ON THE SPECTRUM OF TWO DIFFERENT FRACTIONAL OPERATORS RAFFAELLA SERVADEI AND ENRICO VALDINOCI Abstract. In this paper we deal with two nonlocal operators, that are both well known and widely studied in the literature in connection with elliptic problems of fractional type. Precisely, for a fixed s 0, 1 we consider the integral definition of the fractional Laplacian given by s ux := cn, s 2 R n 2ux ux + y ux y y n+2s dy, x R n, where cn, s is a positive normalizing constant, and another fractional operator obtained via a spectral definition, that is A su = i N a i λ s i e i, where e i, λ i are the eigenfunctions and the eigenvalues of the Laplace operator in Ω with homogeneous Dirichlet boundary data, while a i represents the projection of u on the direction e i. Aim of this paper is to compare these two operators, with particular reference to their spectrum, in order to emphasize their differences. Contents 1. Introduction 1 2. A comparison between the eigenfunctions of A s and s 3 2.1. Poisson kernel of fractional type 4 2.2. Optimal regularity for the eigenfunctions of s 4 3. The spectrum of A s and s 9 4. One-dimensional analysis 9 5. A relation between the first eigenvalue of A s and that of s 14 References 17 1. Introduction Recently in the literature a great attention has been devoted to the study of nonlocal problems driven by fractional Laplace type operators, not only for a pure academic interest, but also for the various applications in different fields. Indeed, many different problems driven by the fractional Laplacian were considered in order to get existence, non-existence and regularity results and, also, to obtain qualitative properties of the solutions. In particular, two notions of fractional operators were considered in the literature, namely the integral one which reduces to the classical fractional Laplacian, see, for instance, [7, 8, 9, 10, 14, 15, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33] and references therein and the Key words and phrases. Fractional Laplace operator, Laplace operator, eigenvalues and eigenfunctions. 2010 AMS Subject Classification: 35R09, 45K05, 35R11, 26A33. The first author was supported by the MIUR National Research Project Variational and Topological Methods in the Study of Nonlinear Phenomena, while the second one by the MIUR National Research Project Nonlinear Elliptic Problems in the Study of Vortices and Related Topics and the FIRB project A&B Analysis and Beyond. Both the authors were supported by the ERC grant ε Elliptic Pde s and Symmetry of Interfaces and Layers for Odd Nonlinearities. 1
2 R. SERVADEI AND E. VALDINOCI spectral one that is sometimes called the regional, or local, fractional Laplacian, see, e.g. [2, 4, 5, 6, 35] and references therein. For any fixed s 0, 1 the fractional Laplace operator s at the point x is defined by 1.1 s ux := cn, s 2 R n 2ux ux + y ux y y n+2s dy, where cn, s is a positive normalizing constant 1 depending only on n and s. A different operator, which is sometimes denoted by A s, is defined as the power of the Laplace operator, obtained by using the spectral decomposition of the Laplacian. Namely, let Ω be a smooth bounded domain of R n, and let λ k and e k, k N, be the eigenvalues and the corresponding eigenfunctions of the Laplacian operator in Ω with zero Dirichlet boundary data on Ω, that is { ek = λ k e k in Ω e k = 0 on Ω, normalized in such a way that e k L 2 Ω = 1. For any s 0, 1 and any u H0 1 Ω with ux = i N a i e i x, x Ω, one considers the operator 1.2 A s u = i N a i λ s i e i. Aim of this paper is to compare the two previous definitions of fractional Laplace operators. First of all, we would like to note that these two fractional operators i.e. the integral one and the spectral one are different in spite of some confusion that it is possible to find in some of the existent literature in which the two operators are somehow freely interchanged. Indeed, the spectral operator A s depends on the domain Ω considered since its eigenfunctions and eigenvalues depend on Ω, while the integral one s evaluated at some point is independent on the domain in which the equation is set. 2 Of course, by definition of A s, it is easily seen that the eigenvalues and the eigenfunctions of A s are respectively λ s k and e k, k N, that is the s-power of the eigenvalues of the Laplacian and the very same eigenfunctions of the Laplacian, respectively. On the other hand, the spectrum of s may be less explicit to describe. We refer to [28, Proposition 9 and Appendix A], [23, 24], [25, Proposition 5] and [30, Proposition 4] for the variational characterization of the eigenvalues and for some basic properties. A natural question is whether or not there is a relation between the spectrum of A s and s and, of course, between the respective eigenfunctions. In the present paper, by using the classical regularity theory for the eigenfunctions of the Laplace operator and some recent regularity results for the fractional Laplace equation see [22, 23, 24, 32], we will show that the eigenfunctions of A s and s are different for more details see Section 2. In particular, we will show that the eigenfunctions of s are, in general, no better than Hölder continuous up to the boundary, differently from the eigenfunctions of A s i.e. of the classical Laplacian that are smooth up to the boundary if so is the domain. 1 Different definitions of the fractional Laplacian consider different normalizing constants. The constant cn, s chosen here is the one coming from the equivalence of the integral definition of s and the one by Fourier transform see, e.g., [7] and [10, 3.1 3.3 and 3.8] and it has the additional properties that lim u = u and lim u = u see [10, Proposition 4.4]. s 1 s s 0 2 + s Also, the natural functional domains for the operators s and A s are different, but this is a minor distinction, since one could consider both the operators as acting on a very restricted class of functions for which they both make sense - e.g., C0 Ω.
