 + β(z)u0.

Since

‖β(x)u0 + β(z)u0‖ ≤ ‖β(x)u0‖+ ‖β(z)u0‖
≤ τ‖x‖+ τ‖z‖
≤ τ‖x‖+ τ(‖y‖+ ‖x‖)
≤ τ‖x‖+ τ(τ‖x‖+ ‖x‖)
≤ τ(τ + 2)‖x‖,

there is a constant σ > 0 such that for any x ∈ E there exist u(x) ∈ P and β′(x) > 0
such that

−u(x) ≤ x ≤ u(x), Lmu(x) ≤ β′(x)u0, ‖β′(x)u0‖ ≤ σ‖x‖.
Therefore, there is a constant β (β > σ

‖u0‖ ) such that for every x ∈ E satisfying

‖x‖ ≤ 1, we can find an element u(x) ∈ P such that

−u(x) ≤ x ≤ u(x) and Lmu(x) ≤ βu0.
Hence

−βu0 ≤ −Lmu(x) ≤ Lmx ≤ Lmu(x) ≤ βu0,
and inductively, for any p ≥ 1,

−βλpu0 ≤ −Lm+npu(x) ≤ Lm+npx ≤ Lm+npu(x) ≤ βλpu0,
from which it follows that

0 ≤ Lm+npx+ βλpu0 ≤ 2βλpu0.

Since P is normal we obtain

‖Lm+npx+ βλpu0‖ ≤ 2Kβλp‖u0‖,
where K is the normal constant of P . Then for every x ∈ E, satisfying ‖x‖ ≤ 1
and p ≥ 1 we have

‖Lm+npx‖ ≤ (2K + 1)βλp‖u0‖.
From which it follows that

‖Lm+np‖ ≤ (2K + 1)βλp‖u0‖.
Then

‖Lm+np‖
1

m+np ≤ (2K + 1)
1

m+np β
1

m+npλ
p

m+np ‖u0‖
1

m+np .

Letting p→∞, for all ε > 0 there exists N , such that for all p ≥ N , we have

r(L) ≤ ‖Lm+np‖
1

m+np ≤ λ 1
n + ε.

By letting ε→ 0 we have

r(L) ≤ λ 1
n .

Now, if λ = 0 then for any ε > 0 we have Lnu0 ≤ εnu0 and from what has already
been proved we have r(L) ≤ ε. This completes the proof. �



EJDE-2022/29 BOUNDS FOR THE SPECTRAL RADIUS 5

Remark 2.3. Theorem 2.2 generalizes Theorem 2 in Zabreiko, Krasnosel’skii and
Stetsenko [9], where the element u0 is supposed to be a “quasi-interior element of
the cone”, a more restrictive condition.

Remark 2.4. If we assume that P is a solid cone and u0 is a quasi interior element
of P (as in [9]), then it follows from [2, Theorem 1.4.1] that u0 belongs to the
interior of the cone P .

As a consequence of Theorem 2.2, by using a result by Krasnosel’skii and Zabreiko
[6, Theorem 49.3, p. 320] (see also [3, Theorem 3.1.14]), we have the following result.

Theorem 2.5. Let (E,P ) be an ordered Banach space with normal reproducing
cone P and A : E → E be an operator. Suppose that there exists an u0-bounded
above linear positive operator L : E → E such that

− L(x− y) ≤ A(x)−A(y) ≤ L(x− y), x, y ∈ E, x ≥ y. (2.1)

If Lnu0 ≤ λu0 for some λ ∈ [0, 1), an integer n and an element u0 of P \ {0}
then A has a unique fixed point x∗ in E and for any x0 ∈ E, if xn = Axn−1(n =
1, 2, 3, . . . .), then xn → x∗ as n→∞.

Proof. It follows from Theorem 2.2 that r(L) < 1. �

It should be remarked above that [6, Theorem 49.3, p. 320] remains valid if we
replace condition (2.1) by the condition

−L1(x− y) ≤ A(x)−A(y) ≤ L2(x− y), x, y ∈ E, x ≥ y,
where L1 and L2 are positive linear operators with σ(L1 + L2) < 1. Here, we
observe that even if A is a linear operator the latter condition cannot be replaced
by the inequalities σ(L1) < 1 and σ(L2) < 1 (See the remark after the proof of
[6, Theorem 49.3, p. 320]). By using Theorem 2.5 we can solve this problem if we
impose new conditions concerning every operators L1 and L2 as follows.

Corollary 2.6. Let (E,P ) be an ordered Banach space with normal reproducing
cone P and A : E → E be an operator. Suppose that there exist linear positive
operators L1, L2 : E → E such that

− L1(x− y) ≤ A(x)−A(y) ≤ L2(x− y), x, y ∈ E, x ≥ y. (2.2)

Assume that L1 + L2 is u0-bounded above, then if L1u0 ≤ λ1u0 and L2u0 ≤ λ2u0
for some λ1, λ2 ≥ 0 satisfying 0 ≤ λ1 + λ2 < 1, then A has a unique fixed point x∗

in E.

