Some inequalities and superposition operator in the space of regulated functions

Definition 2.1

A mapping μ : 𝔐_E → ℝ₊ = [0, ∞) is said to be a measure of noncompactness in E if it satisfies the following conditions:

1. The family ker μ : = {X ∈ 𝔐_E : μ(X) = 0} is nonempty and ker μ ⊂ 𝔑_E.

2. X ⊂ Y ⇒ μ(X) ≤ μ(Y).

3. μ(Conv X) = μ(X).

4. μ(λX + (1 – λ)Y) ≤ λμ(X) + (1 – λ)μ(Y) for λ ∈ [0, 1].

5. If (X_n) is a sequence of closed sets from 𝔐_E such that X_n+1 ⊂ X_n (n = 1, 2, …) and if limn→∞ μ(X_n) = 0, then the intersection X_∞ : = ⋂n=1∞ X_n is nonempty.

Definition 2.2

A function x : [a, b] → E, where E is a topological vector space, is said to be a regulated function if for every t ∈ [a, b) the right-sided limit x(t⁺) : = lims→t+ x(s) exists and for every t ∈ (a, b] the left-sided limit x(t^–) := lims→t− x(s) exists.

Definition 2.3

We will say that the set X ⊂ R(J, E) is equiregulated on the interval J if the following two conditions are satisfied:

Theorem 2.4

[3, 4, 5, 9] A nonempty subset X ⊂ R(J, E) is relatively compact in R(J, E) if and only if X is equiregulated on the interval J and the sets X(t) are relatively compact in E for t ∈ J.

Theorem 2.5

[13] The function μ_m given by formula (2.1) satisfies conditions 1^o – 7^o and 9^o in the space R(J, E).

Remark 2.6

Above construction of the measure (2.1) addresses inaccuracies existent in the construction of measures given in [3, 4].

Theorem 3.1

Let X ⊂ R(J, E) be nonempty, bounded and equiregulated. Then the function J ∋ t ↦ β_E(X(t)) ∈ ℝ₊ is regulated and the following inequality holds

Proof

Let us fix ε > 0. The condition of X being equiregulated implies that

Since the family of intervals {[a, a + δ_a), (b – δb′ , b], (s – δs′ , s + δ_s) : s ∈ (a, b)} is an open cover of compact interval J, then there is a finite subcover. Obviously it contains intervals [a, a + δ_a) and (b – δb′ , b]. From this finite subcover we can choose following subcover

such that a = s₀ < s₁ < … < s_k = b and additionally every two consecutive intervals from (3.3) have nonempty intersection. Now we choose one point si′ belonging to each of those k intersections, i.e. we have the sequence of points

For simplicity let us denote them by t_i, i = 0, …, n where n = 2k i.e.

From (3.1) and (3.2) we yield that

Hence

which means that the function J ∋ t ↦ β_E(X(t)) is regulated, thus Riemann integrable. Let us choose arbitrary s_i ∈ (t_i–1, t_i), i = 1, …, n. Then by (3.5) we have

For each i = 1, …, n there exist vji ∈ E, j = 1, …, m_i, such that

Hence and by (3.4) we have

Let us set functions y_j1,…,jn ∈ R(J, E), j_i ∈ {1, …, m_i}, i = 1, …, n by formulas

We prove that the set of vectors given by ∫ab y_j1,…,jn(t)dt for j_i ∈ {1, …, m_i}, i = 1, …, n is ( ∫ab β_E(X(t))dt + 3ε(b – a))-net for ∫ab X(t)dt. Let x ∈ X. Because of (3.7) there exists such a sequence j₁, …, j_n, that

Then, using (3.8) and (3.6) we have

which means that

which for ε → 0 proves the theorem.□

Theorem 3.2

If {x_n : n ∈ ℕ} ⊂ R(J, E) and there exists Lebesgue integrable function g : J → ℝ₊ such that ∥x_n(t)∥ ≤ g(t) for t ∈ J, n ∈ ℕ, then the function J ∋ t ↦ β_E({x_n(t) : n ∈ ℕ}) is Lebesgue integrable on J and

Remark 3.3

Above theorem is also true given weaker assumption that functions x_n are strongly measurable [16]. The example from [17] shows that factor 2 from the above theorem cannot be replaced by smaller even for the sequence {x_n} of regulated functions.

