PUBLICATIONS DE L’INSTITUT MATHÉMATIQUE Nouvelle série, tome 83(97) (2008), 49–55 DOI: 10.2298/PIM0897049H

APPLICATION OF THE MEAN ERGODIC THEOREM TO CERTAIN SEMIGROUPS

Gerd Herzog and Christoph Schmoeger

Communicated by Stevan Pilipović

Abstract. We study the asymptotic behaviour of solutions of the Cauchy problem u′ = (∑_{j=1}^{n}(A_{j} + A_{j}^{−1}) − 2nI)u, u(0) = x as t → ∞, for invertible isometries A_{1}, …, A_{n}.

1. Introduction

Let E be a complex Banach space, L(E) the Banach algebra of all bounded linear operators on E, and let A_{1}, …, A_{n} ∈ L(E) be invertible, pairwise commuting, and such that ∥A_{k}∥ = ∥A_{k}^{−1}∥ = 1 (k = 1, …, n). Let T_{1}, …, T_{n} ∈ L(E) be defined by T_{k} = A_{k} + A_{k}^{−1} − 2I, and let T = T_{1} + ⋯ + T_{n}. The aim of this paper is to clear the asymptotic behaviour of the Cauchy problem

(1.1)  u′(t) = T u(t), u(0) = u_{0}

that is of t ↦ exp(tT)u_{0} for t → ∞. Such problems occur in a natural way by semidiscretization of the parabolic Cauchy problem v_{t} = Δv, v(0, x) = v_{0}(x): For example, if v_{0} : ℝ^{n} → ℝ is bounded, the longitudinal line method, see for example[4], with step size 1 leads to a linear Cauchy problem of type (1.1) in l^{∞}(ℤ^{n}) with

A_{k}x = (x(j_{1}, j_{2}, …, j_{k−1}, j_{k} + 1, j_{k+1}, …, j_{n}))_{j∈ℤ^{n}}.

The corresponding problem for the heat equation was studied in [1].

2. Notations and preliminaries

For A ∈ L(E) let N(A), A(E), σ(A) and r(A) denote the kernel, the range, the spectrum and the spectral radius of A, respectively. Let 𝔻 denote the complex unit circle {z ∈ ℂ : |z| < 1}.

Proposition 2.1. Let A ∈ L(E), 0 ∉ σ(A) and ∥A∥ = ∥A^{−1}∥ = 1. Then:

(1) A is an isometry; (2) ∥A^{n}∥ = ∥A∥^{n} = 1 (n ∈ ℕ), hence A is normaloid;

2000 Mathematics Subject Classification: Primary 34G10; Secondary 47A35.

(3) r(A) = 1 and σ(A) ⊆ ∂𝔻;

(4) N(A − I) ∩ \overline{(A − I)(E)} = \{0\};

(5) (A − I)(E) = (A^{−1} − I)(E);

(6) N(A − I) = N((A − I)^{2});

(7) N(A − I) ⊕ \overline{(A − I)(E)} is closed;

(8) if (A − I)(E) is closed then E = N(A − I) ⊕ (A − I)(E).

Proof. (1) and (2) are obvious.

(3) From (2) we get r(A) = 1. Next, it is clear that σ(A) ∪ σ(A^{−1}) ⊆ \overline{𝔻}. Since σ(A) = {z ∈ ℂ : z^{−1} ∈ σ(A^{−1})}, we conclude σ(A) ⊆ ∂𝔻.

(4) Let x ∈ N(A − I) ∩ \overline{(A − I)(E)}, let ε > 0 and choose z ∈ E such that ∥x−(A−I)z∥ < ε. According to [2, Satz 102.3], we have ∥x∥ ⩽ ∥x−(A−I)z∥ < ε, hence x = 0.

(5) Follows from (A − I)x = (I − A^{−1})(Ax).

(6) Follows from [2, Satz 102.3].

