ON THE SHAPE OF THE GROUND STATE EIGENFUNCTION FOR STABLE PROCESSES

Transcription

1 ON THE SHAPE OF THE GROUND STATE EIGENFUNCTION FOR STABLE PROCESSES RODRIGO BAÑUELOS, TADEUSZ KULCZYCKI, AND PEDRO J. MÉNDEZ-HERNÁNDEZ Abstract. We prove that the ground state eigenfunction for symmetric stable processes of order α, 2 killed upon leaving the interval, 1 is concave on 1 2, 1 2. We call this property mid concavity. A similar statement holds for rectangles in R d, d > 1. These result follow from similar results for finite dimensional distributions of Brownian motion and subordination. 1. Introduction Let D be a bounded convex domain in R d, d 1, and let ϕ 1 be the first eigenfunction for the Dirichlet Laplacian in D. In their seminal paper 13], Brascamp and Lieb proved that ϕ 1 is log concave in D. That is, logϕ 1 is concave on any segment contained in the domain. This result has led to many interesting applications in analysis, geometry, pde, mathematical physics and probability. For some of these applications, see Borell 1], 11], 12] and the many references therein. In particular, the log concavity of ϕ 1 leads to estimates of the spectral gap λ 2 λ 1 which in tern describe the rate to equilibrium of the Brownian motion conditioned to remain forever in the domain D. We refer the reader to 3], 18], 2] and 21] for some of these applications and additional references. In 4], the first two authors initiated the study of what may be called the fine spectral theoretic properties of symmetric stable processes. Unfortunately, given the nonlocality of the generator of these processes, even the most basic questions seem to be very difficult. It was proved in 4] Theorem 5.1 that the ground state eigenfunction for the 1991 Mathematics Subject Classification. 3C45. Key words and phrases. Symmetric stable processes, ground state eigenfunctions, multiple integrals. R. Bañuelos was supported in part by NSF grant # DMS. T. Kulczycki was supported by KBN grant 2 P3A and RTN Harmonic Analysis and Related Problems, contract HPRN-CT HARP. 1

2 Cauchy process in the interval, 1 is concave. We, of course, expect this to be the case for any symmetric stable process. The purpose of this paper is to prove that for any symmetric stable processes, the ground state eigenfunction is concave in 1, 1. We call this property 2 2 mid concavity. This will follow from a more general result on mid concavity of the finite dimensional distributions of these processes. This mid concavity result is new even for Brownian motion. We first recall some basic definitions. Let Xt α be a d-dimensional symmetric stable process of index < α 2. The process Xt α has stationary independent increments and its transition density p α t x, y = p α t x y, t >, x, y R d, is determined by its Fourier transform exp t ξ α = e iξ y p α t y dy. R d These are Lévy processes with right continuous sample paths. transition densities satisfy the scaling property p α t x, y = t d/α p α 1 t /α x, t /α y, The hence the process has the scaling property of index α. When α = 2, Xt 2 is just Brownian motion B t running at twice the speed and when α = 1, Xt 1 is the Cauchy process. In the first case, p 2 t x, y is the usual Gaussian distribution heat kernel and in the second, p 1 t x, y is the Cauchy distribution Poisson kernel. Our interest here is on symmetric stable processes of index < α < 2 killed upon leaving a domain D. That is, let D R d, d 1, be a nonempty bounded connected open set and let be the first exit time of X α t τ α D = inf{t : X α t / D} from D. Let Tt D fx = E x fxt α, τd α > t, for x D, t > and f L 2 D, be the semigroup of the killed process. The killed process has transition densities p α D t, x, y and 1.1 Tt D fx = p α Dt, x, yfy dy. As with Brownian motion, 1.2 p α Dt, x, y = p α t, x, y r D t, x, y, where 1.3 r D t, x, y = E x p α t τ α D Xα τ α D, y, τ α D < t. D 2

