Riesz-Kolmogorov theorem in variable exponent Lebesgue spaces and its applications to Riemann-Liouville fractional differential equations

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1 . ARTICLES. SCIENCE CHINA Mathematics October 2018 Vol. 61 No. 10: Riesz-Kolmogorov theorem in variable exponent Lebesgue spaces and its applications to Riemann-Liouville fractional differential equations Baohua Dong 1, Zunwei Fu 2,3, & Jingshi Xu 4 1 Department of Mathematics, Nanjing University of Information Science and Technology, Nanjing , China; 2 Department of Mathematics, Linyi University, Linyi , China; 3 Department of Computer Science, The University of Suwon, Hwaseong , Korea; 4 Department of Mathematics, Hainan Normal University, Haikou , China baohua dong@126.com, fuzunwei@lyu.edu.cn, jingshixu@126.com Received December 21, 2017; accepted March 4, 2018; published online August 16, 2018 Abstract In this paper, we obtain the necessary and sufficient condition of the pre-compact sets in the variable exponent Lebesgue spaces, which is also called the Riesz-Kolmogorov theorem. The main novelty appearing in this approach is the constructive approximation which does not rely on the boundedness of the Hardy-Littlewood maximal operator in the considered spaces such that we do not need the log-hölder continuous conditions on the variable exponent. As applications, we establish the boundedness of Riemann-Liouville integral operators and prove the compactness of truncated Riemann-Liouville integral operators in the variable exponent Lebesgue spaces. Moreover, applying the Riesz-Kolmogorov theorem established in this paper, we obtain the existence and the uniqueness of solutions to a Cauchy type problem for fractional differential equations in variable exponent Lebesgue spaces. Keywords Lebesgue space with variable exponent, Riesz-Kolmogorov theorem, Riemann-Liouville fractional calculus, fixed-point theorem MSC(2010) 42B35, 26A33 Citation: Dong B H, Fu Z W, Xu J S. Riesz-Kolmogorov theorem in variable exponent Lebesgue spaces and its applications to Riemann-Liouville fractional differential equations. Sci China Math, 2018, 61: , 1 Introduction Let Ω R n be an open set and p( ) be the measurable function defined on Ω and valued in [1, ). Then the Lebesgue space with the variable exponent, L p( ) (Ω), is defined by L p( ) (Ω) : {f is measurable: ϱ L p( ) (Ω)(f/λ) < for some λ (0, )}, where, for any measurable function f and λ (0, ), ϱ L p( ) (Ω)(f/λ) : f(x)/λ p(x) dx. * Corresponding author Ω c Science China Press and Springer-Verlag GmbH Germany, part of Springer Nature 2018 math.scichina.com link.springer.com

2 1808 Dong B H et al. Sci China Math October 2018 Vol. 61 No. 10 It is also well known that L p( ) (Ω) is a Banach space when it is equipped with the following norm f L p( ) (Ω) for any f L p( ) (Ω): f L p( ) (Ω) : inf{λ > 0 : ϱ L p( ) (Ω)(f/λ) 1}. (1.1) We note that ϱ L (Ω)(f) is a modular if p( ) is finite almost everywhere. It is worth mentioning that p( ) f L p( ) (Ω) < if and only if ϱ L (Ω)(f) <. p( ) Variable exponent Lebesgue spaces have been systemically studied by Kováčik and Rákosník [16] in In [16], Kováčik and Rákosník established an existence theorem for the weak solution to the nonlinear Dirichlet boundary value problem with coefficients of variable growth in the spaces. After that, the theory of variable function spaces has been widely applied in differential equations and image restore (see, for example, [2, 14, 18]). In the past two decades, function spaces with variable exponents have been intensively studied (see, for example, [3, 7, 19, 24, 26, 27, 30, 36] and the references therein). Compared with the properties in the theory of classical Lebesgue spaces, the difficulty is that it fails to be rearrangement-invariant in Lebesgue spaces with variable exponents. The fact means that neither good-λ technique nor rearrangement inequalities may be applied for generalizations of some well-known results in harmonic analysis to the case of variable exponent Lebesgue spaces. In addition, variable exponent Lebesgue spaces also fail to be translation invariable. However, by adding some conditions on the variable exponents p( ), many useful results about the boundedness of operators in classical function spaces can be parallel to the variable exponent setting. One of the important conditions on the variable exponents p( ) is the following log-hölder continuous conditions, which contain two parts: (i) p( ) is said to be log-hölder continuous on R n if there exists c log (0, ) such that p(x) p(y) c log log(e + 1/ x y ), x, y Rn, x y < 1 2. (ii) p( ) is said to satisfy the log-hölder decay condition if there exist p R and a constant c log (0, ) such that c log p(x) p log(e + x ), x Rn. and We use the following notation: p : ess inf{p(x) : x Ω}, p + : ess sup{p(x) : x Ω} P(Ω) : {p( ) is measurable: p (1, ) and p + [p, )}. We denote by P log (R n ) the class of all variable exponents p( ) P(R n ) which satisfy (i) and (ii). Based on the log-hölder continuous conditions, the boundedness of the Hardy-Littlewood maximal operator in the variable exponent Lebesgue spaces has been proved (see [4]). Recall that, for any f L 1 loc (Rn ), the Hardy-Littlewood maximal function Mf is defined by Mf(x) : sup r n f(y) dy, x R n, r>0 B(x,r) where B(x, r) : {y R n : x y < r}. By the boundedness of the Hardy-Littlewood maximal operator in variable exponent spaces, many important results in the classical harmonic analysis and the function space theory can be obtained in the variable exponent function spaces (see, for example, [4, 5, 28, 29, 31, 37, 38]). The Kolmogorov theorem (about pre-compact sets) is an important tool to find the solutions to the differential equations in function spaces. In the past decades, many researchers paid attention to characterizations of pre-compact sets in function spaces (see, for example, [12, 13, 21, 23] and the references

