Theorem 1 {Banach Contraction Principle}. Let X be a complete metric space and f : X⟶X be a contraction mapping, then f has a unique fixed point. Even more, let X be a complete metric space and B⊆X be a closed subset such that f(B)⊆B. If f : B⟶B be a contraction mapping, then f has a unique fixed point in B.

Definition 1. Let $$ be the class of all nondecreasing sequences $$, (n = 1,2, …) of positive numbers such that $$ diverges. Let Λ = (Λ¹, Λ²) where $$ and $$ with $$, such that a < b, c < d and I = [a, b], J = [c, d]. For 1 ≤ p < +∞ and $$,

• (1)

  Fix x₂ ∈ J = [c, d], and let [x₁, y₁] ⊂ [a, b]. The partial Λ_p-variation of f in the sense of Shiba on [x₁, y₁] is defined as follows:

  $$ (3)

•  

  where $$ is a sequence of non-overlapping intervals I_i = [a_i, b_i] ⊂ [x₁, y₁] and f(I_i, x₂): = f(b_i, x₂) − f(a_i, x₂).

• (2)

  Fix x₁ ∈ I = [a, b], and let [x₂, y₂] ⊂ [c, d]. The partial Λ_p-variation of f in the sense of Shiba on [x₂, y₂] is defined by the following:

  $$ (4)

•  

  where $$ is a sequence of non-overlapping intervals J_j = [c_j, d_j] ⊂ [x₂, y₂] and f(x₁, J_j): = f(x₁, d_j) − f(x₁, c_j).

• (3)

  The Hardy–Vitali Λ_p variation of the function $$ in the sense of Shiba is defined by the following:

  $$ (5)

•  

  where $$, $$ are sequences of nonoverlapping intervals I_i = [a_i, b_i] ⊂ [a, b] and J_j = [c_j, d_j] ⊂ [c, d], with

  $$ (6)

• (4)

  The total Λ_p-variation of the function $$ in the sense of Shiba is defined as follows

  $$ (7)

Observation 1. Clearly, $$, $$, $$ are non-negative and hence, $$ too.

Observation 2. We denote by $$ the set of functions with finite total Λ_p-variation in the sense of Shiba, that is,

The next theorem asserts that the set of functions $$ is a vector space, the proof is consequence of the proven properties in [17].

Theorem 2. Let f and g in $$, with 1 ≤ p < +∞, and let α be any real constant. Then,

• (1)

  $$.

• (2)

  $$.

Lemma 1. Let $$ be a function in the space $$. f is a constant function if and only if $$.

Proof. Suppose f is constant. By Definition 1, it is fulfilled that $$.

Suppose now that $$. From Observation 1, it follows that

Let $$ and $$ be sequences of nonoverlapping intervals I_i = [a_i, b_i] ⊂ [a, b] and J_j = [c_j, d_j] ⊂ [c, d], with i = 1, …, m and j = 1, …, n, then

In particular, for the sequences of nonoverlapping intervals, $$ and $$ of [a, b] and [c, d], respectively, we have as follows, by (9) that

Analogously, for any partition of non-overlapping intervals I_i = [a_i, b_i] ⊂ [a, b] and J_j = [c_j, d_j] ⊂ [c, d], we have the following:

In particular, it holds for the partition considered above, i.e., we obtain the following:

Finally, substituting (11) and (12) into (14), we conclude that f(x, y) = f(a, c) for all $$. Thus, f is constant on $$. □

Definition 2. Let $$, where 1 ≤ p < +∞. The function $$ is defined in the space $$ as $$ given by the following:

Theorem 3. Let $$, where 1 ≤ p < +∞. The function $$ defined above is a norm on the space $$.

Proof. It is easy to verify that $$ satisfies the properties of a norm on the space $$.

Lemma 2. Let $$ and $$, let $$ be a closed rectangle in ℝ², and let $$ be a function. If $$, then

• (a)

  there exists a $$ such that $$, for all f.

• (b)

  $$ for all $$,

• (c)

  there exists C > 0 such that $$, for all f.

Proof.