3 From this it follows that the transition function p α D t, x, y is nonnegative, symmetric, jointly continuous in x and y, and that for all x, y D and t >, p α Dt, x, y p α t x, y = t d/α p α 1 t /α x, t /α y Ct d/α, where C = 2π d ω d Γd/α/α and ω d is the surface measure of the unit sphere in R d. In fact, p α D t, x, y is strictly positive for x, y D. These properties and the general theory of heat semigroups as in 15] gives an orthonormal basis of eigenfunctions {ϕ α n} on L 2 D with eigenvalues {λ α n} satisfying < λ α 1 < λ α 2 λ α 3..., and λ α n, as n. That is, Tt D ϕ α nx = e λα n t ϕ α nx, x D. In addition, the first eigenvalue λ α 1 is simple and its corresponding eigenfunction ϕ α 1, which we will refer to as the ground state eigenfunction, is an analytic strictly positive function on D. The infinitesimal generator of the semigroup is α/2. We can think of the eigenfunction and eigenvalues as solutions to the eigenvalue problem α/2 ϕ α nx = λ α nϕ α nx, x D and ϕ α nx = for x D c ; the Dirichlet problem for stable processes. We refer the reader to 5], 7], 9], 14] and 16] where many of the general properties of the α stable semigroup and its generator are established. The following question is motivated from the result of Brascamp and Lieb 13] mentioned above for Brownian motion and by its many applications. Question 1.1. Let D R d, d 1, be a bounded convex domain and < α < 2. Is ϕ α 1 log concave? In other words, is logϕ α 1 concave on any segment contained in D? The only known case is when D =, 1 and α = 1, where the question is answered in the affirmative in 4]. In fact, it is shown in 4] that the ground state eigenfunction for the Cauchy process in, 1 is concave. Because of this case we believe this result should hold for all α stable processes. More precisely, we have Conjecture 1.1. Let ϕ α 1 be the ground state eigenfunction for the symmetric stable processes of index < α < 2 killed upon leaving the interval I =, 1. Then ϕ α 1 is concave on I. There are by now many proofs of the log concavity result for Brownian motion. None of them, as far as we can see, adapt to the case of general symmetric stable processes. However, Brascamp Lieb s proof 3

4 does suggest some related questions which may provide some insight. We briefly recall here their argument based on multiple integrals. Let B t be Brownian motion and let τ D be its first exit time from D. Then one can show, see 1], that ϕ 2 1x = lim t e λ2 1 t P x {τ D > 2t}, uniformly in x D. From this it is enough to prove that P x {τ D > t} is log concave in x for every fixed t >. The latter can be written as the limit as n and k tend to infinity of P x {B jt/n D k ; j = 1, 2,..., n} where D k is a sequence of convex domains strictly increasing D k D k+1 up to D. We then reduce the problem to prove that for any convex domain D, P x {B jt/n D; j = 1, 2,..., n} is log concave on D as a function of x, for all t > and all n. This, however, is a multiple convolution of Gaussians with the indicator function of the set D. Since the Gaussian p 2 t x is log concave for all t > and the indicator function of a convex domain is log concave, the result follows from the fact that convolutions of log concave functions are log concave. Using right continuity of paths, we can try to repeat this argument for α stables processes. However, this time the argument breaks down right at the end. For example, if α = 1 the density for the Cauchy process, p 1 t x, y = p 1 t x y, is not log concave for all t. The obvious variation of this argument using the fact that Xt α = B 2σt, where σ t is a stable subordinator of index α/2 independent of B t, also fails basically due to the fact that the sum of log concave functions is not necessarily log concave. There is however, a substitute for log concavity which gives some insight into the shape of the ground state eigenfunction. We call this property mid concavity. Definition 1.1. Let D R d be a convex domain which is symmetric relative to each coordinate axes. Let J be a line segment in D parallel to the x 1 -axis which intersects the boundary D only at the two points a 1, a 2,..., a d, a 1, a 2,..., a d, a 1 >. We will say that the function F : D R, is mid concave on J if it is concave on the segment half of J from the point a 1 /2, a 2,..., a d to a 1 /2, a 2,..., a d. The function is mid concave along the x 1 axis if it is mid concave on every such segment contained in D which is parallel to the x 1 axis. A similar definition applies for mid concavity along the x 2 -axis,, x d axis. The function is mid concave on D if it is mid concave along each coordinate axes. Our main result in this paper is the following Theorem 1.1. Let Q = a 1, a 1 a 2, a 2 a d, a d, < a i < for all i = 1, 2,..., d, be a rectangle in R d. The ground state eigenfunction ϕ α 1 for the symmetric stable process of index < α < 2 4