3 Dong B H et al. Sci China Math October 2018 Vol. 61 No therein). Recently, one interesting thing is the characterization of pre-compact sets in the variable exponent Lebesgue spaces. Observe that in [21], Rafeiro obtained the necessary and sufficient condition of the pre-compact set in the variable exponent Lebesgue spaces with the domain R n. Later, in [13], Górka and Macios generalized the result to the variable exponent Lebesgue spaces where the domain is a metric measure space. In their proofs, the property of the boundedness of the Hardy-Littlewood maximal operator in variable exponent spaces was used. It is the reason why they need to restrict the variable exponent p( ) P log. It is worth pointing out that, in [13], they also obtained a sufficient condition of the pre-compact set in the variable exponent Lebesgue spaces with the domain R n. In this case, their proof does not rely on the boundedness of the Hardy-Littlewood maximal operator in variable exponent spaces. So they do not need the variable exponent condition p( ) P log in their theorem on the sufficient condition of the pre-compact set in the variable exponent Lebesgue spaces with the domain R n. In Section 2, we give the necessary part of the theorem in [13] (see Theorem 2.1 below, which is called the Riesz-Kolmogorov theorem). Furthermore, in the sufficient part of Theorem 2.1, the main idea of our method is the constructive approximation and is different from the method used in [13], which focuses on the division of the domain. The proof of Theorem 2.1 is also different from [21], because we do not need the tool about the boundedness of the Hardy-Littlewood maximal operator in variable exponent spaces. So we can drop out the condition that the variable exponent p( ) P log. In the rest of the paper, we give some applications of the pre-compact sets in Lebesgue spaces with the variable exponent (namely, the Riesz-Kolmogorov theorem). We are interested in investigating the Riemann-Liouville calculus in variable exponent Lebesgue spaces. In Section 3, we first show that the space L p( ) (Ω) is isomorphic to the space L p( ) (Ω). Second, we prove that the Riemann-Liouville integral operator I0 α is bounded in variable exponent Lebesgue spaces. Third, by truncating I α + 0, we construct a + compact operator in variable exponent Lebesgue spaces. In Section 4, using Theorem 2.1, together with the Schauder fixed-point theorem, we prove that the Cauchy type problem (4.1) has a unique solution in variable exponent Lebesgue spaces. Then we give an example to show that the condition in Theorem 4.1 is only sufficient. Throughout the paper, C stands for a positive constant but it may change from line to line. We sometimes use the notation a b as an equivalent of a Cb. χ Ω represents the characteristic function of set Ω. 2 Pre-compact sets in variable exponent Lebesgue spaces In this section, we give a new proof for the Riesz-Kolmogorov theorem on characterization of pre-compact sets in Lebesgue spaces with the variable exponent. Our method differs from that used in [13,21]. Indeed, we use the method in [1] for constant exponent Lebesgue spaces. Before stating our result, we need to recall some notation. If a function f is defined on Ω R n, a.e., we then let f denote the zero extension of f outside Ω, namely, { f(x), if x Ω, f(x) : 0, if x R n \Ω. We first recall the definition of the pre-compact set in the variable exponent Lebesgue spaces. Definition 2.1. Let Ω be an open subset in R n and p( ) P(Ω). Then a subset of L p( ) (Ω) is said to be pre-compact if its closure is compact. Now we can state our result, which is also called the Riesz-Kolmogorov theorem. Theorem 2.1. Let Ω be an open subset in R n and p( ) P(Ω). If E is a bounded subset in L p( ) (Ω), then E is precompact in L p( ) (Ω) if and only if for every number ϵ (0, ) there exist a number δ (0, ) and a subset G Ω (G is a compact subset in Ω) such that for every f E and h R n with h < δ both of the following inequalities hold: f( + h) f( ) L p( ) (Ω) < ϵ (2.1)