Proof. It is easily seen that

−(L1 + L2)(x− y) ≤ A(x)−A(y) ≤ (L1 + L2)(x− y), x, y ∈ E, x ≥ y.
Then all conditions of Theorem 2.5 are satisfied. �

Remark 2.7. If A satisfies the conditions of any of the theorems above, then for
each y ∈, the equation x = Ax+ y has a unique solution.

Next, we use the following converse to the Banach contraction theorem [5].

Theorem 2.8. Let X be a metrizable topological space and let its topology be gen-
erated by the metric ρ. Then for each λ ∈ (0, 1) there exists a metric ρλ on X,
complete if ρ is complete, such that f is a ρλ- contraction if and only if

(i) for some ξ ∈ X, f(ξ) = ξ;



6 M. S. KHANNOUSSI, A. ZERTITI EJDE-2022/29

(ii) fn(x)→ ξ as n→∞ for all x ∈ X;
(iii) there exists an open neighborhood U of ξ such that fn(U) → {ξ}, which

implies that given any neighborhood V of ξ there exists an integer n(V ) > 0
such that fn(U) ⊂ V for all n ≥ n(V ).

As a consequence of the previous results we obtain the following statement.

Theorem 2.9. Let (E,P ) be an ordered Banach space with normal reproducing
cone P and A : E → E be an operator. Suppose that there exists an u0-bounded
above linear positive operator L : E → E such that

− L(x− y) ≤ A(x)−A(y) ≤ L(x− y), x, y ∈ E, x ≥ y. (2.3)

If Lu0 ≤ λu0 for some λ ∈ [0, 1), then for every µ ∈ (0, 1) there exists a metric dµ
such that A is a dµ−contraction. This means that for x, y ∈ E,

dµ(A(x), A(y)) ≤ λdµ(x, y).

Proof. We shall prove that A satisfies all the hypotheses of Theorem 2.8. Indeed,
the hypotheses (i) and (ii) can easly derived from Theorem 2.5. Then it remains
for us to prove (iii).

Take U = {x : ‖x − ξ‖ < 1} where ξ is the unique fixed point of A. Then by
using the same arguments as in the proof of Theorem 2.2, there exist constants
β > 0 and m > 0 such that for every x ∈ U an element u ∈ P can be found such
that −u ≤ x− ξ ≤ u and Lmu ≤ βu0. It follows from the inequalities

x ≥ 1

2
(x+ ξ − u), ξ ≥ 1

2
(x+ ξ − u)

and from (2.3) that

−L
(x− ξ + u

2

)
≤ A(x)−A

(x+ ξ − u
2

)
≤ L

(x− ξ + u

2

)
, (2.4)

−L
(ξ − x+ u

2

)
≤ ξ −A

(x+ ξ − u
2

)
≤ L

(ξ − x+ u

2

)
. (2.5)

By subtracting (2.5) from (2.4), we have

−Lu ≤ A(x)− ξ ≤ Lu.
By repeating this argument m+n times for any integer n, we obtain the inequality

−Lm+nu ≤ Am+n(x)− ξ ≤ Lm+nu.

By using the hypothesis of the theorem we obtain

−βλnu0 ≤ −Lm+nu ≤ Am+n(x)− ξ ≤ Lm+nu ≤ βλnu0,
from which it follows that

‖Am+n(x)− ξ‖ ≤ (2K + 1)βλn‖u0‖. (2.6)

Since λ < 1, for any neighborhood V of ξ we can choose n(V ) so large that for all
n ≥ n(V ),

{x : ‖x− ξ‖ < (2K + 1)βλn‖u0‖} ⊂ V,
which implies by (2.6) that Am+n(U) ⊂ V . Consequently for every n ≥ n(V ) +m
we have An(U) ⊂ V . Thus (iii) is proved. This completes the proof. �

Conclusion. This article generalizes and improves well-known results by Kras-
nosel’skii, Zabreiko and Stetsenko, and other authors. Note that the present results
can be used for generalizing other results in the literature.



EJDE-2022/29 BOUNDS FOR THE SPECTRAL RADIUS 7

References

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York, 1988.
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Publishers, New York, 2004.

[4] M. S. El Khannoussi, A. Zertiti; Topological methods in the study of positive solutions for
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2016 (2016) No. 171, 1–13.
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(1967), 73–76.

[6] M. A. Krasnosel’skii, P. P. Zabreiko; Geometrical Methods of Nonlinear Analysis, Springer-
Verlag, Berlin 1984

[7] M. G. Krein, M. Rutman; Linear operators leaving invariant a cone in a Banach space, Amer.

Math. Soc. Transl, 10 (1962) , 1–128.
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operators. Translated from Matematicheskie Zametki, 1, (4) (1967), 461–468.

Mohammed Said El Khannoussi
Université Abdelmalek Essaadi, Faculté des sciences, Département de Mathématiques,

BP 2121, Tétouan, Morocco
Email address: said 774@hotmail.com

Abderrahim Zertiti
Université Abdelmalek Essaadi, Faculté des sciences, Département de Mathématiques,

BP 2121, Tétouan, Morocco

Email address: abdzertiti@hotmail.fr


	1. Introduction
	2. Main results
	Conclusion

	References