5 is mid concave on Q. In addition, if x = x 1,..., x n Q, then 1.4 x i ϕ α 1 x, if x i <, and x i ϕ α 1 x, if x i >. Using arguments of multiple integrals as described above, we will show that Theorem 1.1 follows from Theorem 1.2. Let Q be a rectangle in R d. Let < t 1 < t 2 < < t n <. The function 1.5 F x = P x {X α t 1 Q,..., X α t n Q} is mid concave in Q for any < α 2. In addition, if x = x 1,..., x n Q, then 1.6 x i F x, if x i <, and x i F x, if x i >. Remark 1.1. It is important to note here that Theorem 1.2 is new even in the Brownian motion case α = 2. Indeed, as we shall see, the case α = 2 implies the general case by subordination. If we consider the eigenfunction for the Laplacian in the unit disk ID in the plane, one can show, by analysis of the Bessel function, that such a function is not concave in ID but it is mid concave. Also, it may be tempting to conjecture that for any symmetric domain in the plane the eigenfunction is mid concave. This, however, is not the case, even for the Brownian motion, as we will show at the end of the paper. The paper is organized as follows. In 2, we prove that the multiple convolutions of Gaussians in the interval, 1 is mid concave. In 3, we show how this and subordination implies Theorem 1.2. Here we also show that full concavity fails for general multiple integrals and that mid concavity fails in general symmetric domains in the plane. Let 2. Mid concavity for Brownian motion p t x = 1 e x2 2t 2πt be the Gaussian density in one dimension. With the notation of the introduction, we have p 2 t x, y = p 2t x y. Proposition 2.1. Let n = 1, 2,... and let t 1, t 2,..., t n be real numbers in,. For x, 1 define 1 1 n 2.1 Φ n x = p ti x i x i dx 1... dx n, i=1 5

6 where x = x. The function Φ n x is mid concave on, 1. That is, Φ n x is concave on 1 2, 1 2. Clearly Φ n x is a positive even function on, 1]. Integrating by parts we obtain x Φ 1 1 y x nx = 2 2πtn y e Φ n y dy 2.2 = Φ n1 e 1 x2 2πtn 1 1 2πtn Notice that for all t >, 2.3 e 1 x2 2t e 1+x2 2t e y x2 = e x2 2 is a positive increasing function on, 1]. e 1+x2 y Φ ny dy. 1 e 2x t, Lemma 2.1. The function Φ n x is decreasing on, 1 for all n 1. Proof. We argue by induction. If n = 1, then 2.4 x Φ 1x = 1 1 2πt1 = 1 2πt1 <, y x 2 y e 2t 1 dy e 1 x 2 2t 1 + e 1+x2 2t 1 for all x, 1. Thus Φ 1 x is decreasing on, 1. Let us assume that Φ n x is decreasing on, 1. That is, suppose that x Φ nx, for all x, 1. Because of 2.3, it is enough to prove that πtn e y x2 y Φ ny dy. By symmetry y Φ ny = y Φ n y. On the other hand, if x > then e x y2 2t e x+y2 2t, 6

7 for all t, y >. Hence for all y >, e x y2 e x y2 e x y2 y Φ ny e x+y 2 y Φ ny + e x+y 2 + e x+y2 y Φ n y = y Φ ny = y Φ ny. Integrating this inequality on, 1] we obtain 2.5. Notice that x 2 Φ 1x = 1 1 2πt1 2 y x 2 y 2 e 2t 1 dy 1 = xe 1 x 2t 1 t 1 2πt1 <, xe 1+x2 2t 1 for all x, 1. Thus Φ 1 x is concave in, 1. We will know prove that Φ n x is concave in 1 2, 1 2. Lemma 2.2. If x 1, then for all n 1, 2 x Φ nx x Φ n1 x. Proof. By 2.6 the result is true for n = 1. Let us assume that the result is true for n 1. Let ψ n x = 1 2πtn e y x2 Because of 2.3, it is enough to prove that y Φ ny dy. 2.7 ψ n 1 x ψ n x. Let y,, then Thus Since 1 x y x y. e 1 x y2 e x y2. y Φ ny, 7