4 1810 Dong B H et al. Sci China Math October 2018 Vol. 61 No. 10 and f L p( ) (Ω\G) To show Theorem 2.1 we need some preliminaries. < ϵ. (2.2) Lemma 2.1. A set E is pre-compact in a Banach space X if and only if for every positive number ϵ there is a finite subset N ϵ of points of X such that E B ϵ (y), y N ϵ where B ϵ (y) denotes a ball centered at y with radius ϵ. The set N ϵ is called an ϵ-net of E. Lemma 2.2 (Ascoli-Arzela theorem). Let Ω be a bounded domain in R n. A subset E of C(Ω) is pre-compact in C(Ω) if the following two conditions hold: (i) E is uniformly bounded, which means that there exists a positive constant M such that u(x) M holds for any x Ω and u E. (ii) E is equi-continuous, which means that for any ϵ (0, ) there exists δ (0, ) such that if u E, x, y Ω, and x y < δ, then u(x) u(y) < ϵ. Lemma 2.3 (See [6, Lemma 2.2]). Let Ω be an open subset in R n and p( ) P(Ω). Then min{ϱ L p( ) (Ω)(f) 1 p, ϱl p( ) (Ω)(f) 1 p + } f L p( ) (Ω) max{ϱ L p( ) (Ω)(f) 1 p, ϱl p( ) (Ω)(f) 1 p + }. The proof of Theorem 2.1 needs some ideas of the constructive approximation. In what follows, the notation C (R n ) means the set of all infinitely differentiable functions on R n and we also use C 0 (R n ) to denote the set of all C (R n ) functions with compact supports. Let J be a nonnegative, real-valued function belonging to C 0 (R n ) and having the following properties: (i) J(x) 0, x [1, ); (ii) R n J(x)dx 1. Let J ϵ (x) : ϵ n J(x/ϵ) for any x R n and ϵ (0, ). For any suitable function u and x R n, let J ϵ u(x) : J ϵ (x y)u(y)dy. R n It is well known that if u L 1 loc (Rn ), then J ϵ u C (R n ) (see, for example, [1]). Now we prove Theorem 2.1. Proof of Theorem 2.1. We first show that {J η f : f E} is a pre-compact set in C(G). Fix η (0, ). By Hölder s inequality, we conclude that, for any f E and x Ω, J η f(x) J η (x y)f(y)dy R n J η (x y) f(y) dy R n J η (x y) f(y) dy + J η (x y) f(y) dy {y R n : f(y) 1} [ J η (x y)dy R n [ + J η (x y)dy R n [ ] [ 1/p sup J η (y) y R n ] 1/p [ {y Rn: f(y) 1} ] 1/p + [ {y Rn: f(y) <1} {y R n : f(y) 1} {y R n : f(y) <1} J η (x y) f(y) p dy ] 1/p f(y) p dy ] 1/p ] 1/p+ J η (x y) f(y) p + dy

5 where Dong B H et al. Sci China Math October 2018 Vol. 61 No [ ] [ 1/p+ + sup J η (y) y R n [ ] [ 1/p sup J η (y) f(y) p(x) dy y R n R n {y R n : f(y) <1} ] 1/p [ ] [ 1/p+ + sup J η (y) f(y) p(x) dy y R n R n ] 1/p+ f(y) p + dy ] 1/p+ C η {[ϱ p( ) (f)] 1/p + [ϱ p( ) (f)] 1/p + }, (2.3) {[ ] 1/p, [ ] 1/p+ } C η : max sup J η (y) sup J η (y). y R n y R n By the boundedness of E in L p( ) (Ω) and Lemma 2.3, we know that {J η f : f E} is a uniformly bounded set. Let T h f denote the translation of f by h R n, namely, T h f(x) : f(x + h). Similar to the proof of (2.3), for any f E and x Ω, we have J η f(x + h) J η f(x) C η [ϱ L p( ) (Ω)(T h f f) 1/p + ϱ L p( ) (Ω)(T h f f) 1/p+ ]. Because of (2.1), we obtain lim T hf f L h 0 p( ) (Ω) 0, which is uniform for any f E. For any f E, by Lemma 2.3, we know that So, for any f E and x Ω, we have lim ϱ L (Ω)(T h 0 p( ) h f f) 0. lim J η f(x + h) J η f(x), h 0 which means that {J η f : f E} is an equi-continuous set. Thus, {J η f : f E} is pre-compact in C(G). From this and Lemma 2.1, we deduce that there exists a finite set {ψ 1,..., ψ m } of functions in C(G) such that if f E, then, for some j {1, 2,..., m} and any x G, we have ψ j (x) J η f(x) < ( ) p+/p ϵ G 1/p, 3 where, for simplicity, we assume that ϵ (0, ) is sufficiently small such that the right-hand side of the last inequality is less than 1. Then ψ j (x) J η f(x) p(x) dx ψ j (x) J η f(x) p dx G G ( ) p+ ( ) p+ ϵ ϵ G 1 G <. 3 3 By Lemma 2.3, we know that ψ j J η f L p( ) (G) < ϵ 3. Thus, we have f ψ j L p( ) (R n ) f L p( ) (R n \G) + f ψ j L p( ) (G)

6 1812 Dong B H et al. Sci China Math October 2018 Vol. 61 No. 10 Next, we need to prove that < ϵ 3 + f J η f L p( ) (G) + J η f ψ j L p( ) (G) < 2ϵ 3 + f J η f L p( ) (G). f J η f L p( ) (G) < ϵ 3. First, we prove the sufficiency part of Theorem 2.1. It suffices to prove the special case Ω R n, as it follows for any general Ω from its application in this special case to the set Ẽ : { f : f E}, with p( ) extended to the whole R n by p(x) : p for x R n \Ω. For any ϵ (0, ), we pick G R n such that for any f E, f L p( ) (R n \G) < ϵ 3. Let f L p( ) (R n ). Then, for any η (0, ), the function J η f belongs to C (R n ) and, in particular, to C(G). Therefore, if f L p( ) (R n ), we have, for almost every x R n, J η f(x) f(x) J η (y) [f(x y) f(x)] dy. R n By Hölder s inequality and 1/p(x) + 1/p (x) 1, we conclude that J η f(x) f(x) J η (y)[f(x y) f(x)]dy R n [ ] 1/p (x)[ J η (y)dy J η (y) T y f(x) f(x) p(x) dy R n B η [ ] 1/p(x) J η (y) T y f(x) f(x) p(x) dy, B η ] 1/p(x) where B η denotes a ball centered at 0 with radius η. Then, by the Fubini theorem, we have J η f(x) f(x) p(x) dx J η (y) T y f(x) f(x) p(x) dydx R n R n B η J η (y) T y f(x) f(x) p(x) dxdy B η R n sup T h f(x) f(x) p(x) dx. h B η R n So, we conclude that Condition (2.1) implies that uniformly for any f E, so the limit J η f f L p( ) (R n ) sup h B η T h f f L p( ) (R n ). lim T hf f L h 0 p( ) (R n ) 0 lim J η f f L p( ) (R η 0 n ) 0 is uniform for any f E. In other words, for some fixed η (0, ), the inequality holds for any f E. J η f f L p( ) (G) < ϵ 3