ON THE SPECTRUM OF TWO DIFFERENT FRACTIONAL OPERATORS 3 Furthermore, with respect to the eigenvalues of A s and s, we will prove that the first eigenvalue of s is strictly less than the first one of A s. To this purpose we will use some extension results for the fractional operators A s and s see [7, 34]. Summarizing, the results given in this paper are the following: Theorem 1. The operators s and A s are not the same, since they have different eigenvalues and eigenfunctions. In particular: the first eigenvalues of s is strictly less than the one of A s ; the eigenfunctions of s are only Hölder continuous up to the boundary, differently from the ones of A s that are as smooth up the boundary as the boundary allows. For further comments on similarities and differences between the operators A s and s for s = 1/2 see [13, Remark 0.4]. The paper is organized as follows. Section 2 is devoted to a comparison between the eigenfunctions of A s and s. In Section 3 we deal with the spectrum of the two fractional operators we are considering. Section 4 is devoted to the extension of the operator A s, while in Section 5 we discuss the relation between the first eigenvalues of A s and s. 2. A comparison between the eigenfunctions of A s and s This section is devoted to some remarks about the eigenfunctions of the operators A s and s. Precisely, we will consider the following eigenvalue problems in a smooth bounded domain Ω R n, with Dirichlet homogeneous boundary data, driven, respectively, by A s and s, { As u = λu in Ω 2.1 u = 0 on Ω and 2.2 { s u = λu in Ω u = 0 in R n \ Ω. Note that in 2.2 the boundary condition is given in R n \ Ω and not simply on Ω, due to the nonlocal character of the operator s. In what follows we will denote by e k, As and e k, s, k N, the k th eigenfunction of A s and s, respectively. Taking into account the definition of A s, it is easily seen that its eigenfunctions e k, As, k N, are exactly the eigenfunctions of the Laplace operator, i.e. e k, As = e k. Also, since e k C Ω C m Ω for any m N see, for instance, [11], then 2.3 e k, As C Ω C m Ω. Of course, constructing the eigenfunctions of s is more difficult. In spite of this, we have some regularity results for them. Precisely, denoting by δx = distx, Ω, x R n, by [22, Theorems 1.1 and 1.3] and [30, Proposition 4], we have that e k, s /δ s Ω C0,α Ω for some α 0, 1, namely e k, s /δ s Ω has a continuous extension to Ω which is C0,α Ω. In particular, e k, s is Hölder continuous up to the boundary. Aim of this section will be to show that the Hölder regularity is optimal for the eigenfunctions e k, s of s. To this purpose, first of all we recall the notion of Poisson kernel of fractional type and, then, we discuss the optimal regularity of the eigenfunctions e k, s.
4 R. SERVADEI AND E. VALDINOCI 2.1. Poisson kernel of fractional type. Here we recall the notion of Poisson kernels of fractional type and their relation with the Dirichlet problem see [20, Chapter I]. First of all, for any r > 0, x B r that is the ball of radius r centered at the origin and y R n \ B r, we define r 2 x 2 s 1 P r x, y := c o n, s y 2 r 2 x y n, with c o n, s > 0. It is known see [20, Appendix] that, for any fixed x B r the function Ix := P r x, y dy R n \B r is constant in x. Therefore, we normalize c o n, s in such a way that 3 2.4 P r x, y dy = 1. R n \B r The function P r plays the role of a fractional Poisson kernel, namely if g CR n L R n and P r x, y gy dy if x B r R 2.5 u g x := n \B r gx if x R n \ B r, then u g is the unique solution of { 2.6 s u g = 0 in B r u g = g outside B r. For this, see [20, 33]. 2.2. Optimal regularity for the eigenfunctions of s. In this subsection we prove that the C 0, α -regularity of the eigenfunctions e k, s is optimal. Precisely, we show that, in general, the eigenfunctions of s need not to be Lipschitz continuous up to the boundary i.e. the Hölder regularity is optimal. For concreteness, we consider the case 2.7 n > 2s, the domain Ω := B r and the first eigenfunction e 1, s normalized in such a way that e 1, s L 2 R n = 1 and e 1, s 0 in R n, see [28, Proposition 9 and Appendix A] of s in B r, i.e. { 2.8 s e 1, s = λ 1, s e 1, s in B r e 1, s = 0 in R n \ B r. We prove that Proposition 2. The function e 1, s given in 2.8 is such that Proof. The proof is by contradiction. W 1, R n, that is e 1, s W 1, B r. 2.9 e 1, s x + e 1, s x M, x R n for some M > 0. From now on, we proceed by steps. We suppose that e 1, s W 1, B r and so e 1, s Step 1. The function e 1, s is spherically symmetric and radially decreasing in R n. 3 More explicitly, one can choose con, s := Γn/2 sinπs/π n/2+1, see [20, pages 399 400].