Lemma 3.4

[18] If E is a Banach then for each non-empty and bounded set X ⊂ E there exists such countable set X₀ ⊂ X, that β_E(X) ≤ 2β_E(X₀).

Theorem 3.5

For each non-empty and bounded set X ⊂ R(J, E), where E is a finite dimensional Banach space, there exists such countable set X₀ ⊂ X that μ_m(X₀) = μ_m(X).

Proof

Without loss of generality we can assume that max{ω^–(X), ω⁺(X)} = ω⁺(X). Let {t_n} ⊂ J be such a sequence that ω⁺(X) = supn∈N ω⁺(X, t_n). Let us arbitrarily fix n ∈ ℕ. Then for each i ∈ ℕ there exists such a function xin∈X that limi→∞ω+(xin,tn,1i)=ω+(X,tn). Denote X₀ := { xin : i, n ∈ ℕ}. Then based on the above we have ω⁺(X₀) = ω⁺(X), ω^–(X₀) ≤ ω^–(X) and hence μ_m(X₀) = μ_m(X).□

Theorem 3.6

For each non-empty and bounded set X ⊂ R(J, E) and any ε > 0, there exists such countable set X₀ ⊂ X, that μ_m(X) ≤ 2 μ_m(X₀) + ε. In the above estimation factor 2 cannot be replaced by smaller (see Example 3.7).

Proof

Let X̃₀ be a countable set constructed the same way as in the proof of Theorem 3.5. Thus

Let us arbitrarily fix ε > 0. There exists such a number t₀ ∈ J that

supt∈J β_E(X(t)) < β_E(X(t₀)) + ε. Using Lemma 3.4 we get that there exists such a sequence {x_n} ⊂ X that

Hence

Let X₀ := X̃₀ ∪ {x_n : n ∈ ℕ}. Obviously X₀ is countable and

Hence

□

Example 3.7

Let E be a space consisting of all bounded functions x : ℝ₊ → ℝ such that for each of them there exists such countable set T_x ⊂ ℝ₊, that function x(t) tends to 0 as t → ∞ and t ∈ ℝ₊ ∖ T_x. We assume the supremum norm in E. Now we can define a subset A ⊂ E by A := {1_t : t ∈ ℝ₊} and next we define a subset X ⊂ R(J, E) by X := {â : a ∈ A} where â is defined in (3.9). Thus ω^–(X) = ω⁺(X) = 0, supt∈J β_E(X(t)) = β_E(A) = 1 which means that μ_m(X) = 1 and for any countable subset X₀ = { 1tn^ : n = 1, 2, …} ⊂ X we have ω^–(X₀) = ω⁺(X₀) = 0, supt∈J β_E(X₀(t)) = β_E({1_tn : n = 1, 2, …}) = 12 because {1tn:n=1,2,...}⊂BE(12∑n=1∞1tn,12) which means that μ_m(X₀) = 12 and that proves that factor 2 from the above theorem cannot be replaced by smaller.

Theorem 4.1

[12] A superposition operator F_f maps R(J, E) into itself if and only if the function f has the following properties:

1. the limit lim[a,t)×E∋(s,v)→(t,x) exists for every (t, x) ∈ (a, b] × E,

2. the limit lim(t,b]×E∋(s,v)→(t,x) exists for every (t, x) ∈ [a, b) × E.

Theorem 4.2

[2] Suppose that the function f(⋅, u) is regulated on [0, 1] for all u ∈ E, and the function f(t, ⋅) is continuous on E, uniformly with respect to t ∈ J. Then the superposition operator F_f maps R(J, E) into itself and is (norm) bounded.

Theorem 4.3

[12] A superposition operator F_f maps R(J, E) into itself is continuous if and only if a function f̃ : E ∋ x ↦ f(⋅, x) ∈ R(J, E) is continuous.

Theorem 4.4

Let the superposition operator F_f maps R(J, E) into itself. Then the family of functions {f_t}_t∈J ⊂ E^E satisfies the following conditions:

1. The mapping J ∋ t ↦ f_t ∈ E^E is a regulated function.

2. The following limits of pointwise convergence exist and

   1. lims→t+ f_s is continuous in E for t ∈ [a, b),

   2. lims→t− f_s is continuous in E for t ∈ (a, b].