8 for all y <, it follows that e x y2 To simplify notation let y Φ ny dy φy = y Φ ny. e 1 x y2 y Φ ny dy. Let y, 1, and consider ŷ = 1 y. Notice that ŷ 1, 1 and 2 2 By induction, On the other hand, 1 2 y = ŷ 1 2. φy φŷ. e x y2 e x ŷ2 e x y2 = e ˆx ŷ2, = e ˆx y2, e x ŷ2. Thus e x y2 and we conclude that e x y2 e ˆx y2 φy φy + e x ŷ2 e ˆx ŷ2 φŷ e ˆx y2 Integrating over, 1 we obtained that 2 1 e x y2 y Φ ny dy the desired result immediately follows. 1 e x ŷ2 φŷ, φy + e ˆx ŷ2 φŷ. e 1 x y2 Lemma 2.3. If x < u 1, then for all n 1, 2 x Φ nx x Φ nu. y Φ ny dy, Proof. By 2.6 the result is true for n = 1. Let us assume that the result is true for n 1. As in Lemma 2.1, we let ψ n x = 1 2πtn 8 e y x2 y Φ ny dy

9 and φy = y Φ ny. By 2.3, it is enough to prove that 2.8 ψ n u ψ n x. Let y,, then u y x y. Thus Lemma 2.1 implies that e x y2 e u y2 y Φ ny dy e x y2. e u y2 y Φ ny dy. Let m = x+u. For all y, m, define ỹ = x + u y. Notice that 2 ỹ m, x + u and that We can easily check that x y = u ỹ. e x y2 e x ỹ2 e x y2 for all y, m. We claim that = e u ỹ2, = e u y2, e x ỹ2, 2.9 φy φỹ. This follows immediately from the induction hypothesis if ỹ 1 2. On the other hand, if ỹ = x + u y 1 2, then 1 x + u + y 1, and y 1 x + u + y. 2 Lemma 2.2 and the induction hypothesis imply that Thus φy φ1 x + u + y φx + u y = φỹ. e x y2 φy + e x ỹ2 φỹ e u y2 Integrating over, m we obtained that 9 φy + e u ỹ2 φỹ.

10 x+u e x y2 y Φ ny dy Finally if y x + u, 1] then x+u e u y2 y Φ ny dy. Therefore 1 x+u e x y2 e x y2 e u y2. 1 y Φ ny dy e u y2 x+u y Φ ny dy. By symmetry, Proposition 2.1 follows from Lemma 2.3. The following is an immediate corollary of Proposition 2.1. Corollary 2.1. Let B t be one dimensional Brownian motion and set I = a, a, For < t 1 < t 2 < < t n, the function 2.1 F x = P x {B t1 I, B t2 I,..., B tn I} is mid concave in I. In addition, if x I, then 2.11 F x, if x <, and F x, if x >. Proof. By the Markov property, a a n 2.12 F x = p ti t i x i x i dx 1... dx n, a a i=1 where x = x and t =. This is exactly the same expression as in Lemma 2.1 and Proposition 2.1 except for the fact that the interval, 1 has been replaced by the interval a, a. The proof of the proposition is the same for this case and the corollary follows. Corollary 2.2. Let B t be Brownian motion in R d and let Q = I 1 I 2 I d where I i = a i, a i, be a rectangle in R d. For < t 1 < t 2 < < t n, the function 2.13 F x = P x {B t1 Q, B t2 Q,..., B tn Q} is mid concave in Q. In addition, if x = x 1, x 2,..., x d Q, then 2.14 x i F x, if x i <, and 1 x i F x, if x >.

11 Proof. With x = x 1, x 2,..., x d and B t = Bt 1, Bt 2,..., Bt d, it follows by independence that d F x = P xi {Bt i 1 I i, Bt i 2 I i,..., Bt i n I i } i=1 and the conclusion of the corollary follows from Corollary 2.1 and our definition of mid concavity for domains in R d. 3. Mid concavity for stable processes In this section we prove Theorems 1.1 and 1.2. First, let us recall that for < α < 2 the symmetric stable process Xt α in R d has the representation 3.1 X α t = B 2σt, where σ t is a stable subordinator of index α/2 independent of B t see 6]. Thus 3.2 p α t x y = p 2 sx yg α/2 t, sds, where g α/2 t, s is the transition density of σ t and p 2 t/2x y = 1 x y 2 e 2t. 2πt d/2 Now, let Q and t 1, t 2,..., t n be as in the statement of Theorem 1.2. Set x = x and t =. Using the Markov property of the stable process X α t, the subordination formula 3.2, Fubini s theorem, and the Markov property of the Brownian motion, in this order, we obtain, F x = P x {Xt α 1 Q,..., Xt α n Q} n = p α t i t i x i x i dx 1... dx n = = Q... Q i=1 Q Q n p 2 s i x i x i dx 1... dx n i=1 n g α/2 t i t i, s i ds 1... ds n i=1... P x {B 2s1 Q, B 2s1 +s 2 Q,..., B 2s1 +s 2 + +s n Q} n g α/2 t i t i, s i ds 1... ds n. i=1 11