7 Dong B H et al. Sci China Math October 2018 Vol. 61 No Therefore, we obtain f ψ j L p( ) (R n ) < 2ϵ 3 + f J η f L p( ) (G) < ϵ, which means that E has a finite ϵ-net in L p( ) (R n ). By Lemma 2.1, we know that E is a pre-compact set. Now we turn to prove the necessary part of Theorem 2.1. Suppose E is a pre-compact set in L p( ) (Ω). Without loss of generality, we may assume ϵ (0, 1). We have the following two properties. By Lemma 2.1, E has a finite ϵ/6-net. Since C 0 (Ω) is dense in L p( ) (Ω) (see [5, Theorem ]), it follows that there exists a finite set F of continuous functions with compact supports in Ω such that for each f E, there exists a ϕ F satisfying f ϕ L p( ) (Ω) < (ϵ/3) p +/p. Note that the number of the functions in F is finite, and let G be the union of the supports of the functions in F. Then G Ω and f L p( ) (Ω\G) f ϕ L p( ) (Ω) < (ϵ/3) p +/p < ϵ, which shows (2.2). To prove (2.1), we choose a closed ball B r centered at the origin with radius r and containing G. Note that, for any x R n, (T h ϕ ϕ)(x) ϕ(x + h) ϕ(x) is uniformly continuous and vanishes outside B r+1 provided h [0, 1). Thus, lim T hϕ ϕ L p( ) (R h 0 n ) 0. Because F is finite, the convergence is uniform for any ϕ F. Thus, for h sufficiently small, we have T h ϕ ϕ L p( ) (Ω) < ϵ/3. Besides, let ϕ F satisfy By Lemma 2.3, we know that f ϕ L p( ) (Ω) < (ϵ/3) p +/p. ϱ L p( ) (Ω)(f ϕ) < (ϵ/3) p +. From the Lebesgue dominated convergence theorem, it follows that Let h R n be sufficiently small. We then have By Lemma 2.3 again, we conclude that lim ϱ L (Ω)(T h 0 p( ) h f Th ϕ) ϱ L (Ω)(f ϕ). p( ) ϱ L p( ) (Ω)(T h f Th ϕ) < (ϵ/3) p +. T h f Th ϕ L p( ) (Ω) < ϵ/3. Therefore, for sufficiently small h R n (independent of f E), we obtain T h f f L p( ) (Ω) T h f Th ϕ L p( ) (Ω) + T h ϕ ϕ L p( ) (Ω) + f ϕ L p( ) (Ω) < ϵ/3 + ϵ/3 + ϵ/3 ϵ, which shows (2.1). This finishes the proof of Theorem 2.1.

