Mem. Differential Equations Math. Phys. 11(1997), 163–167

L. F. Rakhmatullina

THE UPPER ESTIMATE OF THE SPECTRAL RADIUS OF THE ISOTONIC OPERATOR IN THE SPACE OF CONTINUOUS FUNCTIONS

(Reported on February 24, 1997)

For the isotonic compact integral operator

(Ax)(t) =^{def} ∫_{a}^{b} K(t, s)x(s)ds (K(t, s) ≥ 0, (t, s) ∈ [a, b] × [a, b])

in the space ℂ[a, b] of continuous on [a, b] functions the following assertion holds: the spectral radius ρ(A) of A : ℂ[a, b] → ℂ[a, b] is less than 1 if and only if there exists a v ∈ ℂ[a, b] such that

v(t) ≥ 0, r(t) =^{def} v(t) − (Av)(t) ≥ 0, t ∈ [a, b].

Besides, the set of zeros of r is at most countable. This assertion plays an important role in the theory of differential equations. In the theory of functional differential equations, there arises the necessity in the estimate ρ(A) < 1 for the isotonic operator A : ℂ[a, b] → ℂ[a, b] which is not integral [1]. The above assertion is a corollary of G.G. Islamov’s theorem [2, 3]. In accordance with this theorem, the inequality ρ(A) < 1 for a general isotonic compact linear operator A : ℂ[a, b] → ℂ[a, b] holds if and only if there exists a v ∈ ℂ[a, b] such that

v(t) ≥ 0, r(t) =^{def} v(t) − (Av)(t) ≥ 0, t ∈ [a, b],

the set of zeros of r being at most countable, and besides r(t) > 0 at some special points of [a, b], the so-called ”singular points”.

The refusal from the compactness of A and the weakening of the demand concerning r became possible at the expense of some properties of A. We offer some development of the ideas proposed in [4].

Let T ⊂ ℝ^{1} be a Lebesgue-measurable set, mes T ≤ +∞, ℂ be the Banach space of continuous bounded functions x : T → ℝ^{1}, ∥x∥_{ℂ} = sup_{t∈T}|x(t)|. Let further γ : T → ℝ^{1} be

continuous, γ(t) > 0, t ∈ T, ℂ^{γ} be a Banach space of the functions x : T → ℝ^{1} such that \frac{x}{γ} ∈ ℂ, ∥x∥_{ℂ^{γ}} = sup_{t∈T} \frac{|x(t)|}{γ(t)} . The linear operator A : ℂ^{γ} → ℂ^{γ} is said to be isotonic, if

(Ax)(t) ≥ 0, t ∈ T, for any x ∈ ℂ^{γ} such that x(t) ≥ 0, t ∈ T.

Lemma. Let A : ℂ^{γ} → ℂ^{γ} be linear, bounded and isotonic. ρ(A) < 1 if and only if there exists v ∈ ℂ^{γ} such that

inf_{t∈T} \frac{v(t)}{γ(t)} > 0, inf_{t∈T} \frac{v(t) − (Av)(t)}{γ(t)} > 0.

1991 Mathematics Subject Classification. 47A10.

Key words and phrases. Isotonic operator, Spectral radius, boundary value problem.

Note that for the case T = [a, b], γ(t) ≡ 1 this assertion is well known. Proof. The necessity is obtained by taking the solution of the equation x − Ax = γ in the capacity of v.

To prove the sufficiency, let us introduce in the space ℂ^{γ} a new norm ∥x∥_{v} = sup_{t∈T} \frac{|x(t)|}{v(t)} .

Then for the norm ∥A∥_{v} of A with respect to ∥ · ∥_{v} we have ∥A∥_{v} = ∥Av∥_{v}. Since ∥Av∥_{v} < 1, by the assertion we obtain ρ(A) ≤ ∥A∥_{v} < 1. ■

The demands concerning v and r = v − Av might be weakened at the expense of additional assumptions on the properties of A. One of such properties is

Property M. We will say that a linear operator A : ℂ^{γ} → ℂ^{γ} has Property M, if

inf_{t∈T} \frac{(Ax)(t)}{γ(t)} > 0 for any x ∈ ℂ^{γ} such that x(t) ≥ 0, x(t) ≢ 0, t ∈ T.

Theorem 1. Let a linear bounded A : ℂ^{γ} → ℂ^{γ} have Property M. Let further there exist v ∈ ℂ^{γ} such that

inf_{t∈T} \frac{v(t)}{γ(t)} > 0, r(t) =^{def} v(t) − (Av)(t) ≥ 0, r(t) ≢ 0, t ∈ T.