12 Since the function P x {B 2s1 Q, B 2s1 +s 2 Q,..., B 2s1 +s 2 + +s n Q} is mid concave and satisfies the monotonicity property 2.14, by Corollary 2.2, so is the integral against the densities g α/2 t i t i, s i and this completes the proof of Theorem 1.2. With Theorem 1.2 proved, we argue as in the proof of the log concavity for Brownian motion discussed in the introduction. Recall that ϕ α 1 is the ground state eigenfunction for the stable process of index α, α, 2, killed upon leaving Q and λ α 1 is its eigenvalue. Let τq α be the first exit time of the symmetric stable process from Q. Since Q is certainly intrinsically ultracontractive, see 17], we have that 3.3 ϕ α 1 x = lim t e λα 1 t P x {τ α Q > t}. The convergence is uniform for x Q. Thus to prove mid concavity for ϕ α 1 x it is enough to prove mid concavity for P x {τq α > t}. By the right continuity of the sample paths, we have, 3.4 P x { τ α Q > t } = P z { X α s Q, s t } = lim P x { X α it n n Q, i = 1,..., n }. Theorem 1.1 now follows from this and Theorem 1.2. We remark that in the case of Brownian motion, there is an extra approximation by an increasing sequence of domains in passing from the first equality to the second in 3.4. This is not needed for our stable processes since, as explain in 8], Lemma 6, for any domain D R d with Lipschitz boundary, P x {X α τ D D} = for x D. The above argument applies not only to symmetric stable processes but also to any other process which is obtained by subordination of Brownian motion. In particular, the above results hold for the so called relativistic Brownian motion and relativistic α stable processes studied in 19]. It is of course natural to ask if the function of Proposition 2.1 is concave in the whole interval, 1 for all n and all t i. Notice that, thanks to the proof of Lemma 2.2, this is the case for n = 1. If this were the case, it would show that the same is true for the function P x { τ α Q > t } and hence for the function ϕα 1, as desired. Unfortunately, this is not the case. 12

13 Proposition 3.1. Let 3.5 Φ n x = 1 1 n p ti x i x i dx 1... dx n, i=1 where x = x. Then there exist a positive integer n and real numbers t 1, t 2,..., t n in, such that the function Φ n x is not concave on, 1. Proof. We may replace, to simplify certain notation below, the interval, 1 by the interval, π. Fix t and s both positive. Let t 1 = t and t 2 = = t n = s. If the function Φ n nx is concave on, π for all n with these chosen t 1, t 2,..., t n, letting n we see that the function 3.6 π p t x yp y { τ,π > s } dy is also concave on, π. Here we have used τ,π to denote the first exit time of Brownian motion from the interval. We have 3.7 lim s eλ 1s P y { τ,π > s } = c siny, uniformly for y, π, where c > and λ 1 = 1 the first eigenvalue for, π. It follows that for each t >, the function 3.8 F t x = π p t x y siny dy must also be concave on, π. We will now show that the function F t x is not concave. Without any difficulty we may differentiate under the integral to obtain that 3.9 F t x = 1 π x y 2 t ] e x y2 2t siny dy. 2πt 5/2 Taking x = and using the elementary inequality y y3 3! siny y 13

14 valid for all y >, we see that F t is equal to 1 π y 2 t e y2 2t siny dy 2πt 5/2 π = = = and Since 1 2πt 5/2 1 2πt 5/2 1 2πt 5/2 π π 1 t 2πt t 2 π t y 2 e y2 2t y 3 e y2 1 2t dy 3! y 3 e y2 2 siny dy t π π y 3 e y2 2 dy t 3 t 3! 3! y 5 e y2 2t π t t t π dy y 5 e y2 2 3! π t t π π t y 5 e y2 2 dy, y 3 e y2 2 dy 2, y e y2 2 dy 1, e y2 2t siny dy dy t π y 5 e y2 2 dy t 2 dy π t ye y2 2t dy π t y e y2 2 dy ye y2 2 dy as t +, we see that F t is positive for sufficiently small t. By continuity, we have that F t x > for sufficiently small x, π and sufficiently small t. This, of course, contradicts the concavity of the function and shows that Φ n x is not concave. Of course, it may still be the case that the function Φ n x is concave on the whole interval when we restrict to a sequence of times satisfying t 1 = t 2 = = t n and substitute p ti x by p α t i x, which is what is needed for our applications Conjecture 1.1. That is, the following conjecture may still be true. Conjecture 3.1. Let I =, 1 and let n be a positive integer. If t i = it for 1 i n, then the function n { 3.1 F x = P x X α t1 I,..., Xt α n I } is concave on I. A natural question is whether ϕ α 1 is mid concave for any symmetric, convex domain in the plane. We will now show that for a large enough 14.