8 1814 Dong B H et al. Sci China Math October 2018 Vol. 61 No Bounded and compact operators in variable exponent Lebesgue spaces In the rest of this paper, for simplicity, we work on a subspace L p( ) (Ω) of the variable exponent Lebesgue space L p( ) (Ω), which is defined by L p( ) (Ω) : {f L p( ) (R) : f 0 and p p a.e. on R\Ω}, since both L p( ) (Ω) and L p( ) (Ω) are isomorphic. Proposition 3.1. Proof. it is obvious that Let Ω be a subset in R and p P(Ω). Then L p( ) (Ω) is isomorphic to L p( ) (Ω). By the definition of the norm in variable exponent Lebesgue spaces and the equation ϱ L (Ω)(f) f(x) p(x) dx f(x)χ p( ) Ω p(x) dx ϱ Lp( ) (Ω) (f), This finishes the proof of Proposition 3.1. Ω f L p( ) (Ω) f Lp( ) (Ω). The next theorem is to prove the boundedness of Riemann-Liouville fractional integral operators in variable exponent Lebesgue spaces. First of all, we present some lemmas that will be used later. Lemma 3.1 (See [5, Lemma 3.2.4]). Let Ω be a subset in R n and p( ) P(Ω). Then f L p( ) (Ω) 1 if and only if ϱ L p( ) (Ω)(f) 1. Let f L p( ) (Ω). Then (a) if f L p( ) (Ω) 1, then ϱ L (Ω)(f) f p( ) L (Ω); p( ) (b) if 1 < f L (Ω), then f p( ) L p( ) (Ω) ϱ L (Ω)(f). p( ) A measure µ is atom-less if for any measurable set A with µ(a) > 0 there exists a measurable subset A of A such that µ(a) > µ(a ) > 0. Now we introduce the following lemma. Lemma 3.2 (See [5, Corollary 3.3.4]). Let Ω be a subset in R n, p( ), q( ) P 0 (Ω), and the measure µ be atom-less with µ(ω) <. Then L p( ) (Ω) L q( ) (Ω) if and only if q(x) p(x) for almost every x Ω. The embedding constant is less than or equal to the minimum between 2 max{µ(ω) (1/q 1/p) +, µ(ω) (1/q 1/p) } and 2[1 + µ(ω)]. A function f L p( ) (Ω) has an absolutely continuous norm if for every decreasing sequence {G n } n N of subsets of Ω satisfying µ(g n ) 0 as n, there exists fχ Gn L p( ) (Ω) 0 as n. We say that L p( ) (Ω) has an absolutely continuous norm if every f L p( ) (Ω) has an absolutely continuous norm. The following lemma shows that L p(x) (Ω) has an absolutely continuous norm under a suitable condition. Lemma 3.3 (See [20, Lemma 3.1]). Let Ω be a subset in R n and p P 0 (Ω). Then the following statements are equivalent: (i) p + < ; (ii) L p( ) (Ω) has an absolutely continuous norm. In the remainder of this section, let Ω : (0, δ) with a positive constant δ. We recall the notions of the Riemann-Liouville integral and its derivatives. Let δ be a positive constant. For any α (0, 1), the Riemann-Liouville derivative of order α is defined by D0 α +f(t) : 1 d Γ(1 α) dt R (0,t) f(τ) dτ, t [0, δ]. (t τ) α Correspondingly, the Riemann-Liouville integral of order α is given as I0 α 1 f(τ) +f(t) : dτ, t [0, δ]. Γ(α) (t τ) 1 α (0,t) The fractional calculus has a long history. It appeared in a letter from L Hôpital to Leibniz in Up to now, it has been developed for more than three hundred years. Many famous mathematicians

9 Dong B H et al. Sci China Math October 2018 Vol. 61 No paid their attention to it, such as, Euler, Fourier, Abel, Liouville, Riemann, Hadamard and so on (see, for example, [15]). After 1900, the fractional calculus experiences a fast development because more and more non-classical phenomena in science and engineering can be described by it. The theory not only provides a useful tool for problems in elasticity, electromagnetic theory and signal processing (see [15]), but also indicates potential ways for solving integral, differential and some related problems (see [8 10, 17, 25, 32, 34, 35]). Now we establish the boundedness of Riemann-Liouville integral operators in the variable exponent Lebesgue spaces. Theorem 3.1. Let δ (0, ), p( ) P(0, δ) and α (1/p, 1). Then I α 0 + is bounded in L p( ) (0, δ). Proof. For any f L p( ) (0, δ) with f L p( ) 1, we need to prove that there exists a positive constant C, depending on δ, such that I0 α f + L p( ) C. Let A : {x (0, δ) : I0 α +f(x) 1}. Then, by Hölder s inequality and Lemma 3.2, we have ϱ L (I α p( ) 0 +f) : I0 α +f(x) p(x) dx \A \A \A I0 α +f(x) p(x) dx + 1 Γ(α) f p + L p( ) By Lemma 3.1, we further conclude that which completes the proof of Theorem 3.1. (0,x) f p+ L p (0,x) A f(t) (x t) {[ I α 0 +f(x) p(x) dx 1 α dt (x t) (0,x) dx + δ p+ x (αp 1)p+/p dx + δ 1. I α 0 +f L p( ) 1 <, (α 1)p p 1 dt] (p 1)/p } p+ dx + δ The following theorem is to prove the compactness of truncated Riemann-Liouville integral operators in the spaces L p( ) (0, δ). Theorem 3.2. Let δ (0, ), p( ) P(0, δ) and max{1/p, 1/2(1 + 1/p 1/p + )} < α < 1. Then T u(t) : χ (t) Γ(α) t 0 u(τ) dτ, t [0, δ], (t τ) 1 α is a compact operator in L p( ) (0, δ). Proof. Let K be an arbitrary bounded set in L p( ) (0, δ). We need to prove that T (K) is a pre-compact set in L p( ) (0, δ). By Theorem 2.1, we only need to show the following two things: (i) For every ϵ (0, ) there exists δ (0, ) such that T u( + y) T u( ) Lp( ) < ϵ, where u K and y (0, δ) with sufficiently small y. (ii) For every ϵ (0, ) there exists a subset M (0, δ) such that T u Lp( ) (\M) < ϵ,

