Space of Functions with Some Generalization of Bounded Variation in the Sense of de La Vallée Poussin

Proposition 1. A function f is of bounded second variation if and only if f can be written as the difference of two convex functions.

Definition 2. Let f be a real-valued function defined on [a, b] and let Π be a type I partition of [a, b]. Let

where where the supremum is taken over all partitions of type I of [a, b]. The space $$ is called (2, α)-variation in the sense of de La Vallée Poussin of type I. If $$ the function f is said to be of (2, α)-variation in the sense of de La Vallée Poussin of type I. The set of all such functions is denoted by $$.

Definition 3. Let f be a real-valued function defined on [a, b] and let Π be a type II partition of [a, b]. Let

where the supremum is taken on all type II partition of [a, b]. The number $$ is called the (2, α)-variation in the sense of de La Vallée Poussin of type II in [a, b]. If $$, f is said to be a function of (2, α)-variation in the sense of de La Vallée Poussin of type II. The set of all such functions is denoted by $$.

Definition 4. A function $$ is said to be an α-Lipschitz function if there exists a constant M > 0 such that

for all x, y ∈ [a, b]. We define the space $$ as the space of all α-Lipschitz functions. This space is normable, via the norm

Definition 5. One said that a function f is α-convex in [a, b] if, for a ⩽ x ⩽ z ⩽ y ⩽ b, one has

Definition 6. A function $$ is said to be absolutely continuous with respect to α if, for every ε > 0, there exists some δ > 0 such that if $$ are disjoint open subintervals of [a, b], then

All functions in α-$$ are bounded and form an algebra of functions under pointwise defined standard operations.

Definition 7. Suppose f and α are real-valued functions defined on the same open interval I and let x₀ ∈ I. One says that f is α-differentiable at x₀ if the following limit exists:

If the limit exists we denote its value by $$, which we call the α-derivative of f at x₀.

Theorem 8. Let f ∈ α-$$; then $$ exists and is finite μ_α a.e. on [a, b]. Moreover $$ is integrable in the Lebesgue-Stieltjes sense and

Theorem 9. Let p be an increasing function and α a continuous function in [a, b]. Then one has that

is an α-convex function on [a, b].

Lemma 10. Let v be an α-convex function and a ⩽ λ ⩽ ξ ⩽ μ ⩽ b; then

that is, v_α[λ, ξ] ⩽ v_α[λ, μ] ⩽ v_α[ξ, μ] with v_α[p, q] = (v(q) − v(p))/(α(q) − α(p)).

Lemma 11. Let f be defined on [a, b] and let Π₁ = {a = x₁ < x₂ < x₃ = b} be a partition of [a, b]. Then there exists λ and μ in [0,1] with λ + μ = 1 such that

Proof. Observe that

where λ and μ are straightforward from the equation.

Theorem 12. Let f be defined on [a, b] and let Π = {a = x₁ < ⋯<x_s−1 < x_s < ⋯<x_n−1 < x_n = b} be a partition of type I and y ∈ (x_s−1, x_s) with s = 2, …, n. Let one consider

Then

Proof.   

  

Case  1 (2 < s < n). Let us consider all the terms of σ_(2,α)(f, Π) on which y play a role; we have to estimate the expression

Applying Lemma 11 in order to introduce the point y in (x_s−1, x_s), we obtain that there exists positive λ and μ with λ + μ = 1 such that replacing in (21) and using the fact that λ + μ = 1 we have From the triangle inequality in (23) we obtain That is, the expression in (21) is less than or equal to (24). Since the remaining terms in σ_(2,α)(f, Π) and σ_(2,α)(f, Π^′) coincide, we get (20).

Case  2 (s = 2). Let y ∈ (x₁, x₂); in this case we have a unique term of σ_(2,α)(f, Π), where the interval (x₁, x₂) plays a role; applying Lemma 11 we obtain f_α[x₁, x₂] = λf_α[x₁, y]+(1 − λ)f_α[y, x₂] with 0 < λ < 1. Hence

and since the remaining terms of σ_(2,α)(f, Π) and σ_(2,α)(f, Π^′) coincide, we obtain (20).

Case  3 (s = n). This case is similar to Case 2.

Corollary 13. Let f be a function defined in [a, b] and let Π^′ be a refinement of partition Π. Then

Proof. The proof follows from Theorem 12 by induction.

Theorem 14. Let f be defined on [a, b]. Then

Therefore

Proof. Consider the following.