15 rhombus and α = 2 Brownian motion, this is not the case. Below we use λ 1 D and ϕ D to denote the first eigenvalue for the domain D and its corresponding eigenfunction, respectively, for the Brownian motion. We also denote the first exit time of the Brownian motion from a domain D by τ D. Proposition 3.2. For n 1, set Dn = { x 1, x 2 R 2 : x 1 n, n, x 2 + x 1 n, 1 x } 1. n There exists an n large enough such that ϕ Dn is not mid concave on Dn. Proof. The rectangle Rn = n, n is a subset of Dn and therefore, λ 1 Dn < λ 1 Rn = + 1, 1 1 n n π 2 2 2/ n + π2 2 2 n π , n for n large enough. Now, for any a, 1/2, consider the subset of Dn define by { Qa, n = x 1, x 2 R 2 : x 1 an, n, x 2 + x 1 n, 1 x } 1. n Since ϕ Dn + λ 1 Dnϕ Dn = in Qa, n and λ 1 Dn < λ 1 Qa, n. That is, ϕ Dn is a q-harmonic function with q = λ 1 Dn. The Feynman Kac formula gives that for any x Qa, n, ϕ Dn x = E x e λ 1 Dn τ Qa,n ϕ Dn Bτ Qa,n ] 3.11 Of course, ϕ Dn E x e λ 1 Dn τ Qa,n ; Bτ Qa,n Dn \ Qa, n ]. ϕ Dn = max { ϕ Dn x : x D }, by symmetry. Let pa = 1 a/2 2 /1 a 2 and qa be such that 1/pa + 1/qa = 1. Note that pa > 1 so qa >. By Hölder s inequality the expression in 3.11 is bounded above by ϕ Dn E x e λ 1 Dn τ Qa,n pa ] 1/pa P x B τqa,n Dn \ Qa, n ] 1/qa. 15

16 Since Qa, n, + a, 1 a we have that for any x Qa, n, E x e λ 1 Dn τ Qa,n pa ] E exp π 2 / / n pa τ +a,1 a ] = E exp π 2 / / n pa 1 a 2 τ,1 ] = E exp π 2 / / n 1 a/2 2 τ,1 ]. By a simple calculation we see that 1 + 3/ n1 a/2 1 when n 6 3a 2 /a 2. For such n, we have Ex e λ 1 Dn τ Qa,n pa ] 1/pa E exp π 2 /4 1 a/2 τ,1 ] 1/pa = C1 a. Using the fact that π2 is the eigenvalue for the interval, 1, we 4 have that for any c, 1, E expc τ,1 π 2 /4 ] <. Thus for any a, 1/2 we have C 1 a <. By standard results for Brownian motion or the trivial estimate of the harmonic measure in the strip obtained by conformal mapping to the disk, for any b and x 1 > b we have Bτb,,1, b, 1 ] C 2 e π 2 x1 b, P x1, where C 2 > is an absolute constant. Note that x = 2an, Qa, n. It follows that P 2an, BτQa,n Dn \ Qa, n ] C 2 e π 2 an. Now choose a = 1/8. For such a we have 2an, = n/4,. For n 6 3a 2 /a 2 we have 3.12 ϕ Dn n/4, ϕ Dn, C 1 1/8 C 2 e π 16 n ] 1 q1/8. If ϕ Dn were mid concave, we would have ϕ Dn n/4, 1 2 ϕ Dn, + ϕ Dn n/2, ] 1 2 ϕ Dn,. However, by 3.12 for large enough n we have that ϕ Dn n/4, is smaller than ϕ Dn, /2. Thus ϕ Dn is not mid concave. Indeed, the same argument shows that for any c, 1 there exists an n large enough such that ϕ Dn is not concave on the interval with endpoints cn,, cn,. 16

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