(7) Choose a sequence (x_{n}) in N(A − I) ⊕ \overline{(A − I)(E)} with x_{n} → x_{0} and corresponding decompositions x_{n} = y_{n} + z_{n}. According to [2, Satz 102.3] we have

∥y_{n} − y_{m}∥ ⩽ ∥x_{n} − x_{m}∥ (n, m ∈ ℕ),

hence (y_{n}) is convergent to a vector y_{0} ∈ N(A − I). Thus

z_{n} = x_{n} − y_{n} → x_{0} − y_{0} ∈ \overline{(A − I)(E)},

and therefore x_{0} ∈ N(A − I) ⊕ \overline{(A − I)(E)}.

(8) Follows from [2, Satz 72.4 and 102.4]. □

Proposition 2.2. Let A ∈ L(E) be as in Proposition 2.1, let T = A+A^{−1}−2I, and let c : [0, ∞) → ℝ denote the function

c(t) = exp(−t)(1 + ∑_{n=0}^{∞} \frac{t^{n}}{n!} |1 − \frac{t}{n + 1}|).

We have (1) ∥ exp(tT)∥ ⩽ 1 (t ⩾ 0);

(2) t ↦ \sqrt{t}c(t) is bounded on [0, ∞) and

∥ exp(tT)(A − I)x∥ ⩽ c(t)∥x∥ (t ⩾ 0, x ∈ E);

(3) lim_{t→∞} exp(tT)y = 0 (y ∈ (A − I)(E));

(4) if y ∈ E then

lim_{t→∞} exp(tT)y = 0 ⟺ y ∈ \overline{(A − I)(E)};

(5) N(A − I) = {x ∈ E : exp(tT)x = x (t ⩾ 0)}.

Proof. (1) For each t ⩾ 0 we have

∥ exp(tT)∥ = ∥ exp(−2t) exp(tA) exp(tA^{−1})∥
⩽ exp(−2t) exp(t∥A∥) exp(t∥A^{−1}∥) = 1.

(2) Since T = (A^{−1} − I) + (A − I) we have

exp(tT)(A − I) = exp(t(A^{−1} − I)) exp(t(A − I))(A − I) (t ∈ ℝ),

and

exp(t(A − I))(A − I)x = exp(−t) ∑_{n=0}^{∞} \frac{t^{n}}{n!} (A^{n+1} − A^{n})x

= exp(−t)(∑_{n=0}^{∞} \frac{t^{n}}{n!} (1 − \frac{t}{n + 1}))A^{n+1}x − exp(−t)x.

Hence, since ∥A∥ = 1,

∥ exp(tT)(A − I)x∥ ⩽ ∥ exp(t(A^{−1} − I))∥ ∥ exp(t(A − I))(A − I)x∥ ⩽ c(t)∥x∥ (t ⩾ 0, x ∈ E).

To see that t ↦ \sqrt{t}c(t) is bounded on [0, ∞) let N ∈ ℕ and N ⩽ t ⩽ N + 1. Then

∑_{n=0}^{∞} \frac{t^{n}}{n!} |1 − \frac{t}{n + 1}| = ∑_{n=0}^{N−1} \frac{t^{n}}{n!} (\frac{t}{n + 1} − 1) + ∑_{n=N}^{∞} \frac{t^{n}}{n!} (1 − \frac{t}{n + 1}) = 2\frac{t^{N}}{N!} − 1,

and therefore

\sqrt{t} c(t) = \sqrt{t} exp(−t)(1 + ∑_{n=0}^{∞} \frac{t^{n}}{n!} |1 − \frac{t}{n + 1}|)

⩽ \sqrt{N + 1} exp(−N)(1 + 2\frac{(N + 1)^{N}}{N!} − 1) = \frac{2(N + 1)^{N+1/2} exp(−N)}{N!},

which is bounded according to Stirling’s formula.

(3) Follows from (2).

(4) The implication ⇐ follows from (3). Now suppose that exp(tT)y → 0 as t → ∞. Since

exp(tT)y = y + ∑_{n=1}^{∞} \frac{t^{n}}{n!} T^{n}y = y + (A − I) ∑_{n=1}^{∞} \frac{t^{n}}{n!} (A − I)^{n−1}(I − A^{−1})^{n}y

we conclude y ∈ \overline{(A − I)(E)}.