Case  1. Let $$ and Π be a type II partition of [a, b]. First we assume that x_j,2 < x_j,3 for all j. On each σ^(2,α)(f, Π) we may apply the triangle inequality for j = 1, …, n to obtain

Then Adding all of those terms we have This last sum is bounded by σ_(2,α)(f, Π) if we consider Π as a type I partition; thus In the case x_j,2 = x_j,3 for some j, we do not need to apply the triangle inequality, since |f_α[x_j,1, x_j,2] − f_α[x_j,3, x_j,4]| = |f_α[x_j,1, x_j,2] − f_α[x_j,2, x_j,4]|. σ^(2,α)(f, Π) is bounded by where Σ^′ is the summation with the terms x_j,2 = x_j,3 and Σ is the summation where x_j,2 < x_j,3.

Since (32) holds for all type II partitions, we conclude that $$, implying that $$.

Case  2. Conversely, let $$ and Π be a type I partition of [a, b]; that is, Π = {a = x₁ < x₂ < ⋯<x_n = b}. Eventually adding one or two points we can suppose that n = 3p, since the addition of points in the partition does not decrease the sum σ_(2,α) (Theorem 12). Hence

where we group the terms in threes.

On each interval (x_j, x_j+1) we introduce two points: x_j < x_j,1 < x_j,2 < x_j+1 for j = 3, …, n − 1. On each term from the sum we apply the triangle inequality (for the last term we give a different treatment) to obtain

for j = 1, …, n − 3. We introduce these terms in σ_(2,α)(f, Π) to get A partition of type II is obtained in each summation in parenthesis and therefore it is bounded by $$; accordingly The term |f_α[x_n−2, x_n−1] − f_α[x_n−1, x_n]| is bounded by $$. Finally $$. Since this holds for all partitions of type I we conclude that $$ and this $$.

Remark 15. This result guarantees that the $$ functions satisfy all the properties verified by the $$ functions and reciprocally. In the coming demonstrations we choose the partition that will be more appropriate.

Lemma 16. If $$, then there exists M > 0 such that, for all x₁, x₂ with a ⩽ x₁ < x₂ ⩽ b, it holds that

Proof. Fix a₁ and a₂ such that a < a₁ < a₂ < b and

Case  1 (x₂ < b). Let x₃ and x₄ be points such that max{x₂, a₂} < x₃ < x₄ ⩽ b; then a₁ < a₂ < x₃ < x₄ and x₁ < x₂ < x₃ < x₄ are points in partitions of type II of [a, b]. Therefore $$ and $$. Using the triangle inequality we have

Case  2 (x₂ = b and a < x₁). Let x₃, x₄ be points such that a ⩽ x₃ < x₄ < min{a₁, x₁}; then x₃ < x₄ < a₁ < a₂ and x₃ < x₄ < x₁ < x₂ are partitions of type II of [a, b]. Therefore $$ and $$. Similarly as in Case 1 we obtain |f_α[x₁, x₂]| ⩽ M₁.

Case  3 (x₁ = a and x₂ = b). Consider

Taking M = max{M₁, M₂} we deduce that for all x₁, x₂ such that a ⩽ x₁ < x₂ ⩽ b we have |f_α[x₁, x₂]| ⩽ M.

Theorem 17. If f is a (2, α)-variation function in [a, b], then f is an α-Lipschitz function.

Proof. If $$ from Lemma 11 there exists M > 0 such that for all a ⩽ x < y ⩽ b we have that |f_α[x, y]| ⩽ M holds. Then by the definition of f_α, we get |f(y) − f(x)| ⩽ M | α(y) − α(x)|; that is, f is an α-Lipschitz function in [a, b].

Corollary 18. If $$, then one has that f is α-absolutely continuous in [a, b].

Corollary 19. If $$, then $$.

Corollary 20. If $$, then f is (uniformly) continuous in [a, b] and thus bounded in [a, b].

Theorem 21. The set $$ equipped with the sum of functions and the scalar product is a linear space.

Proof. Let $$ and Π be a type I partition of [a, b]. Then by the triangle inequality, we obtain σ_(2,α)(f + g, Π) ⩽ σ_(2,α)(f, Π) + σ_(2,α)(g, Π) and thus $$ and from this we conclude that $$. Moreover σ_(2,α)(λf, Π) = |λ | σ_(2,α)(f), $$, and $$ from which we obtain that if $$, then $$, $$.

Corollary 22. One has that

• (1)

  $$ is a real linear space;

• (2)

  $$ is a real linear space.

Lemma 23. Let $$. If a < c < b

Theorem 24. One has that $$ if and only if f = f₁ − f₂, where f₁ and f₂ are α-convex functions.