• (a)

  Suppose $$. For I = [a, b] and J = [c, d], we have the following:

  $$ (16)
  $$ (17)
  $$ (18)

•  

  Let us compute the value of $$,

  $$ (19)

•  

  By the inequalities (16), (17), and (18), we have the following:

  $$ (20)

• (b)

  Let $$ and $$. Let us consider $$ a sequence of nonoverlapping intervals I_i = [a_i, b_i] ⊂ [a, x₁], and $$ be a sequence of nonoverlapping intervals J_j = [c_j, d_j] ⊂ [c, x₂]. By adding [x₁, b] to the sequence of nonoverlapping intervals $$ and [x₂, d] to the sequence of nonoverlapping intervals $$, we obtain that the following inequality holds

  $$ (21)

•  

  Raising to the power (1/p) yields  $$    and analogously,

  $$ (22)

•  

  Let us consider the sequences of nonoverlapping intervals $$ and $$ of [a, x₁] and [c, x₂], respectively, and by an argument analogous to that above, we can add [x₁, b] and [x₂, d] to nonoverlapping interval sequences $$ and $$, respectively, to get the sequences of nonoverlapping intervals of [a, b] and [c, d]. Then,

  $$ (23)

•  

  Raising to the power (1/p) yields$$.

•  

  Hence,

  $$ (24)

•  

  Therefore, $$ for all $$.

• (c)

  It follows from Part (a) and (b) that there exists a $$ such that

  $$ (25)

with $$. Thus,

Taking supremum over all $$ leads to the following:

Lemma 3. Let $$, $$ and $$ be a closed rectangle in ℝ², and let $$ be a function. If $$, then f is bounded.

Proof. The proof is an immediate consequence of part (c) of Lemma 2.

Theorem 4. Let $$ be a closed rectangle in ℝ². Then, $$ is a Banach space.

Proof. Theorem 3 shows that $$ is a norm on the space $$. We need to prove that $$ is a complete space. Suppose $$ is a Cauchy sequence in $$, which is to say that there exists N > 0 such that

That is,

Thus,

Since $$, then $$ for every s ≥ 1. It follows from Part (c) of Lemma 2 that there exists C > 0 such that $$ for all s, r ≥ 1, so that the sequence $$ is Cauchy uniformly on $$. Because this space is complete, there exists a function $$ with ‖f‖_∞ < +∞ such that f_s⟶f uniformly on $$. To see that $$, let us consider the sequences of non-overlapping intervals $$ and $$ of [a, b] and [c, d], respectively, where I_i = [a_i, b_i] ⊂ [a, b] and J_j = [c_j, d_j] ⊂ [c, d], then

Now, we know that

Therefore,

Taking supremum over $$, we have the following:

For every s ≥ N. Similarly

By the inequalities (34) and (35), for every s ≥ N we have the following:

Thus, $$ for every s ≥ N. Since $$ and $$ is a vector space, $$. Now, we can conclude that $$ converges to f in the norm $$. Indeed, for every s ≥ N,

On the other hand, by the inequality (29),

So, fixing s and taking limit respect to r, we have as follows:

Using (36) and (39) into (37)

Therefore, $$ is a Banach space.

Lemma 4. Suppose the hypotheses $$ and $$ hold. Let the integral function F be defined by $$ for all $$ and all $$, and let A be the area of the rectangle $$. Then,

Proof. Let $$ be as in hypothesis $$, $$ and $$, then H(x) = f(x, u(x)) is measurable and bounded in $$, hence H(x) is Lebesgue integrable. By the hypothesis $$, K(x, ·) is Lebesgue integrable for all $$. Thus, F(u) is well defined. Let $$ and $$ be sequences of nonoverlapping intervals of [a, b] and [c, d], respectively, where I_i = [a_i, b_i] ⊂ [a, b] and J_j = [c_j, d_j] ⊂ [c, d]. We know that the following:

Let us study each of these variations separately.

Because p ≥ 1, the function x^p is convex on ℝ. Moreover, by the hypothesis, K is Lebesgue integrable and also f(·, u(·)) is bounded, and so

is a Lebesgue integrable. Applying Jensen’s inequality yields as follows:

Raising both sides of the inequality above to the power (1/p) yields as follows:

Taking supremum over $$ in the inequality above, and using the hypothesis $$, i.e., that $$ where M is L_p integrable, we obtain the following:

Analogously, it is shown that

Now, we compute $$:

where ΔK((I_i, J_j), y) = K((a_i, c_j), y) + K((b_i, d_j), y) − K((a_i, d_j), y) − K((b_i, c_j), y). Applying Jensen’s inequality, we obtain the following:

Raising both sides of the inequality to the power (1/p) yields the following:

Taking supremum over {I_i × J_j}, and by the hypothesis $$, we have the following:

It follows then from the inequalities (47), (48), and (52) that

which completes the proof.