Then ρ(A) < 1.

Proof. The proof is needed only in the case inf_{t∈T} \frac{r(t)}{γ(t)} = 0. Applying A to the both

parts of the equality v − Av = r, we get Av − A^{2}v = Ar. From this and the inequality v(t) − (Av)(t) ≥ 0 we have

r_{1}(t) =^{def} v(t) − (A^{2}v)(t) ≥ (Ar)(t).

Consequently, inf_{t∈T} \frac{r_{1}(t)}{γ(t)} > 0. Because of Lemma, ρ(A^{2}) < 1. Thus

ρ(A) = \sqrt{ρ(A^{2})} < 1. ■

Remark 1. It is impossible to weaken the condition of Lemma about v in the presence of Property M. Indeed, from r(t) ≥ 0, there follow

\frac{v(t)}{γ(t)} ≥ \frac{(Av)(t)}{γ(t)} and inf_{t∈T} \frac{v(t)}{γ(t)} ≥ inf_{t∈T} \frac{(Av)(t)}{γ(t)} > 0,

if v(t) ≥ 0, v(t) ≢ 0.

Property N. We will say that a linear operator A : ℂ^{γ} → ℂ^{γ} has Property N, if there exist a measurable set Δ ⊂ T and an element φ ∈ ℂ^{γ} such that

φ(t) ≥ 0, φ(t) ≢ 0, t ∈ T, inf_{t∈Δ} \frac{φ(t) − 2(Aφ)(t)}{γ(t)} > 0.

This property is common for some operators arising in studying multipoint boundary value problems and makes it possible to weaken the conditions of Lemma with respect to v as one can see by the following assertion.

Theorem 2. Let a linear bounded isotonic A : ℂ^{γ} → ℂ^{γ} have Property N. Let further there exist v ∈ ℂ^{γ} such that

v(t) ≥ 0, t ∈ T, inf_{t∈T \Δ} \frac{v(t)}{γ(t)} > 0;

r(t) =^{def} v(t) − (Av)(t) ≥ 0, t ∈ T, inf_{t∈T \Δ} \frac{r(t)}{γ(t)} > 0.

Then ρ(A) < 1.

The proof consists in constructing the bases of v and φ of a function satisfying the conditions of Lemma. Such will be the function v_{ε} = v + ε(φ − aφ) with an ε > 0.

Property MN. We will say that a linear A : ℂ^{γ} → ℂ^{γ} has Property MN, if it has Property N and inf_{t∈T \Δ} \frac{(Ax)(t)}{γ(t)} > 0 for any x ∈ ℂ^{γ} such that x(t) ≥ 0, x(t) ≢ 0, t ∈ T.

Theorem 3. Let a linear bounded isotonic A : ℂ^{γ} → ℂ^{γ} have Property MN. Let further there exist v ∈ ℂ^{γ} such that

v(t) ≥ 0, t ∈ T, inf_{t∈T \Δ} \frac{v(t)}{γ(t)} > 0;

r(t) =^{def} v(t) − (Av)(t) ≥ 0, r(t) ≢ 0, t ∈ T.

Then ρ(A) < 1.

The proof can be obtained by using the scheme of the proof of Theorem 1 and by replacing T by T \Δ and substituting the reference to Theorem 2.

Remark 2. Due to Lemma, the conditions of Theorems 1, 2 and 3 with respect to v and r are necessary for the estimate ρ(A) < 1.

Corollary follows from Theorem 2 of [4].

Let T = [a, b] and A : ℂ^{γ} → ℂ^{γ} be linear, bounded and isotonic. Let further the following conditions be satisfied: there exist the points t_{1}, …, t_{k} ∈ [a, b] such that (Ax)(t_{i}) = 0, i = 1, …, k, for any x ∈ ℂ^{γ}. Then ρ(A) < 1 if and only if there exists v ∈ ℂ^{γ} such that v(t) > 0 and r(t) > 0 for t ∈ [a, b]\{t_{1}, …, t_{k}}.

In this case, the operator A has Property N. Really, if we take as Δ the union of neighborhoods of the points t_{1}, …, t_{k} such that in these neighborhoods the inequality \frac{(Aγ)(t)}{γ(t)} ≤ q < \frac{1}{2} holds, then

inf_{t∈Δ} \frac{γ(t) − 2(Aγ)(t)}{γ(t)} > 0.