Proof. Observe that the set $$ has measure zero. For x ∈ [a, b]∖E let us define

then $$. Using Lemma 23 we observe that for y < x we have and a similar calculation can be made regarding the function q and therefore p and q are increasing in [a, b]∖E. Since f ∈ α-$$, then $$ is integrable in the sense of Lebesgue-Stieltjes and $$, for x ∈ [a, b]. In E we define p and q such that p and q are increasing and bounded in [a, b] and therefore LS-integrable in [a, b] with respect to α; from Theorem 8 we obtain Since each integral is α-convex (see Theorem 9), the result follows.

Conversely, suppose that f = f₁ − f₂ where f₁ and f₂ are α-convex functions. We are going to prove that $$. To do so, it is enough to verify that all α-convex functions have second bounded α-variation.

Let $$ be α-convex; then by Lemma 10 we have $$. Let Π = {a = x₁ < x₂ < ⋯<x_n = b} be a partition of type I of the interval [a, b]. Since v_α[·, ·] is increasing we obtain

This amount is part of a telescopic series; one more time by Lemma 10 we have Hence $$ and $$. Since f₁ and f₂ are α-convex, therefore $$.

Corollary 25. If $$, then f can be written as

where g is a bounded variation function in [a, b].

Definition 26. Let $$. One defines $$ as

When there is no danger of confusion, one will denote $$.

Lemma 27. If $$ and $$, then f(x) = λα(x) + μ, where $$.

Proof. Note that $$ if and only if σ_(2,α)(f) = 0 for all partitions Π of [a, b]. Consider the particular partition Π₀ = {a < x < b}. Then from the fact that σ_(2,α)(f, Π₀) = 0 we deduce

Taking the result follows.

Theorem 28. The functional $$ is a norm in the space $$.

Proof. Let $$. We have

Let $$, $$; we get ‖λf‖_(2,α) = |λ|‖f‖_(2,α). Finally ‖f‖_(2,α) = 0 if and only if |f(a)| = 0 and $$ and $$. Since $$, from Lemma 27 we have f(x) = λα(x) + μ, $$. Moreover $$, but $$; then λ = 0. Thus f(x) = μ, but $$; then μ = 0. Hence f = 0.

In this way we have shown that $$ is a normed space.

Lemma 29. Let $$; then

Proof. Let us consider a < a + h < s < t < b; then from Theorem 14 we obtain

Letting h → 0 we have thus from this we get That is, $$.

Theorem 30. $$ is a Banach space.

Proof. Let $$ be a Cauchy sequence in $$. Given ε > 0 there exists $$ such that if p, q > N_ɛ then $$. By the definition of the norm, we obtain |f_p(a) − f_q(a)| < ɛ, $$ and $$. Invoking Lemma 29 it follows that $$ and thus $$ is a Cauchy sequence in α-$$. In other words

which implies that |f_p(a) − f_q(a)| < ɛ and |((f_p − f_q)(x)−(f_p − f_q)(y))/(α(x) − α(y))| < ɛ for x ∈ [a, b]. By the triangle inequality we have |f_p(x) − f_q(x)| < ɛ(1 + α(b) − α(a)), for x ∈ [a, b]. This tells us that $$ is a Cauchy sequence in $$. Since $$ is complete we can define a function f in the following way: $$ given by f(x) = lim_p→∞f_p(x), for x ∈ [a, b]. Next we need to prove that (I) $$ and (II) $$ converges to f in ‖·‖_(2,α)-norm.

(I) Indeed, let Π = {a = x₁ < x₂ < ⋯<x_n = b} be a partition of [a, b]. Then

for all $$ and partition Π (the sequence $$ is bounded since $$ is a Cauchy sequence).

Now,

for all partitions Π; therefore $$; thus $$. Also note that $$ exists.

(II) Let Π = {a = x₁ < x₂ < ⋯<x_n = b} be a partition of [a, b]. For p, q > N_ɛ we obtain

Taking limit as q goes to infinity, we have which holds for any partition of [a, b]; then $$ if p > N_ɛ. Since |f_p(a) − f_q(a)| < ɛ we have |f_p(a) − f(a)| ⩽ ɛ, for p > N_ɛ, if q → ∞.

Next, for a < a + h < s < t < b, we have

if p > N_ɛ. Taking limit as h goes to zero if p > N_ɛ. Since $$ if p > N_ɛ, by virtue of $$. Finally, for p > N_ɛ we have $$, that is, $$.

Theorem 31. The space $$ equipped with the norm

is a Banach algebra and the norms $$ and $$ are equivalent.