ON THE SPECTRUM OF TWO DIFFERENT FRACTIONAL OPERATORS 5 Proof. For this, since e 1, s 0 in R n, we consider its symmetric radially decreasing rearrangement e 1, s see, e.g., [19, Chapter 2] for the basics of such a rearrangement. We observe that e 1, s vanishes outside B r, since so does e 1, s. Moreover, we recall that the L 2 - norm is preserved by the rearrangement, while the fractional Gagliardo seminorm decreases, see, e.g. [1, 3, 21]. Then, by this and since λ 1, s is obtained by minimizing the fractional Gagliardo seminorm under constraint on the L 2 -norm for functions that vanish outside B r see [28, Proposition 9], we conclude that the minimum is attained by e 1, s as well as by e 1, s. Since λ 1, s is a simple eigenvalue see [28, Proposition 9 and Appendix A], it follows that e 1, s = e 1, s and Step 1 is proved. Now, let Q be the fractional fundamental solution given by Qx := c 1 n, s x 2s n, x R n \ {0}. Here the constant c 1 n, s > 0 is chosen in such a way that s Q is the Dirac s delta δ 0 centered at the origin see, e.g., [20, page 44] for the basic properties of fractional fundamental solutions. We define 2.10 ṽx := λ 1, s Q e 1, s x = λ 1, s c 1 n, s y 2s n e 1, s x y dy, R n x R n and 2.11 vx := e 1, s x ṽx, x R n. First of all, notice that ṽ 0 in R n, since λ 1, s > 0, Q > 0 and e 1, s 0 in R n. Step 2. The function ṽ is spherically symmetric and radially decreasing in R n. Proof. Indeed, if R is a rotation, we use Step 1 and the substitution ỹ := Ry to obtain for any x R n ṽx = λ 1, s c 1 n, s y 2s n e 1, s x y dy = R n = λ 1, s c 1 n, s y 2s n e 1, s Rx y dy R n = λ 1, s c 1 n, s ỹ 2s n e 1, s Rx ỹ dỹ = ṽrx, R n that shows the spherical symmetry of ṽ. As for the fact that ṽ is radially decreasing in R n, we take ρ > 0 and define 2.12 v ρ := λ 1, s c 1 n, s 1ṽ0,..., 0, ρ = R n y 2s n e 1, s y, ρ y n dy, where we used the notation y = y, y n R n 1 R for the coordinates in R n. The goal is to show that for any ρ > 0 2.13 v ρ 0. For this, first note that v ρ = R n { ρ y n r} y 2s n e 1, s y, ρ y n dy R n { ρ y n >r} y 2s n e 1, s y, ρ y n dy = y 2s n e 1, s y, ρ y n dy, R n { ρ y n r}
6 R. SERVADEI AND E. VALDINOCI since { ρ y n > r } { y, ρ y n > r } and e 1, s vanishes outside B r. Also, since the function e 1, s is spherically symmetric and radially decreasing in R n by Step 1, we write e 1, s x = E x with E 0 in R +. Thus, v ρ = y 2s n E y, ρ y n dy and so 2.14 v ρ = R n { ρ y n r} R n { ρ y n r} Now, let us consider the following change of variables { 2.15 ỹ := y ỹ n := 2ρ y n. y 2s n E y ρ y n, ρ y n y, ρ y n dy. First of all, note that if ỹ n ρ 0, then ỹ n 2ρ ỹ n ỹ n, so that and 2ρ ỹ n 2 ỹ 2 n ỹ, 2ρ ỹ n = ỹ 2 + 2ρ ỹ n 2 ỹ 2 + ỹ 2 n = ỹ. As a consequence of this and recalling that n 2s, we obtain that 2.16 ỹ, 2ρ ỹ n 2s n ỹ 2s n. Therefore, by 2.15 and 2.16 we get y 2s n E y ρ y n, ρ y n R n {0 ρ y n r} y, ρ y n dy = ỹ, 2ρ ỹ n 2s n E ỹ ỹ n ρ, ρ ỹ n R n {0 ỹ n ρ r} ỹ, ρ ỹ n dỹ ỹ 2s n E ỹ ỹ n ρ, ρ ỹ n ỹ, ρ ỹ n dỹ, R n {0 ỹ n ρ r} due to the fact that E 0 in R +. Hence, recalling 2.14, we get v ρ = y 2s n E y ρ y n, ρ y n R n { ρ y n r} y, ρ y n dy = y 2s n E y ρ y n, ρ y n R n {0 ρ y n r} y, ρ y n dy + y 2s n E y ρ y n, ρ y n R n {0 y n ρ r} y, ρ y n dy ỹ 2s n E ỹ ỹ n ρ, ρ ỹ n R n {0 ỹ n ρ r} ỹ, ρ ỹ n dỹ + ỹ 2s n E ỹ ρ ỹ n, ρ ỹ n ỹ, ρ ỹ n dy = 0, R n {0 ỹ n ρ r} due to the fact that ỹ, ρ ỹ n = ỹ, ρ ỹ n. Hence, 2.13 is proved. Then, by 2.12, the spherical symmetry of ṽ and the fact that λ 1, s and c 1 n, s are positive constants, we get that ṽ is radially decreasing in R n. This concludes the proof of Step 2. Next step will exploit assumption 2.9 taken for the argument by contradiction.