Conversely, if additionally E is a finitely dimensional Banach space and conditions (a) and (b) are satisfied then the superposition operator F_f maps R(J, E) into itself.

Proof

(⇒) Let us fix t ∈ [a, b). Using condition (2) of Theorem 4.1 and based on notation (4.1) we have the following equality

First we prove (b1). Let us fix x ∈ E and ε > 0. Because of (4.2) we have the existence of δ > 0 and τ > 0 such that

Now going from s → t⁺ we have ∥g_t(x) – g_t(v)∥ ≤ ε which proves continuity of g_t in x and thereby on E. Analogically we can prove (b2).

We will now prove (a) i.e. that f_s converges to g_t almost uniformly on E when s → t⁺. Let us fix non-empty and compact set K ⊂ E and ε > 0. Then, because of (4.2) and already proven continuity of g_t, we have that for each x ∈ K there exist δ_x > 0, τ_x > 0 such that concurrently

and

Out of family {B_E(x, δ_x)}_x∈K covering compact set K we choose a finite subcover {BE(xi,δxi)}i=1n. Let τ := min{τ_i : i = 1, …, n}. Let us fix arbitrary v ∈ K. Then there exists such i that v ∈ B_E(x_i, δ_i). Thus for s ∈ (t, t + τ) based on (4.3) and (4.4) we have the following estimation

for any v ∈ K, i.e. we have uniform convergence on K. Similarly we can prove the existence of the limit lims→t− f_s in the topology of almost uniform convergence.

(⇐) Assume that E has a finite dimension and fix t ∈ [a, b). Condition (a) assures the existence of the limit g_t := lims→t+ f_s which, based on (b) is continuous on E. Let us fix x ∈ E and ε > 0. Continuity of g_t means that for some r > 0 we have

Moreover (a) implies that for a compact set B_E(x, r) there exists τ > 0 such that

When we combine it with (4.5), for v ∈ B_E(x, r), s ∈ (t, t + τ) we have

i.e. condition (4.2) is satisfied and thereby (2) in Theorem 4.1 holds. Similarly we can prove (1) in Theorem 4.1, so actually F_f acts from R(J, E) into R(J, E).□

Corollary 4.5

If E is a finitely dimensional Banach space then superposition operator F_f acts from R(J, E) into R(J, E) if and only if both conditions (a) and (b) in Theorem 4.4 are satisfied.

Theorem 4.6

Superposition operator F_f acts from R(J, E) into R(J, E) if and only if the following four conditions are satisfied:

1. ∀_v∈E f(⋅, v) ∈ R(J, E).

2. For each v ∈ E set D_v is finite or countable.

3. For each v ∈ E, if the set D_v = {t_n} is infinite then

   lim(n,δ)→(∞,0+)ν(f¯tn,v,δ)=0.

4. The mapping E ∋ u ↦ f(⋅, u) ∈ R(J, E) is continuous in every point v ∈ E in regard to pseudonorm ∥⋅∥_J∖Dv, i.e. for each v ∈ E and each sequence v_n → v we have ∥f(⋅, v) – f(⋅, v_n)∥_J∖Dv → 0 when n → ∞.

Lemma 4.7

If there exist a number ε₀ > 0, a sequence {t_n} ⊂ J convergent to a point t₀ ∈ J from one side and a sequence {v_n} ⊂ E convergent to vector v ∈ E, such that

then superposition operator F_f does not act from R(J, E) into R(J, E).

Proof

We give a proof by contradiction. Let us assume that F_f acts from R(J, E) into R(J, E). Hence we have

Let assume that {t_n} is convergent to some t₀ ∈ J from one side, for example t_n → t0+ , and moreover {t_n} is strictly decreasing (we can have that choosing a proper subsequence). Let us put

Obviously y ∈ R(J, E). Since F_fy ∈ R(J, E), then there exists a limit

Moreover, by (4.9), there exists a limit limn→∞ f(t_n, v). Hence, by (4.8) we get

Now let us define next function z ∈ R(J, E) as follows

Since F_fz ∈ R(J, E) then there is a limit limt→t0+ (F_fz)(t) which means that

However it is in contradiction with (4.10).□

Proof of Theorem 4.6

(⇒) Take an arbitrary v ∈ E and put x(t) ≡ v, t ∈ J. Since F_fx ∈ R(J, E), then (a) is satisfied.