(5) The inclusion

N(A − I) ⊆ {x ∈ E : exp(tT)x = x (t ⩾ 0)}

is obvious. Now suppose that x ∈ E and exp(tT)x = x (t ⩾ 0). By differentiation 0 = T exp(tT)x (t ⩾ 0), thus A^{−1}(A − I)^{2}x = T x = 0. Part (6) of Proposition 2.1 now shows that x ∈ N(A − I). □

3. The asymptotic behaviour of exp (tT)

Theorem 3.1. Let A and T be as in Proposition 2.1. For x ∈ E the following assertions are equivalent:

(1) lim_{t→∞} exp(tT)x exists in E [resp. lim_{t→∞} exp(tT)x = 0];

(2) x ∈ N(A − I) ⊕ \overline{(A − I)(E)} [resp. x ∈ \overline{(A − I)(E)}];

(3) the sequence

(\frac{x + Ax + ⋯ + A^{m}x}{m + 1})_{m∈ℕ}

is convergent [resp. is convergent with limit 0].

Proof. That (2) implies (1) follows from Proposition 2.2. Now, assume that (1) holds, and let z = lim_{t→∞} exp(tT)x. As in the proof of

part (4) of Proposition 2.2

exp(tT)x = x + (A − I) ∑_{n=1}^{∞} \frac{t^{n}}{n!} (A − I)^{n−1}(I − A^{−1})^{n}x,

hence x − z ∈ \overline{(A − I)(E)}. [In particular, if z = 0 then x ∈ \overline{(A − I)(E)}.] From part (3) of Proposition 2.2 we obtain

(A − I)z = lim_{t→∞} exp(tT)(A − I)x = 0,

and therefore x = z + (x − z) ∈ N(A − I) ⊕ \overline{(A − I)(E)}.

The equivalence of (2) and (3) is proved in [3, Ch.2, Theorem 1.3]. □

According to part (8) of Proposition 2.1 the following corollary shows, that lim_{t→∞} exp(tT)x exists for each x ∈ E if T (E) is closed:

Corollary 3.1. We have

(1) T (E) = (A − I)^{2}(E) ⊆ (A − I)(E) ⊆ \overline{T (E)};

(2) T (E) = \overline{T (E)} ⟺ (A − I)^{2}(E) = (A − I)(E)

⟺ (A − I)(E) = \overline{(A − I)(E)}.

Proof. (1) Part (5) of Proposition 2.1 gives

T (E) = (A − I)^{2}(E) ⊆ (A − I)(E).

As in the proof of Theorem 3.1 we obtain (A − I)(E) ⊆ \overline{T (E)}. Now, (2) follows by [2, Satz 102.4]. □

4. The general case

Now, let A_{1}, …, A_{n}, T_{1}, …, T_{n} and T be as in section 1. Moreover we introduce the following subspaces of E:

X_{1} = ⋂_{j=1}^{n} N(A_{j} − I), X_{2} = \overline{∑_{j=1}^{n}(A_{j} − I)(E)}, X = X_{1} + X_{2}.

Theorem 4.1. Under the assumptions above

(1) X_{2} = {x ∈ E : lim_{t→∞} exp(tT)x = 0};
(2) X_{1} = {x ∈ E : exp(tT)x = x (t ⩾ 0)};
(3) X = {x ∈ E : lim_{t→∞} exp(tT)x exists in E};
(4) X_{1} ∩ X_{2} = {0}, and X is closed.

Proof. (1) Let x ∈ X_{2}. Then x = lim_{m→∞} x_{m}, where x_{m} ∈ ∑_{j=1}^{n}(A_{j}−I)(E). By part (1) and part (3) of Proposition 2.2 we obtain

lim_{t→∞} exp(tT)x_{m} = 0 (m ∈ ℕ).