ON THE SPECTRUM OF TWO DIFFERENT FRACTIONAL OPERATORS 7 Step 3. The function ṽ is such that ṽ W 1, B 2r. Proof. To check this, we observe that for any x R n ṽx = λ 1, s c 1 n, s y 2s n e 1, s x y dy = λ 1, s c 1 n, s R n B rx y 2s n e 1, s x y dy, since e 1, s vanishes outside B r by 2.8. Here, B r x denotes the ball of radius r centered at x. Now, we notice that if x B 2r then B r x B 3r. As a consequence, recalling also 2.9, we obtain that for any x B 2r ṽx + ṽx λ 1, s c 1 n, s B rx λ 1, s c 1 n, sm y 2s n dy, B 3r y 2s n e 1, s x y + e 1, s x y dy which is finite being s > 0. Hence, Step 3 is established. Now we can conclude the proof of Proposition 2. For this, note that, from 2.9 and Step 3, we get i.e. there exists M > 0 such that v = e 1, s ṽ W 1, B 2r, 2.17 vx vy M x y for any x, y B 2r. Also, by 2.10 and the choice of Q and so, by 2.8 and 2.11 s ṽ = λ 1, s e 1, s s Q = λ 1, s e 1, s δ 0 = λ 1, s e 1, s s v = s e 1, s s ṽ = λ 1, s e 1, s λ 1, s e 1, s = 0 in B r. Therefore, we can reconstruct v by its values outside B r via the fractional Poisson kernel, that is, for any x B r, 2.18 vx = P r x, yvy dy, R n \B r for this see 2.5 and 2.6. Since 2.11 holds true and e 1, s = 0 outside B r, by 2.18 we deduce vx = P r x, yvy dy R n \B r 2.19 = P r x, ye 1, s y dy P r x, yṽy dy R n \B r R n \B r = P r x, yṽy dy. R n \B r
8 R. SERVADEI AND E. VALDINOCI By 2.11, 2.17, 2.18 and 2.19 we get P r x, yṽy dy ṽ0,..., 0, r R n \B r = P r x, yvy dy + v0,..., 0, r e 1, s 0,..., 0, r R n \B r 2.20 = P r x, yvy dy v0,..., 0, r R n \B r = v0,..., 0, r vx M 0,..., 0, r x for any x B r. If in 2.20 we take x := 0,..., 0, r ε B r for a small ε 0, r, recalling 2.4, we deduce that Mε = M 0,..., 0, r x P r x, yṽy dy ṽ0,..., 0, r R n \B r = P r x, y ṽy ṽ0,..., 0, r dy R n \B r r 2 x 2 s ṽ0,..., 0, r ṽy 2.21 = c o n, s y 2 r 2 x y n dy where y >r = c o n, s r 2 x 2 s c o n, s r s r x s = ε s y >r f ε y dy, y >r y >r ṽ0,..., 0, r ṽy y 2 r 2 s x y n dy ṽ0,..., 0, r ṽy y 2 r 2 s y 2 + y n r + ε 2 n/2 dy f ε y := c o n, s r s ṽ0,..., 0, r ṽy y 2 r 2 s y 2 + y n r + ε 2 n/2. We remark that f ε y 0 for any y > r, since 2.22 ṽ0,..., r ṽy for any y > r, thanks to Step 1. Moreover lim f εy = c o n, s r s ṽ0,..., 0, r ṽy ε 0 + y 2 r 2 s y 2 + y n r 2 n/2. So, we divide by ε s the inequality obtained in 2.21 and we use Fatou s Lemma: we conclude that 0 = lim inf Mε 1 s lim inf f ε y dy ε 0 + ε 0 + y >r = c o n, s r s ṽ0,..., 0, r ṽy y 2 r 2 s y 2 + y n r 2 n/2 dy. y >r This and 2.22 yield that ṽy is constantly equal to ṽ0,..., 0, r for any y > r, so that, in particular, if x := 0,..., 2r we have that 2.23 n ṽx = 0.