10 1816 Dong B H et al. Sci China Math October 2018 Vol. 61 No. 10 where u K. With Lemma 3.3, it is easy to know that (ii) is obviously true by choosing a suitable M such that the measure of set (0, δ)\m is small enough. For (i), by Lemma 2.3, we only need to show that is sufficiently small when y approaches to zero. Let u Lp( ) 1 for any u K and It follows that ϱ Lp( ) (T u( + y) T u( )) B : {x (0, δ) : T u(x + y) T u(x) 1}. ϱ Lp( ) (T u( + y) T u( )) \B \B : I 1 + I 2. T u(x + y) T u(x) p(x) dx T u(x + y) T u(x) p(x) dx + B T u(x + y) T u(x) p(x) dx T u(x + y) T u(x) p+ dx + T u(x + y) T u(x) p dx B To proceed, we estimate T u(x + y) T u(x) first. For any x (0, δ), y (0, δ) being sufficiently small such that x + y (0, δ) and α (0, 1), by Hölder s inequality and Lemma 3.2, we have (0,x) x α y α x y α, T u(x + y) T u(x) 1 u(t) u(t) Γ(α) dt (0,x+y) (x + y t) 1 α (0,x) (x t) 1 u(t) Γ(α) (0,x) (x + y t) 1 α u(t) dt (x t) 1 α + 1 u(t) Γ(α) dt (x,x+y) (x + y t) 1 α 1 Γ(α) u(t) (x t)1 α (x + y t) 1 α (0,x) (x t) 1 α dt (x + y t) 1 α + 1 u(t) Γ(α) dt (x,x+y) (x + y t) 1 α y 1 α u(t) (0,x) (x t) 1 α dt (x + y t) 1 α + 1 u(t) Γ(α) dt (x,x+y) (x + y t) 1 α { [ y 1 α 1 u Lp (0,x) (x t) 1 α (x + y t) 1 α + u Lp (x,x+y) [ y 1 α u Lp( ) [ (x,x+y) (0,x) (x + y t) (x t) 1 α dt ] p /(p 1) } (p 1)/p dt ] (p 1)/p 1 dt (1 α)p /(p 1) ] (p 1)/p 1 dt 2(1 α)p /(p 1)

11 Dong B H et al. Sci China Math October 2018 Vol. 61 No u Lp( ) From above, we deduce that [ (x,x+y) y 1 α x (2αp p 1)/p + y (αp 1)/p. ] (p 1)/p 1 (x + y t) dt (1 α)p /(p 1) T u(x + y) T u(x) χ (x)[y 1 α x (2αp p 1)/p + y (αp 1)/p ]. (3.1) Now, we first estimate I 1. Putting (3.1) in I 1, we have I 1 y 1 α x (2αp p 1)/p + y (αp 1)/p p+ dx \B \B [y (1 α)p + x (2αp p 1)p + /p + y (αp 1)p + /p ]dx y (1 α)p + + y (αp 1)p + /p. Then, we estimate I 2. By a similar way to the above, we obtain Thus, we obtain I 2 y (1 α)p + y αp 1. ϱ Lp( ) (T u( + y) T u( )) Therefore, we can draw a conclusion that y (1 α)p+ + y (αp 1)p+/p + y (1 α)p + y αp 1. ϱ Lp( ) (T u( + y) T u( )) is sufficiently small when y approaches to zero. This finishes the proof of Theorem Nonlinear fractional differential equations in variable exponent Lebesgue spaces A famous application about fractional calculus is to consider the existence and the uniqueness of solutions to the classical Cauchy problem for the following nonlinear fractional differential equations: { D α 0 +u(t) f(t, u(t)), (4.1) I 1 α 0 u(0) 0. + Recently, many researchers are interested in studying the classical Cauchy problem (4.1) in some spaces. For example, Zhang [33] gave the positive solutions of the Cauchy problem (4.1) with a continuous function f. Kilbas et al. [15] established the existence and the uniqueness of solutions to the equation in integrable spaces L p (a, b), and weighted spaces of continuous functions C n α [a, b] and Cn α[a, α b], respectively. In [11], by using a fixed-point theorem, Fu et al. obtained the existence and the uniqueness of solutions to the Cauchy problems for nonlinear fractional ordinary differential equations in Morrey spaces. In this section, applying the Riesz-Kolmogorov theorem, we obtain the existence and the uniqueness of solutions to a Cauchy type problem for fractional differential equations in the variable exponent Lebesgue space. Let L 1 (a, b) be the set of all measurable functions defined on (a, b) satisfying f(x) dx <, (a,b) and AC[a, b] be the space of functions which are absolutely continuous on [a, b]. Before stating the main theorem of this section, we first recall the following two lemmas.