Let us assume that the set D_v is uncountable. Since

then there exists such k ∈ ℕ, that the set {t ∈ J : ν(f_t, v) > 1k } is uncountable. Let us put ε₀ := 1k and choose an arbitrary injective sequence {t_n} ⊂ {t ∈ J : ν(f_t, v) > 1k }. Since ν(f_tn, v) > ε₀ for n = 1, 2, … it follows that for each n ∈ ℕ we can choose such v_n ∈ E, that ∥f_tn(v) – f_tn(v_n) ∥ ≥ ε₀ and ∥v – v_n∥ ≤ 1n . Choosing from the sequence {t_n} a subsequent (also denoted as {t_n}) which converges from one side to some t ∈ J we have v_n → v and

However then, by Lemma 4.7 operator F_f would not act from R(J, E) into R(J, E) which proves (b).

Analogically negating conditions (c) and (d) we would have the existence of ε₀ > 0, a sequence {t_n} ⊂ D_v for condition (c) or {t_n} ⊂ J ∖ D_v for condition (d), convergent from one side to some t ∈ J, a vector v ∈ E and a sequence {v_n} ⊂ E, such that v_n → v and (4.11) holds which, by Lemma 4.7 contradicts the hypothesis and thereby prove conditions (c) and (d).

(⇐) Let us fix x ∈ R(J, E). To prove that F_fx ∈ R(J, E) we will show that for fixed t ∈ [a, b) there exists a limit lims→t+ (F_fx)(s) (we omit the proof of the existence of left-hand side limit as similar to the following). Let us fix such a sequence {t_n} ⊂ (t, b], that t_n → t⁺ and define v_n := x(t_n). Since x ∈ R(J, E), then there exists a limit v := limn→∞ v_n. By (a) there exists also a limit limn→∞ f(t_n, v). To prove the existence of the limit lims→t+ (F_fx)(s) it is enough to show the existence of the limit limn→∞ f(t_n, v_n). Additionally we show that

Let us consider 3 cases:

Case (i): the sequence {t_n}, outside a finite number of terms is contained in J ∖ D_v.

Therefore, by (d), for each ε > 0 and for sufficiently large n we have ∥f(t_n, v_n) – f(t_n, v) ∥ ≤ ε. When n → ∞ we have

which considering that ε > 0 is arbitrary implies (4.12).

Case (ii): The sequence {t_n}, outside a finite number of terms is contained in D_v.

Let us fix ε > 0. By (b) and (c) there exists such n₀ ∈ ℕ and δ₀ > 0, that for n ≥ n₀ and for 0 < δ < δ₀ we have ν(f_tn, v, δ) < ε, i.e. for large enough n we have ∥f(t_n, v_n) – f(t_n, v) ∥ ≤ ε. When n → ∞ using similar reasoning as in case (i) we also get that (4.12) holds.

Case (iii): Infinitely many terms of the sequence {t_n} is contained in D_v as well as infinitely many terms of the sequence {t_n} is contained in J ∖ D_v.

Dividing the sequence {t_n} into two adequate subsequences the case can be reduced to previous cases (i) and (ii). □

Remark 4.8

The above Theorems 4.4 and 4.6 improve some mistake contained in [13] (Theorem 5.1).

Theorem 4.9

Let E be a separable Banach space and let the superposition operator F_f : R(J, E) → R(J, E) be continuous and compact (i.e. F_f transforms bounded sets in relatively compact sets). Then

1. there exists a function g ∈ R(J, E),

2. there exists a countable or finite set T = {t_n} ⊂ J,

3. there exist a countable or finite sequence of functions h_n : E → E, n = 1, 2, … that are continuous, compact and when it is infinite

   ∀r>0limn→∞∥hn(BE(r))∥=0 (4.13)

   such that

   f(t,x)=g(t)+∑n=1∞1tn(t)hn(x),t∈J,x∈E. (4.14)

Conversely, if the conditions (H1)-(H3) are satisfied and E is a Banach space then the formula (4.14) gives such a function f(t, x), that operator F_f : R(J, E) → R(J, E) and it is continuous and compact.