ON THE SPECTRUM OF TWO DIFFERENT FRACTIONAL OPERATORS 9 On the other hand, by 2.10, 1 λ 1,s c 1 n, s nṽx = x z 2s n e 1,s z dz x n B r x=x = 2s n x z 2s n 2 x n z n e 1,s z dz B r = 2s n z 2 + 2r z n 2 2s n 2/2 2r zn e 1,s z dz, B r which is strictly negative, by 2.7. This is a contradiction with 2.23 and hence Proposition 2 is proved. 3. The spectrum of A s and s In this section we focus on the spectrum of the operators A s and s. In what follows, we will denote by 0 < λ 1 < λ 2... λ k... the divergent sequence of the eigenvalues of the Laplace operator in Ω with Dirichlet homogeneous boundary data, while by λ k,as the sequence of eigenvalues of problem 2.1 and, finally, by λ k, s the eigenvalues of 2.2. By definition of A s, it easily follows that the eigenvalues λ k,as are exactly the s power of the ones of the Laplacian, that is 3.1 λ k,as = λ s k, k N. As for λ k, s, we refer to [28, Proposition 9 and Appendix A], [25, Proposition 5] and [30, Proposition 4] for their variational characterizations and some basic properties. In particular, we recall that for k N 3.2 λ k, s = where and cn, s 2 min u P k, s \{0} Rn Rn ux uy 2 x y n+2s dx dy Ω ux 2 dx P 1, s = X 0 Ω := { u H s R n s.t. u = 0 a.e. in R n \ Ω} 3.3 P k, s := { u X 0 Ω s.t. u, e j,s X0 Ω = 0 j = 1,..., k 1 }, k 2 with ux uy vx vy u, v X0 Ω = R n R n x y n+2s dx dy. In what follows we will show that A s and s have different eigenvalues. Of course, at this purpose we will use properties 3.1 and 3.2, but the main ingredient will be the extension of the operator A s, carried on in the forthcoming Section 4. 4. One-dimensional analysis In this section we perform an ODE analysis related to the extension of the operator A s, as it will be clear in the forthcoming Section 5. This analysis is not new in itself see also [7, Section 3.2] and [34, Section 3.1]: similar results were obtained, for instance, in [34] by using a conjugate equation and suitable special functions such as different kinds of Bessel and Hankel functions. Here, we use an elementary and self-contained approach. Given a 1, 1 in what follows we denote by Wa 1,2 R + the following Sobolev space { } Wa 1,2 R + := g W 1,1 loc R+ : t a gt 2 dt < + and t a ġt 2 dt < + R + R +,
10 R. SERVADEI AND E. VALDINOCI endowed with the norm 4.1 g W 1,2 a R + := R + t a gt 2 dt + 1/2 t a ġt 2 dt. R + Here, as usual, we used the notation R + := 0, +. We also denote by W 1,2 1,a R+ the closure, with respect to the norm in 4.1, of the set of all functions g C R + C 0 R + with bounded support and g0 = 1. It is useful to point out that Wa 1,2 R + and W 1,2 1,a R+ are contained in a classical Sobolev space. Precisely, denoting by W 1,p 0, κ, p 1 and κ > 0, the classical Sobolev space endowed with the norm g W 1,p 0,κ = the following result holds true: Lemma 3. Fix a 1, 1 and κ > 0. Then, for any p [ 1, a, with a = Moreover, there exists C κ > 0 such that for any g W 1,2 a R +. g p L p 0,κ + ġ p L p 0,κ 1/p, Wa 1,2 R + W 1,p 0, κ { 2/a + 1 if a 0, 1 2 if a 1, 0]. g W 1,p 0,κ C κ g W 1,2 a R + Proof. Let a 1, 1, g Wa 1,2 R + and p [ 1, a. We use the Hölder Inequality with exponents 2/2 p and 2/p note that both these exponents are greater than 1, thanks to the choice of p to see that g p L p 0,κ = κ 0 [ κ 0 t pa/2 t pa/2 gt p dt ] 2 p/2 [ κ t pa/2 p dt 0 ] p/2 t a gt 2 dt [ ] 2 p 2 p/2 [ ] p/2 = 2 p1 + a κ2 p1+a/2 p t a gt 2 dt < +, R + again since p < a. A similar inequality holds if we replace g with ġ, and this proves the desired result. Hence, as a consequence of Lemma 3, the functions in W 1,2 1,a R+ are uniformly continuous in any interval, by the standard Sobolev embedding, and have a distributional derivative which is well-defined a.e. Now, for any λ > 0 and any g W 1,2 1,a R+, we consider the functional G λ g := gt 2 dt + λ ġt 2 dt. R + t a The minimization problem of G λ is described in detail by the following result: Theorem 4. There exists a unique g λ W 1,2 1,a R+ such that 4.2 m λ := inf G λ g = G λ g λ, g W 1,2 1,a R+ that is, the above infimum is attained.