12 1818 Dong B H et al. Sci China Math October 2018 Vol. 61 No. 10 Lemma 4.1 (See [15, Lemma 2.5]). Let α (0, 1), f 1 α (x) I 1 α 0 + f(x) for any x (a, b) and I α 0 +(Lp (a, b)) : {f : f I α a+φ, φ L p (a, b)}. (i) If p [1, ] and f I α 0 + (L p (a, b)), then, for any x (a, b), (I0 α +Dα 0 +f)(x) f(x). (ii) If f L 1 (a, b) and f 1 α AC[a, b], then the equality holds almost everywhere on [a, b]. (I0 α +Dα 0 +f)(x) f(x) f 1 α(0) Γ(α) xα 1 Lemma 4.2. Let Ω be a convex and closed subset of a Banach space. Then any continuous and compact mapping S : Ω Ω has a fixed point. Lemma 4.2 is the famous Schauder fixed point theorem proved by Schauder [22] in Now, applying the above fixed point theorem, we can obtain the existence and the uniqueness of solutions to the Cauchy problem (4.1) in Lebesgue space with the variable exponent. Theorem 4.1. Let δ (0, ), p( ) P(0, δ), the operator F : u f(t, u(t)) be bounded and continuous in L p( ) (0, δ), and f(t, 0) 0. If max{1/p, 1/2(1 + 1/p 1/p + )} < α < 1, then the Cauchy problem (4.1) has solutions in the spaces L p( ) (0, δ) for a suitable δ (0, ). Moreover, if there exists a positive constant C F, related to the operator F, such that F u 1 F u 2 Lp( ) C F u 1 u 2 Lp( ), (4.2) where u 1, u 2 L p( ) (0, δ), then the Cauchy problem (4.1) has a unique solution u L p( ) (0, δ) for a suitable δ (0, ). Proof. For any u L p( ) (0, δ), from Lemma 3.2, it is easy to deduce u L 1 (0, δ). By Lemma 4.1, the equation (4.1) is equivalent to the following integral equation: 1 f(t, u(t)) dt, x (0, δ), u(x) Γ(α) (0,x) (x t) 1 α 0 a. e., x [σ, ), χ (x) f(t, u(t)) dt T (f(, u)). Γ(α) (x t) 1 α (0,x) Let Su : (T F )(u) T (f(, u)). Then the equation (4.1) has a solution in the space L p( ) (0, δ) if and only if the operator S has a fixed point in L p( ) (0, δ). It is easy to show that S is completely continuous. Since, by Theorem 3.2, T is a compact operator in L p( ) (0, δ), together with F (u) : f(t, u(t)) being bounded and continuous in L p( ) (0, δ) and Su T F (u) T (f(, u)), we deduce that S is a compact and continuous operator in L p( ) (0, δ). Therefore, S is completely continuous in L p( ) (0, δ).

13 Dong B H et al. Sci China Math October 2018 Vol. 61 No For any positive constant r, denote the set E by E : {u L p( ) (0, δ) : u Lp( ) r}. We know that E is a bounded and closed convex set, and want to prove that S : E E by choosing a suitable δ (0, ). This means that S has a fixed point in E by Lemma 4.2. Thus, (4.1) has a solution in L p( ) (0, δ). The next of the proof is to obtain r. We prove this by estimating ϱ Lp( ) (Su). Let Su Lp( ) G : {x (0, δ) : Su(x) r}. Then Su(x)/r 1 for any x G. Furthermore, we have ϱ Lp( ) ( ) Su r p(x) Su(x) r dx p(x) Su(x) r dx + Su(x) G r Su(x) dx + δ p+ r \G \G : I 3 + δ. For I 3, by Hölder s inequality and Lemma 3.2, we conclude that I 3 χ (x) f(t, u(t)) dt dx \G rγ(α) (0,x) (x t) 1 α p+ 1 r p + Γ p +(α) f(, u( )) p+ L p (0,x) \G [ (p 1)p + /p (x t) (α 1)p /(p 1)] dtdx (0,x) [ 2 max{δ (1/p 1/p) +, δ (1/p 1/p) ] p+ } f(, u( )) p + L p( ) ( p 1 αp 1 rγ(α) ) (p 1)p +/p x (αp 1)p+/p dx ] p+ F p+ u p + L p( ) [ 2 max{δ 1/p 1/p +, 1} rγ(α) ( ) (p 1)p p 1 + /p δ 1+(αp 1)p + /p αp (αp 1)p + /p [ 2 F max{δ 1/p 1/p + ] p+, 1} Γ(α) ( ) (p 1)p p 1 + /p δ 1+(αp 1)p +/p. αp (αp 1)p + /p From above, we deduce that ( ) [ Su 2 F max{δ 1/p 1/p + ] p+, 1} ϱ Lp( ) δ + r Γ(α) ( ) (p 1)p p 1 +/p δ 1+(αp 1)p + /p. αp (αp 1)p + /p p(x) dx

14 1820 Dong B H et al. Sci China Math October 2018 Vol. 61 No. 10 We choose a calculated δ (0, ) such that ϱ Lp( ) ( ) Su 1. r Then, by (1.1), we obtain Su Lp( ) r. From above, it follows that (4.1) has a solution u in L p( ) (0, δ). Now we turn to show the uniqueness of the solution. Suppose that u 1, u 2 L p( ) (0, δ) are two solutions of (4.1). For any given constant C (0, 1), let H : {x (0, δ) : Su 1 (x) Su 2 (x) C u 1 u 2 Lp( ) }. Then ϱ Lp( ) ( Su 1 Su 2 C u 1 u 2 Lp( ) ) p(x) Su 1 (x) Su 2 (x) dx C u 1 u 2 Lp( ) p(x) Su 1 (x) Su 2 (x) dx + C u 1 u 2 Lp( ) Su 1 (x) Su 2 (x) dx + δ p+ C u 1 u 2 Lp( ) \H \H : I 4 + δ. H Su 1 (x) Su 2 (x) C u 1 u 2 Lp( ) p(x) dx For I 4, similar to the estimation of I 3, we have I 4 χ (x) \H Γ(α) C u 1 u 2 Lp( ) 1 C p + Γ p +(α) u1 u 2 p + L p( ) [ (0,x) \H (p 1)p + /p (x t) (α 1)p /(p 1)] dtdx (0,x) [ 2 max{δ (1/p 1/p) +, δ (1/p 1/p) } CΓ(α) u 1 u 2 Lp( ) ( ) (p 1)p p 1 + /p x (αp 1)p + /p dx. αp 1 f(t, u 1 (t)) f(t, u 2 (t)) (x t) 1 α dt p+ dx f(, u 1 ( )) f(, u 2 ( )) p + L p (0,x) ] p+ f(, u 1 ( )) f(, u 2 ( )) p+ L p( ) By (4.2), we further obtain [ 2 max{δ 1/p 1/p + ] p+, 1} I 4 C p+ F CΓ(α) u 1 u u 1 u 2 p+ L p( ) 2 Lp( ) ( ) (p 1)p p 1 + /p δ 1+(αp 1)p +/p αp (αp 1)p + /p [ 2CF max{δ 1/p 1/p + ] p+, 1} CΓ(α) ( ) (p 1)p p 1 + /p δ 1+(αp 1)p + /p. αp (αp 1)p + /p Then we choose a sufficiently small δ (0, ) such that ( ) Su 1 Su 2 ϱ Lp( ) 1. C u 1 u 2 Lp( )