Lemma 5. Suppose the hypotheses $$ and $$ hold. Let F(u)(x) be the integral function for all $$ and all $$, defined as in the preceding lemma. Then, for all $$,

Proof. Consider $$,  $$ sequences of nonoverlapping intervals of [a, b] and  [c, d], respectively, where I_i = [a_i, b_i] ⊂ [a, b] and J_j = [c_j, d_j] ⊂ [c, d]. Let $$ and let r > 0 such that $$. L_r denotes the Lipschitz constant of f corresponding to the cube $$. We know that

Let us study again each of these variations separately. To calculate the variation $$ we can proceed similarly as in the proof of Lemma 4. So, we obtain the following:

Raising to the power (1/p), by part (c) of Lemma 2, and by the hypothesis $$, we have the following:

Taking supremum over $$,

It is shown analogously as follows:

Let us then compute

where

Note that

with

By Jensen’s inequality, and by the fact that f is locally Lipschitz, we have the following:

By raising both sides of the inequality above to the power (1/p), and by part (c) of Lemma 2,

Taking supremum over {I_i × J_j}, and by the hypothesis $$, we obtain as follows:

So, we conclude from the inequalities (58) and (59), and (66) that

and the proof is complete.

The following lemma is a direct consequence of the triangle inequality.

Lemma 6. Suppose the hypotheses $$, $$, $$ hold. Let $$ be the function defined by G(u)(x) = v(x) + λF(u)(x), where F(u) is as in Lemma 4 and λ ∈ I where I is a bounded interval. Then,

Lemma 7. Let $$, and let K : T⟶ℝ be a function with $$, $$ and p ≥ 1. Then, for

we have that

where

Proof. Let {I_i × J_j}, where $$ and $$ sequences of nonoverlapping intervals of [0, b] and [0, d], respectively, where I_i = [a_i, b_i] ⊂ [0, b] and J_j = [c_j, d_j] ⊂ [0, d]. Let us consider $$. We know that

First, we determine $$. Since $$ is a sequence of nonoverlapping intervals I_i = [a_i, b_i] ⊂ [0, b], let t^′ = (b_r, 0) and s₁ ∈ [a_r, b_r] for some r, with 1 ≤ r ≤ m. Let us consider s = (s₁, 0) ∈ [0, b] × {0}, then,

By raising to the power (1/p), taking supremum over $$, and applying the property (a + b)^(1/p) ≤ a^(1/p) + b^(1/p) if (1/p) < 1, we obtain as follows:

It is shown analogously that

To compute $$, let t^′ = (b_r, d_l) for some r, l with 0 ≤ r ≤ m, 0 ≤ l ≤ n, and let s ∈ [a_r, b_r] × [c_l, d_l] be an arbitrary fixed point. So

Now, we can display the terms of summation and substitute the values of $$ according to (70) to obtain as follows:

Raising to the power of (1/p), taking supremum over {I_i × J_j}, and applying the property (a + b)^(1/p) ≤ a^(1/p) + b^(1/p), we have as follows:

It follows from (75) and (76), and (78) that

taking

yields

which completes the proof.

Theorem 5. Suppose the hypotheses $$, $$, and$$ hold. Then, there exists a number ρ > 0 such that for every |λ| < ρ, equation (1) has a unique solution in $$ defined on $$.

Proof. Let {I_i × J_j}, where $$ and $$ be sequences of nonoverlapping intervals of [a, b] and [c, d], respectively, where I_i = [a_i, b_i] ⊂ [a, b] and J_j = [c_j, d_j] ⊂ [c, d]. Let r > 0 such that $$, and L_r denotes the Lipschitz constant of f corresponding to the cube $$. $$ denotes the closed ball with center 0 and radius r in the space $$, that is,

Using part (c) of Lemma 2, choose a number ρ > 0 such that

and where C is guaranteed by part (c) of Lemma 2, and A is the area of the rectangle $$. In order to prove this theorem, the Banach contraction principle is needed (Theorem 1). Fix |λ| < ρ and define the operator 4$$ by

It can be verified that it is well defined, as in the proof of Lemma 4. We show first that $$. Indeed, for every $$, it follows from Lemma 6 that

For the second term in the right hand side of the inequality above, we have as follows

By Lemma 4, the inequality (88) can be rewritten as follows:

Substituting the inequality (89) into (87) yields the following:

Therefore, $$. To show now that G is a contraction mapping, let $$, that is, $$. So,

It is straightforward to compute that

By the fact that f is locally Lipschitz, and by part (c) of Lemma 2, we have the following:

Lemma 5 implies as follows:

Moreover, it follows from the inequality (94) that

Thus, G is a contraction mapping, by the inequality (86). Theorem 1 implies that G has a unique fixed point in $$, which is to say that there exists a unique $$ such that

Therefore, u is a unique solution of (1).