Let ε > 0, and choose N ∈ ℕ such that ∥x − x_{N}∥ < ε/2. Next, choose t_{0} ∈ [0, ∞) such that ∥ exp(tT)x_{N}∥ < ε/2 (t ⩾ t_{0}). Then

∥ exp(tT)x∥ = ∥ exp(tT)(x − x_{N}) + exp(tT)x_{N}∥
⩽ ∥x − x_{N}∥ + ∥ exp(tT)x_{N}∥ < ε (t ⩾ t_{0}).

Thus lim_{t→∞} exp(tT)x = 0.

Now suppose that x ∈ E and lim_{t→∞} exp(tT)x = 0. Set

h(t) = ∑_{n=1}^{∞} \frac{t^{n}}{n!} T^{n}x.

Since T_{j} = (A_{j} − I)(I − A_{j}^{−1}) (j = 1, …, n), we have

T x = ∑_{j=1}^{n}(A_{j} − I)(I − A_{j}^{−1})x ∈ ∑_{j=1}^{n}(A_{j} − I)(E),

hence

h(t) ∈ ∑_{j=1}^{n}(A_{j} − I)(E).

Thus, exp(tT)x = x + h(t) and lim_{t→∞} exp(tT)x = 0 imply x ∈ X_{2}.

(2) The inclusion ⊆ is obvious. For the reversed inclusion let x ∈ E be such that exp(tT)x = x (t ⩾ 0). Then by part (1) we obtain

(A_{j} − I)x = exp(tT)(A_{j} − I)x → 0 (t → ∞) (j = 1, …, n),

hence x ∈ X_{1}.

(3) Here, the inclusion ⊆ follows from parts (1) and (2) directly. Now, assume that x ∈ E is such that lim_{t→∞} exp(tT)x = z. As in the proof of part (1)

exp(tT)x = x + h(t), h(t) ∈ X_{2}.

Therefore x − z ∈ X_{2}. From part (1) we derive

(A_{j} − I)z = lim_{t→∞} exp(tT)(A_{j} − I)x = 0 (j = 1, …, n).

Thus z ∈ X_{1}, and so x = z + (x − z) ∈ X_{1} ⊕ X_{2} = X.

(4) Let x ∈ X_{1} ∩ X_{2}. Then, by parts (1) and (2), we have

exp(tT)x = x (t ⩾ 0), exp(tT)x → 0 (t → ∞),

thus x = 0. Next, if (x_{m}) is a sequence in X with limit x_{0}, then there exist sequences (y_{m}) and (z_{m}) in X_{1} and X_{2}, respectively, with x_{m} = y_{m} + z_{m}. From part (1) and part (2) we obtain

exp(tT)(x_{m} − x_{k}) → y_{m} − y_{k} (t → ∞).

Since ∥ exp(tT)(x_{m} − x_{k})∥ ⩽ ∥x_{m} − x_{k}∥ (t ⩾ 0), we have ∥y_{m} − y_{k}∥ ⩽ ∥x_{m} − x_{k}∥, thus (y_{m}) is convergent. Let y_{0} = lim_{m→∞} y_{m}. Then z_{m} = x_{m} − y_{m} → x_{0} − y_{0}. Hence we have y_{0} ∈ X_{1}, x_{0} − y_{0} ∈ X_{2}, and therefore x_{0} ∈ X_{1} ⊕ X_{2} = X. □

The following result provides sufficient conditions for the convergence of exp(tT)x.

Theorem 4.2. Let (k_{1}, …, k_{n}) ∈ ℕ_{0}^{n}, and set B = A_{1}^{k_{1}} … A_{n}^{k_{n}}. We have

(1) ⋂_{j=1}^{n}(N(A_{j} − I) ⊕ \overline{(A_{j} − I)(E)}) ⊆ X;

(2) \overline{(B − I)(E)} ⊆ X_{2};

(3) if x ∈ E and if the sequences

(\frac{x + A_{j}x + ⋯ + A_{j}^{m}x}{m + 1})_{m∈ℕ}

are convergent (j = 1, …, n), then lim_{t→∞} exp(tT)x exists in E;

(4) if x ∈ E and if the sequence

(\frac{x + Bx + ⋯ + B^{m}x}{m + 1})_{m∈ℕ}

is convergent to 0 in E, then lim_{t→∞} exp(tT)x = 0.