15 Dong B H et al. Sci China Math October 2018 Vol. 61 No By (1.1) again, we obtain Thus, we have Su 1 Su 2 Lp( ) C u 1 u 2 Lp( ). u 1 u 2 Lp( ) Su 1 Su 2 Lp( ) C u 1 u 2 Lp( ), which is not possible because C (0, 1). So, we have u 1 u 2, which means (4.1) has a unique solution u in L p( ) (0, δ). This finishes the proof of Theorem 4.1. The next example will show that the condition is sufficient and not necessary for Theorem 4.1. Example 4.1. equation max{1/p, 1/2(1 + 1/p 1/p + )} < α < 1 Let α (0, 1), t (0, ) and λ, β R. As [15, Example 3.2], the fractional differential D0 α u(x) f(x, u(x)), + f(x, u(x)) λx β [u(x)] 1 2, D α 1 u(0) 0 has the exact solution [ ] 2 λγ(α + 2β + 1) x 2(α+β), x (0, δ), u(x) Γ(2α + 2β + 1) 0, x [δ, ). In this case, we have λ 2 Γ(α + 2β + 1) f(x, u(x)) λx β [u(x)] 1 2 Γ(2α + 2β + 1) xα+2β, x (0, δ), 0, x [δ, ). Let p( ) P(0, δ) and 1 α < β 1 ( ) 1 2p 2 p + + α. We show that u L p( ) (0, δ), but f(, u( )) / L p( ) (0, δ). Actually, for any u L p( ) (0, δ), we only need to prove the equation ϱ Lp( ) (u) u(x) p(x) dx [ ] 2 λγ(α + 2β + 1) p(x) x 2(α+β) dx Γ(2α + 2β + 1) is finite almost everywhere. For α+β [0, ), it is easy to see that ϱ Lp( ) (u) <. If α+β (, 0) and δ (0, 1], we have ϱ Lp( ) (u) x 2(α+β)p dx If α + β (, 0) and δ (1, ), we obtain ϱ Lp( ) (u) x 2(α+β)p dx + (0,1) δ1+2(α+β)p 1 + 2(α + β)p <. [1,δ) x 2(α+β)p + dx

16 1822 Dong B H et al. Sci China Math October 2018 Vol. 61 No δ1+2(α+β)p+ < (α + β)p 1 + 2(α + β)p + While, for we show that f(, u( )) / L p( ) (0, δ), ϱ Lp( ) (f(, u( ))) f(x, u(x)) p(x) dx λ 2 Γ(α + 2β + 1) Γ(2α + 2β + 1) xα+2β p(x) dx is infinite almost everywhere. When δ (0, 1], we have [f(, u( ))] { λ 2 Γ(α + 2β + 1) min Γ(2α + 2β + 1) ϱ Lp( ) and, when δ (1, ), we obtain p+, λ 2 p Γ(α + 2β + 1) } Γ(2α + 2β + 1) { ϱ Lp( ) [f(, u( ))] min λ 2 Γ(α + 2β + 1) Γ(2α + 2β + 1) [ x (α+2β)p+ dx + (0,1) { λ 2 Γ(α + 2β + 1) min [ > min Γ(2α + 2β + 1) p+, p+, x (α+2β)p + dx λ 2 Γ(α + 2β + 1) Γ(2α + 2β + 1) ] x (α+2β)p dx [1,δ) λ 2 Γ(α + 2β + 1) Γ(2α + 2β + 1) ] x (α+2β)p + dx + δ1+(α+2β)p 1 (0,1) 1 + (α + 2β)p { λ 2 Γ(α + 2β + 1) Γ(2α + 2β + 1) (0,1) x (α+2β)p + dx. p+, λ 2 Γ(α + 2β + 1) Γ(2α + 2β + 1) p } p } p } By the condition we know that 1 α < β 1 ( ) 1 + α, 2p 2 p + (α + 2β)p + 1, which means (0,1) x (α+2β)p + dx and x (α+2β)p + dx. Thus, ϱ Lp( ) [f(, u( ))] is infinite almost everywhere no matter what the value δ is. This shows our claim, which completes the proof of Example 4.1. Acknowledgements Baohua Dong was supported by the Startup Foundation for Introducing Talent of Nanjing University of Information Science and Technology (Grant No. 2017r098). Zunwei Fu was supported by National Natural Science Foundation of China (Grant Nos and ) and National Science Foundation of Shandong Province (Grant No. ZR2017MA041). Jingshi Xu was supported by National Natural Science Foundation of China (Grant No ).

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