ON THE SPECTRUM OF TWO DIFFERENT FRACTIONAL OPERATORS 11 Moreover, g λ C R + C 0 R + and it satisfies g λ t + a t ġλt λg λ t = 0 for any t R + 4.3 g λ 0 = 1, and 4.4 lim t 0 + ta ġ λ t = m 1 λ a+1/2. Finally, 4.5 g λ t [0, 1] and ġ λ t 0 for all t R + and lim t + g λt = 0. Proof. By plugging a smooth and compactly supported function in G λ, we see that m λ [0, +, so we can take a minimizing sequence g j in W 1,2 1,a R+, that is a sequence g j such that G λ g j m λ as j +. In particular, G λ g j m λ + 1. As a consequence of this, g j W 1,2 is bounded a R + uniformly in j. Hence, there exists g λ W 1,2 1,a R+ such that g j g λ weakly in W 1,2 1,a R+ as j +. Also, for any k N, k 2, we have that C k k g j t 2 + ġ j t 2 k dt t a g j t 2 + ġ j t 2 dt m λ + 1, where C k = a if a 0, while C k = k a if a < 0. Namely, g j W 1,2 [,k] is bounded uniformly in j. Now, we perform a diagonal compactness argument over the index k. Namely, we take an increasing function φ k : N N and we use it to extract subsequences. We have a subsequence g φ2 j that converges a.e. in [1/2, 2] to g λ with ġ φ2 j converging to ġ λ weakly in L 2 [1/2, 2] as j +. Then, we take a further subsequence g φ3 φ 2 j that converges a.e. in [1/3, 3] to g λ with ġ φ3 φ 2 j converging to ġ λ weakly in L 2 [1/3, 3] as j +. Iteratively, for any k N, we get a subsequence g φk...φ 2 j that converges a.e. in [, k] to g λ with ġ φk φ 2 j converging to ġ λ weakly in L 2 [, k] as j +. Hence we look at the diagonal sequence g j := g φj φ 2 j. By construction g j converges to g λ a.e. in R + as j + and therefore, by Fatou Lemma, 4.6 lim inf t a g j t 2 dt t a g λ t 2 dt. j + R + R + On the other hand, by the weak convergence of g j to ġ λ in L 2 [, k] as j +, we have that ġ λ L 2 [, k] and so ψt := t a ġ λ t is also in L 2 [, k], which gives that is k k lim g j tψt dt = ġ λ tψt dt, j + k k lim t a g j tġ λ t dt = t a ġ λ t 2 dt j +
12 R. SERVADEI AND E. VALDINOCI for any k N, k 2. As a consequence of this, we obtain that 4.7 0 lim inf j + = lim inf j + = lim inf j + k k k t a g j t ġ λ t 2 dt k t a g j t 2 dt + t a g j t 2 dt k for any k N, k 2. By 4.6, 4.7 and the positivity of λ we get k t a ġ λ t 2 dt 2 t a g j t ġ λ t dt t a ġ λ t 2 dt m λ = lim G λg j j + = lim t a g j t 2 dt + λ t a g j t 2 dt j + R + R + k lim inf t a g j t 2 dt + λ t a g j t 2 dt j + R + k t a g λ t 2 dt + λ t a ġ λ t 2 dt R + for any k N, k 2. By taking k +, we deduce that m λ G λ g λ. This proves that the infimum in 4.2 is attained at g λ. The uniqueness of the minimizer is due to the convexity of the functional G λ. This completes the proof of 4.2. Now, notice that, since g λ W 1,2 1,a R+, then g λ 0 = 1 and g λ W 1,p 0, κ for any p [1, a and any κ > 0, by Lemma 3. Hence, it is uniformly continuous on 0, κ for any κ > 0, by the standard Sobolev embedding, and so it can be extended with continuity at 0, that is the function g λ C 0 R +. Moreover, by taking standard perturbation of the functional G λ at g λ + εφ, with φ C0 R+ and ε R small, one obtains that 4.8 R + t a g λ tφt + λġ λ t φt dt = 0. Hence, g λ satisfies weakly an ODE and so g λ C R + by uniformly elliptic regularity theory see for instance 4 [16, Theorem 8.10]. Moreover, integrating by parts in 4.8 it easily follows that g λ solves problem 4.3. Now, we prove 4.4. For this, it is convenient to reduce to the case λ = 1, by noticing that if g λ t := gt/ λ, we have that and therefore G λ g λ = λ a+1/2 G 1 g 4.9 m λ = λ a+1/2 m 1 and g λ t = g 1 t/ λ. Let us fix φ C 0 [ 1, 1] with φ0 = 1 and let γt := t a g 1 tφt + ġ 1 t φt. By the Cauchy Schwarz Inequality, we have that R + γt dt G 1 g 1 G 1 φ < +, 4 In further detail, gλ satisfies [16, Equation 8.2] with n = 1, a ij = a 11 = λt a, b i = b 1 = 0, c i = c 1 = 0, d = t a and this equation is uniformly elliptic in bounded domains separated from 0: so we can apply [16, Theorem 8.10] with f = 0 and obtain that g λ W k,2 b 1, b 2 for any b 2 > b 1 > 0 and any k N.