Theorem 6. Suppose the hypotheses $$, $$, and $$ hold. Then, there exists a rectangle $$ such that the equation (2) has a unique solution in Λ_pBV(ℛ).

Proof. Let r and L_r be denoted as in the proof of the preceding theorem. Let us denote $$ with $$ and $$, $$ the closed ball with center 0 and radius r in the space $$, that is,

Choose $$ with $$ such that

where the existence of $$ are assured by part (c) of Lemmas 2 and 7, respectively, and A is the area of the rectangle $$. Let {I_i × J_j}, where $$ and $$ are sequences of nonoverlapping intervals of $$ and $$, respectively, with $$ and $$. We define the functions $$ and the operator $$    by $$, where

With $$, it can be verified that it is well defined, as in the proof of Lemma 4.

As in the proof of Theorem 5, we will use the Banach contraction principle (Theorem 1) to prove this theorem. We begin by showing that $$. So, let $$,

It can easily be shown that

Thus,

However,

In addition,

To compute $$, we proceed as in the proof of Lemma 4. Let t^′ = (b_r, 0),   $$ for some r, with 1 ≤ r ≤ m and let us consider $$, $$. Then,

Raising both sides of the inequality above to the power (1/p) yields the following:

Now,

By taking supremum over {I_i} in the inequality (107), by Lemma 7 and the Hypothesis $$, we conclude as follows:

It is shown analogously as follows:

and

By the inequalities (109)–(111), we obtain the following:

Substituting the inequality (113) into (104) leads to the following:

Next, we show that $$ is a contraction mapping. For $$, we have as follows:

Now we compute the variation $$. Let t^′ = (b_i, 0) and $$ for some i, then, for $$

By part (c) of Lemma 2, the inequality (108), the Hypothesis $$, and raising both sides of the inequality above to the power (1/p), we obtain the following:

An argument analogous to the one we used above shows the following:

Therefore,

The inequality (98) implies that $$ is a contraction mapping; and by Theorem 1, $$ has a unique fixed point in $$, that is, there exists a unique $$ such that

Hence, $$ is a unique solution of the (2).

Example 1. Let $$, and let the functions $$, $$ and $$ be defined by v(x₁, x₂) = cos(x₁ + x₂), $$ and $$, respectively. We want to show that the nonlinear Hammerstein–Volterra integral equation

has a unique solution $$.

In order to solve this problem, we verify the hypotheses of Theorem 6.

• (1)

  v(x₁, x₂) = cos(x₁ + x₂) is measurable in $$, to show that v(x₁, x₂) belongs to $$, let {I_i × J_j}, where $$ and $$ be sequences of nonoverlapping intervals of [0,1] and [0,1], respectively, where I_i = [a_i, b_i] ⊂ [0,1] and J_j = [c_j, d_j] ⊂ [0,1]. To determine $$, $$ and $$, let us first compute

  $$ (121)

•  

  From Lemma 5 in [24] we have the following:

  $$ (122)

•  

  Hence,

  $$ (123)

•  

  Thus, $$. An argument entirely analogous to that above leads to $$. On the other hand,

  $$ (124)

•  

  Now, using the inequality |a + b|^p ≤ 2^p(|a|^p + |b|^p), the mean value theorem, and the inequality (122) to obtain

  $$ (125)

•  

  Then,

  $$ (126)

•  

  Thus, $$ and in consequence, $$.

• (2)

  We need to show now that $$ is locally a Lipschitz on $$. Indeed, let r > 0,  $$, and $$. Then

  $$ (127)

•  

  So there exists L_r = max {2,2r} such that

  $$ (128)

•  

  Moreover, therefore f is locally a Lipschitz constant on $$.

  $$ (129)

Hence, K(x, (·, ·)) is Lebesgue integrable. On the other hand,

We proceed as in Part 1 of this example to show that $$, and so it is evident that M(y) is L_p integrable. Thus, all the conditions of Theorem 6 are satisfied, therefore, there exists a rectangle $$ such that the (120) has a unique solution in $$.