Remark 4.10

Obviously the case when all h_n functions in the previous theorem are equal to θ, that is when f(t, x) = g(t), or only a finite number of them is not equal to θ is also allowed.

Lemma 4.11

If F_f : R(J, E) → R(J, E) is compact, then for each x, y ∈ R(J, E) the set supp (F_fx – F_fy) is countable or finite and for each injective sequence {t_n} ⊂ J, we have

Proof

If the set supp(F_fx – F_fy) was uncountable or if (4.15) was not satisfied, then there would exist a injective sequence {s_n} ⊂ J and a number ε₀ > 0, such that

Let us define the sequence of functions u_n : J → E, n = 1, 2, … by

Obviously u_n ∈ R(J, E). For n ≠ m we have s_n ≠ s_m and therefore

i.e. the sequence {F_fu_n} is ε₀-separable, hence F_f is not compact, which ends the proof.□

Lemma 4.12

If E is a separable Banach space and F_f : R(J, E) → R(J, E) is compact and continuous, then there exists a countable or finite set T = {t_n} ⊂ J, such that

Proof

Let A = {a_n : n ∈ ℕ} ⊂ E be a countable dense subset of E. Let us put T := ∪n=1∞supp(Ffan^−Ffθ^). By the previous Lemma 4.11, the set T is countable or finite. If there existed x ∈ E such that (4.16) did not hold then there would exist s ∈ J such that s ∈ supp(F_fx̂ – F_fθ̂)∖T. Thus ∥(F_fx̂ – F_fθ̂)(s) ∥ = ε₀ for some ε₀ > 0 and additionally (F_f an^ – F_fθ̂)(s) = θ for n ∈ ℕ. If we took such a subsequence {a_kn}, that a_kn → x in E we would have

which is in contradiction with continuity F_f.□

Proof of Theorem 4.9

Let us assume first that E is separable and F_f is compact and continuous. Let the set T = {t_n} be like in Lemma 4.12. We define the function g : J → E and the sequence of functions h_n : E → E, n = 1, 2, … by

By Lemma 4.12, for each x ∈ E the mapping J ∋ t → f(t, x) – f(t, θ) can be non-zero only on the set T and its formula is given by ∑n=1∞1tn(t)hn(x). Therefore

Since F_f is continuous the functions h_n must be continuous and since F_f is compact also the functions h_n must be compact. Moreover it results from Lemma 4.11 that limn→∞ f(t_n, x) – f(t_n, θ) = θ for x ∈ E i.e.

We have only (4.13) left to prove. Let us assume that it is not satisfied. Then there would exist r > 0, a number ε₀ > 0 and such a subsequence of the sequence {h_n}, (also denoted by {h_n}), that

Let k₁ = 1. We choose x₁ ∈ B_E(r), such that ∥h_k1(x₁)∥ ≥ ε02 . By (4.17) we know that there exists k₂ ∈ ℕ, such that k₂ > k₁ i ∥h_i(x₁)∥ ≤ ε04 for i ≥ k₂. By (4.18) we conclude that there exists x₂ ∈ B_E(r), such that ∥h_k2(x₂)∥ ≥ ε02 . By (4.17) we know that there exists k₃ ∈ ℕ, such that k₃ > k₂ and ∥h_i(x₂)∥ ≤ ε04 for i ≥ k₃. Continuing this procedure we get an strictly increasing sequence {k_n} ⊂ ℕ and the sequence {x_n} ⊂ B_E(r) such that

Let n > m. Then

i.e. {F_f xn^ } is positively separated and F_f is not compact which contradicts the assumptions.

Let us assume now that the conditions (H1)-(H3) are satisfied and the function f(t, x) is given by the formula (4.14). First we prove that F_f : R(J, E) → R(J, E). Let us fix x ∈ R(J, E), t ∈ J and the sequence { tj′ } convergent to t from one side, e.g. tj′ → t⁺. Since the sequence {x( tj′ )} is bounded, then by (4.13) we have limj→∞∑n=1∞1tn(tj′)hn(x(tj′))=θ and in virtue of (4.14) we get limj→∞(Ffx)(tj′)=limj→∞g(tj′)=g(t+) i.e. F_fx ∈ R(J, E).