ON THE SPECTRUM OF TWO DIFFERENT FRACTIONAL OPERATORS 13 so that γ L 1 R +. Therefore, by the absolute continuity of the Lebesgue integral, for any fixed ε > 0 there exists δ ε > 0 such that if 0 < t 1 < t 2 < δ ε then t2 γτ dτ < ε. t 1 As a consequence, the function Γt := + t γτ dτ is uniformly continuous in 0, 1 and therefore it may be extended with continuity at 0 as follows + + Γ0 = γτ dτ = τ a g 1 τφτ + ġ 1 τ φτ 4.10 dτ. 0 By 4.3 with λ = 1 it is easy to see that for any t R + 0 t a g 1 t = d dt taġ 1 t. So, by this and recalling that φ0 = 1 and φt = 0 if t 1, we get 1 Γ0 = τ a g 1 τφτ + ġ 1 τ φτ dτ 0 1 [ d ] = τ a ġ 1 τ φτ + τ a ġ 1 τ 0 dτ φτ dτ 4.11 1 d = lim τ a ġ 1 τφτ dτ t 0 + t dτ = lim t 0 ta ġ 1 tφt + = lim t 0 ta ġ 1 t. + Note that the computation carried on in 4.11 has also shown that the above limit exists. Now, to compute explicitly such limit, we consider the perturbation g 1,ε := g 1 + εφ/1 + ε with ε R small. First of all, notice that g 1,ε = g 1 + εφ εg 1 + oε and so g 1,ε 2 = g 1 2 + 2εg 1 φ 2ε g 1 2 + oε, and similarly if we replace g 1,ε with ġ 1,ε. It follows that G 1 g 1,ε = G 1 g 1 + 2ε g 1 τφτ g 1 τ 2 + ġ 1 τ φτ ġ 1 tτ 2 dt + oε. R + τ a Then, the minimality condition implies that g 1 τφτ g 1 τ 2 + ġ 1 τ φτ ġ 1 τ 2 dτ = 0. R + τ a Hence, by this, 4.10 and the definition of m 1 we deduce 0 = τ a g 1 τφτ g 1 τ 2 + ġ 1 τ φτ ġ 1 τ 2 dτ R + = g 1 τφτ + ġ 1 τ φτ dτ g 1 τ 2 + ġ 1 τ 2 dτ R + τ a = Γ0 m 1. R + τ a This and 4.11 prove 4.4 for λ = 1. In general, recalling 4.9, we obtain lim t 0 + ta ġ λ t = λ 1/2 lim t 0 + ta ġ 1 tλ 1/2 = λ a+1/2 lim t 0 +tλ 1/2 a ġ 1 tλ 1/2 = m 1 λ a+1/2,
14 R. SERVADEI AND E. VALDINOCI thus establishing 4.4 for any λ > 0. Now, let us prove 4.5 For this we first observe that G 1 g 1 = G 1 g 1, which implies, by the uniqueness of the minimizer, that g 1 = g 1 and so g 1 0 in R +. We start showing that 4.12 ġ 1 0 in the whole of R +. By contradiction, if g 1 was increasing somewhere, there would exist t 2 > t 1 0 such that 0 g 1 t 1 < g 1 t 2. Let b := g 1 t 1 + g 1 t 2 /2 g 1 t 1, g 1 t 2. Notice that there exists t 3 > t 2 such that gt 3 = b: otherwise, by continuity, we would have that gt > b > 0 for any t > t 2 and so, using that a 1, 1, + + t a g t 2 dt b 2 G 1 g 1 t a dt = +, t 2 1 t 2 which is against our contraction. Having established the existence of the desired t 3, we use the Weierstrass Theorem to obtain t [t 1, t 3 ] in such a way that Note that, by definition of b, g 1 t = max t [t 1,t 3 ] g 1t. g 1 t g 1 t 2 > b > g 1 t 1. Hence, t t 1 and also t t 3, being g 1 t 3 = b. Thus, t is an interior maximum for g 1. Accordingly ġ 1 t = 0 and g 1 t 0. Thus, by 4.3, 0 = g 1 t + a t ġ 1 t g 1 t 0 + 0 b = b < 0. This is a contradiction and it proves 4.12. A consequence of 4.12 is also that g 1 t g 1 0 = 1 for any t R +. Moreover, it implies that the limit l := lim g 1t [0, 1] t + exists. Necessarily, it must be l = 0. Otherwise, if l > 0, it would follow that gt l/2 for any t t o, for a suitable t o > 0. This yields that using also that a 1, 1 G 1 g 1 + t o t a g 1 t 2 dt l/2 2 + t o t a dt = +, which is against our contraction. All these considerations imply 4.5 for λ = 1, and thus for any λ > 0, thanks to 4.9. 5. A relation between the first eigenvalue of A s and that of s This section is devoted to the study of the relation between the first eigenvalue of A s and of s, that is between λ 1, As and λ 1, s. Precisely, in this framework our main result is the following: Proposition 5. The relation between the first eigenvalue of s and the one of A s is given by λ 1, s < λ 1, As. Proof. Let us take a := 2s 1 1, 1 and for any x, t Ω R +, set E 1 x, t := g λ1 te 1 x, where the setting of Theorem 4 is in use, λ 1 is the first eigenvalue of the Laplacian and e 1 = e 1, As is the first eigenfunction of the operator A s see Section 2.