Example. Consider the boundary value problem

\begin{matrix}x^{(n)}(t) + ∫_{a}^{b} x(s)d_{s}r(t, s) = f(t), n ≥ 2, t ∈ [a, b], \\ x^{(i)}(a) = 0, i = 0, …, n − 2, x(b) = 0 \end{matrix}  (1)

under the assumption that r(t, ·) does not decrease on [a, b] for almost all t ∈ [a, b], r(·, s) is summable on [a, b] for any s ∈ [a, b] and f(·) is summable on [a, b]. A solution of (1) is understood to be a function x with absolutely continuous derivative of the (n−1)-th order which satisfy both the boundary value conditions and the equation almost everywhere on [a, b].

We write

(Ax)(t) = − ∫_{a}^{b} G_{0}(t, s) ∫_{a}^{b} x(τ)d_{τ} r(s, τ)ds,  (2)

g(t) = ∫_{a}^{b} G_{0}(t, s)f(s)ds,

where G_{0}(t, s) is the Green function of the problem

x^{(n)}(t) = z(t), x^{(i)}(a) = 0, i = 0, …, n − 2, x(b) = 0.

The operator A : ℂ[a, b] → ℂ[a, b] defined by (2) is isotonic since G_{0}(t, s) < 0 in the square (a, b) × (a, b). Besides, (Ax)(a) = (Ax)(b) = 0 for any x ∈ ℂ[a, b]. The function g and the values of A on continuous functions are functions with absolutely continuous derivative of the (n − 1)-th order. Thus the equation

x = Ax + g

in the space ℂ[a, b] is equivalent to the problem (1). Therefore the inequality ρ(A) < 1 guarantees unique solvability of the problem (1) for any summable f.

Let

v(t) = (t − a)^{n−1}(b − t) = −n! ∫_{a}^{b} G_{0}(t, s)ds.

Then

r(t) = v(t) − (Av)(t) = − ∫_{a}^{b} G_{0}(t, s)[n! − ∫_{a}^{b} (τ − a)^{n−1}(b − τ)d_{τ} r(s, τ)]ds.

Thus r(t) > 0, t ∈ (a, b), if almost everywhere on [a, b]

∫_{a}^{b} (τ − a)^{n−1}(b − τ)d_{τ} r(t, τ) ≤ n!  (3)

and besides, the inequality is strict on a set of positive measure. Consequently, because of Corollary of Theorem 2 we have the estimate ρ(A) < 1.

The solution x of the problem (1) has the representation

x(t) = ∫_{a}^{b} G(t, s)f(s)ds,

where G(t, s) is the Green function of this problem [1]. From the equality

∫_{a}^{b} G(t, s)f(s)ds = g(t) + (Ag)(t) + (A^{2}g)(t) + ⋯

it follows that x(t) does not admit positive values if f(t) ≥ 0. Therefore the inequality (3) guarantees the inequality G(t, s) ≤ 0 in the square (a, b) × (a, b).

In the case of the equation with concentrated deviation of the argument

x^{(n)}(t) + p(t)x[h(t)] = f(t),
x(ξ) = 0, if ξ ∉ [a, b],

under the assumption that p(t) is bounded, p(t) ≥ 0, and h(t) is measurable, the inequality (3) takes the form

p(t)σ_{h}(t)[h(t) − a]^{n−1}[b − h(t)] ≤ n!,

where

σ_{h}(t) = \begin{cases}1, & if h(t) ∈ [a, b], \\ 0, & if h(t) ∉ [a, b]. \end{cases}

Acknowledgement

The research described in this publication was possible in part by Grant No 96-01-01613 of the Russian Foundation for Basic Research.

References

1. N. V. Azbelev and L. F. Rakhmatullina, Theory of linear abstract functional differential equations and applications. Mem. Differential Equations Math. Phys. 8(1996), 1–102.

2. G. Islamov, On an estimate of the spectral radius of the linear positive compact operator. (Russian) In: Funktsional’no-differentsial’nye Uravneniya i Kraevye Zadachi. Perm, 1977, 119–122.

3. G. Islamov, On an upper estimate of the spectral radius. (Russian) Dokl. Akad. Nauk SSSR 322(1992), No. 5, 836–838.

4. N. Azbelev and L. Rakhmatullina, On an upper estimate of the spectral radius of the linear operator in the space of continuous functions. (Russian) Izv. Vyssh. Uchebn. Zaved. Mat. 1996, No. 11.

Author’s address: Porm Politechnical Institute 29^{a}, Komsomolsky ave., GSP-45, Perm 614600 Russia