Proof. According to Theorem 3.1 we see that (3) follows from (1), and (4) follows from (2).

For the proof of (1) we use induction. If n = 1 the result follows by Theorem 3.1. Suppose that n ∈ ℕ and that (1) holds. In the case of n + 1 operators T_{1}, …, T_{n+1} we write T_{0} = T_{1} + ⋯ + T_{n}, so T = T_{0} + T_{n+1}. Let

x ∈ ⋂_{j=1}^{n+1}(N(A_{j} − I) ⊕ \overline{(A_{j} − I)(E)}).

Then

x ∈ ⋂_{j=1}^{n}(N(A_{j} − I) ⊕ \overline{(A_{j} − I)(E)}), x ∈ N(A_{n+1} − I) ⊕ \overline{(A_{n+1} − I)(E)},

and therefore the limits lim_{t→∞} exp(tT_{0})x and lim_{t→∞} exp(tT_{n+1})x exist in E. From

∥ exp(tT)x − exp(sT)x∥ = ∥ exp(tT_{0}) exp(tT_{n+1})x − exp(sT_{0}) exp(sT_{n+1})x∥

= ∥ exp(tT_{0})(exp(tT_{n+1}) − exp(sT_{n+1}))x + exp(sT_{n+1})(exp(tT_{0}) − exp(sT_{0}))x∥

⩽ ∥ exp(tT_{n+1})x − exp(sT_{n+1})x∥ + ∥ exp(tT_{0})x − exp(sT_{0})x∥

we see that lim_{t→∞} exp(tT)x exists.

Next, we prove (2) for (k_{1}, …, k_{n}) ∈ ℕ^{n}, without loss of generality. Let p(z) = z_{1}^{k_{1}} … z_{n}^{k_{n}} − 1 (z = (z_{1}, …, z_{n})), and note that there are polynomials q_{1}, …, q_{n} ∈ ℂ[z_{1}, …, z_{n}] such that

p(z) = (z_{1} − 1)q_{1}(z) + ⋯ + (z_{n} − 1)q_{n}(z).

Hence

(B − I)x ∈ ∑_{j=1}^{n}(A_{j} − I)(E) (x ∈ E),

and therefore \overline{(B − I)(E)} ⊆ X_{2}. □

5. Example

Let us return to the semidiscretization of v_{t} = Δv in ℝ^{2}, that is we consider E = l^{∞}(ℤ^{2}) and

A_{1}x = (x(i + 1, j))_{(i,j)∈ℤ^{2}}, A_{2}x = (x(i, j + 1))_{(i,j)∈ℤ^{2}}.

Let k_{1}, k_{2} ∈ ℕ, and assume that x ∈ l^{∞}(ℤ^{2}) is such that the sequence

((\frac{x(i, j) + x(i + k_{1}, j + k_{2}) + ⋯ + x(i + mk_{1}, j + mk_{2})}{m + 1})_{(i,j)∈ℤ^{2}})_{m∈ℕ}

tends to 0 as m → ∞ in l^{∞}(ℤ^{2}). Then

exp(tT)x → 0 (t → ∞)

(apply part (4) of Theorem 4.2 with B = A_{1}^{k_{1}}A_{2}^{k_{2}}).

References

[1] G. Herzog, R. Lemmert, A note on discrete heat conduction, Arch. Math. 75 (2000), 147–152.

[2] H. Heuser, Funktionalanalysis, 2. Auflage. Mathematische Leitfäden. Stuttgart: Teubner (1986).

[3] U. Krengel, Ergodic theorems. De Gruyter Studies in Mathematics, 6. Berlin-New York: Walter de Gruyter. VIII (1985).

[4] W. Walter, Some new aspects of the line method for parabolic differential equations, Lect. Notes Math. 280 (1972), 181–189.

Institut für Analysis (Received 26 10 2006) Universität Karlsruhe (TH) 76133 Karlsruhe Kaiserstrasse 89 Germany

            Gerd.Herzog@math.uni-karlsruhe.de
    Christoph.Schmoeger@math.uni-karlsruhe.de