ON THE SPECTRUM OF TWO DIFFERENT FRACTIONAL OPERATORS 15 Notice that E 1 may be thought as an extension of e 1 in the half-space R n R + that vanishes in R n \ Ω R +. However, we point out that E 1 does not verify div E 1 = 0 in the whole of R n R +. Also, note that the function E 1 C Ω R + CΩ R +, since e 1 C Ω C m Ω for any m N see formula 2.3 and g λ1 C R + C 0 R + by Theorem 4. Also, lim t 0 + ta t E 1 x, t = lim t 0 + ta ġ λ1 te 1 x = m 1 λ a+1/2 1 e 1 x, thanks to 4.4. Furthermore, since G λ1 g λ1 is finite by Theorem 4, we have that L 1 R + t a g λ1 t 2 + t a ġ λ1 t 2 2t a g λ1 tġ λ1 t and, therefore, there exists a diverging sequence of R for which 5.1 lim R + Ra g λ1 Rġ λ1 R = 0. Now, note that, using 5 the definition of E 1, the fact that e 1 is the first eigenfunction of for this see Section 2, for any x, t Ω R + we have 5.2 t a E 1 x, t 2 = div t a E 1 x, t E 1 x, t at a 1 E 1 x, t t E 1 x, t t a E 1 x, t E 1 x, t = div t a E 1 x, t E 1 x, t at a 1 E 1 x, tġ λ1 te 1 x t a E 1 x, tg λ1 t x e 1 x t a E 1 x, t g λ1 te 1 x = div t a E 1 x, t E 1 x, t at a 1 E 1 x, tġ λ1 te 1 x + λ 1 t a E 1 x, tg λ1 te 1 x t a E 1 x, t g λ1 te 1 x = div t a E 1 x, t E 1 x, t + t a E 1 x, te 1 x at 1 ġ λ1 t + λ 1 g λ1 t g λ1 t = div t a E 1 x, t E 1 x, t, thanks to 4.3. By 5.2 and the Divergence Theorem, we have that 5.3 t a E 1 x, t 2 dx dt = Ω R + = lim div R + = lim R + Ω Ω 0,R lim R + Ω 0,R t a E 1 x, t E 1 x, t t a E 1 x, t t E 1 x, t t a E 1 x, t 2 dx dt t=r dx dt t a E 1 x, t t E 1 x, t t=0 dx, 5 We remark that here is the vector collecting all the derivatives, both in x and it t. Similarly, = x + 2 t.
16 R. SERVADEI AND E. VALDINOCI since for any t R +, E 1, t = 0 on Ω, being e 1 = 0 outside Ω. Hence, by 5.3 and using again the definition of E 1 we deduce that t a E 1 x, t 2 dx dt Ω R + [ ] = lim t a E 1 x, t t E 1 x, t t=r t a E 1 x, t t E 1 x, t t=0 dx R + = lim R + = lim R + = lim R + = lim R + Ω Ω Ω = m 1 λ a+1/2 1 = m 1 λ s 1, [ ] R a E 1 x, R t E 1 x, R t a E 1 x, t t E 1 x, t t=0 dx [ R a g λ1 Rġ λ1 R e 1 x 2 t a g λ1 tġ λ1 t e 1 x 2 ] t=0 dx R a g λ1 Rġ λ1 R e 1 x 2 + m 1 λ a+1/2 1 e 1 x 2 dx Ω R a g λ1 Rġ λ1 R + m 1 λ a+1/2 1 e 1 2 L 2 Ω thanks to 4.4, the fact that g λ 0 = 1, 5.1 and the choice of a. As a consequence, t a Ux, t 2 dx dt t a E 1 x, t 2 dx dt R n R + R n R + 5.4 inf U CR n R + U,0 L 2 Ω =1 U,0=0 outside Ω = t a E 1 x, t 2 dx dt Ω R + = m 1 λ s 1, since E 1, t = e 1 g λ t = 0 in R n \ Ω for any t R +. Now, we use a result in [7] to relate the first term in 5.4 to λ 1, s which, roughly speaking, says the optimal U is realized by the so called a-harmonic extension of u := U, 0. Namely, by [7, formula 3.7 and its proof at page 1250] and [10, Proposition 3.4], we get inf U CR n R + U,0 L 2 Ω =1 U,0=0 outside Ω R n R + t a Ux, t 2 dx dt = m 1 = m 1 cn, s 2 m 1 cn, s 2 = m 1 λ 1, s, inf u CR n u L 2 Ω =1 u=0 outside Ω min u X 0 Ω u L 2 Ω =1 inf u CR n u L 2 Ω =1 u=0 outside Ω R n ξ 2s ûξ 2 dξ Rn Rn ux uy 2 x y n+2s dx dy, Rn Rn ux uy 2 x y n+2s dx dy thanks also to the variational characterization of λ 1, s given in 3.2. Here û denotes the Fourier transform of u. Thus, This and 5.4 give that inf U CR n R + U,0 L 2 Ω =1 U,0=0 outside Ω R n R + t a Ux, t 2 dx dt m 1 λ 1, s. λ 1, s λ s 1.
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