A sharp multidimensional Hermite-Hadamard inequality
aa r X i v : . [ m a t h . C A ] M a y A SHARP MULTIDIMENSIONAL HERMITE–HADAMARDINEQUALITY
SIMON LARSON
Abstract.
Let Ω ⊂ R d , d ≥
2, be a bounded convex domain and f : Ω → R be anon-negative subharmonic function. In this paper we prove the inequality1 | Ω | Z Ω f ( x ) dx ≤ d | ∂ Ω | Z ∂ Ω f ( x ) dσ ( x ) . Equivalently, the result can be stated as a bound for the gradient of the Saint Venanttorsion function. Specifically, if Ω ⊂ R d is a bounded convex domain and u is the solutionof − ∆ u = 1 with homogeneous Dirichlet boundary conditions, then k∇ u k L ∞ (Ω) < d | Ω || ∂ Ω | . Moreover, both inequalities are sharp in the sense that if the constant d is replacedby something smaller there exist convex domains for which the inequalities fail. Thisimproves upon the recent result that the optimal constant is bounded from above by d / due to Beck et al. [2]. Introduction and main results
Introduction.
In this note we make several observations concerning multidimensionalHermite–Hadamard inequalities. These inequalities have been the subject of study inseveral recent articles, we refer to [2, 6, 10, 14, 19]. In particular, we are interested in c d (Ω)the optimal constant in the inequality1 | Ω | Z Ω f ( x ) dx ≤ c d (Ω) | ∂ Ω | Z ∂ Ω f ( x ) dσ ( x ) , (1)where Ω ⊂ R d is convex and bounded, and f : Ω → R is non-negative and subharmonic. If Ωis a ball B the inequality with constant c d ( B ) = 1 is an easy consequence of the maximumand mean value principles, and (1) can in a sense be regarded as a rigidity statementthereof. The case d = 1 is the classical Hermite–Hadamard inequality [4, 5]. Since theone-dimensional case is completely understood we assume throughout that d ≥ Mathematics Subject Classification.
Key words and phrases.
Hermite–Hadamard inequality, subharmonic functions, Saint Venant elasticitytheory, torsion, gradient bounds, convex geometry.c (cid:13)
Our main result concerns the smallest constant c d such that c d (Ω) ≤ c d for all convexdomains Ω ⊂ R d , c d = sup { c d (Ω) : Ω ⊂ R d , convex and bounded } . (2)Specifically, we prove that c d = d and the supremum is not attained. In other words, forany convex Ω ⊂ R d , d ≥
2, the strict inequality c d (Ω) < d is valid. Moreover, in thetwo-dimensional case a result of M´endez-Hern´andez [11] allows us to quantify this gap byshowing that for any convex Ω ⊂ R c (Ω) < q − ce − π D (Ω) − r (Ω) r (Ω) , (3)where D (Ω) and r (Ω) denote the diameter and inradius of Ω, respectively, and c is a positiveconstant. In particular, we see that for c (Ω) to be close to c = 2 the eccentricity of Ω,i.e. D (Ω) /r (Ω), needs to be sufficiently large. We suspect that similar estimates are validalso when d ≥ Main results.
In [19] Steinerberger proved that (1) is valid for any convex Ω ⊂ R d and all non-negative convex functions f with a constant c d (Ω) ≤ π − / d d +1 (note that allconvex functions are subharmonic). More recently it was proven by Beck et al. [2] that theconstant c d in (2) satisfies d − ≤ c d ≤ ( d / if d is odd , d ( d +1) √ d +2 if d is even . (4)Our main result is that up to a slight change the proof of the upper bound in [2] in factyields that c d ≤ d . The new ingredient crucial for this improvement is an isoperimetric-type inequality for solutions of the heat equation on convex domains of fixed inradiusproved by Ba˜nuelos and Kr¨oger [1]. Furthermore, tracking equality cases throughout theproof enables us to construct a sequence of convex domains for which the inequality isasymptotically tight. Theorem 1.1.
Let Ω ⊂ R d , d ≥ , be a bounded convex domain. For all non-negativesubharmonic functions f : Ω → R it holds that | Ω | Z Ω f ( x ) dx ≤ d | ∂ Ω | Z ∂ Ω f ( x ) dσ ( x ) . (5) Equality holds in (5) if and only if f ( x ) ≡ . Moreover, if the constant d in the right-handside were replaced by something smaller there exists a bounded convex domain Ω ⊂ R d anda non-negative subharmonic function f : Ω → R for which the inequality fails. Our approach to proving Theorem 1.1, which follows closely that of [2], is based on thefollowing observation (see also [2, 3, 6, 10, 12]). Let u Ω : Ω → R denote the torsion functionof Ω, that is the solution to the boundary value problem (cid:26) − ∆ u Ω ( x ) = 1 in Ω ,u Ω ( x ) = 0 on ∂ Ω . (6) SHARP HERMITE–HADAMARD INEQUALITY 3
Note that u Ω ( x ) ≥ f ∈ H (Ω) Z Ω f ( x ) dx = Z Ω ( − ∆ u Ω ( x )) f ( x ) dx = Z Ω u Ω ( x )( − ∆ f ( x )) dx − Z ∂ Ω ∂u Ω ∂ν ( x ) f ( x ) dσ ( x ) , where ν denotes the outward pointing unit normal. Since u Ω ≥ f non-negative and subharmonic, ∆ f ( x ) ≥
0, that Z Ω f ( x ) dx ≤ (cid:13)(cid:13)(cid:13) ∂u Ω ∂ν (cid:13)(cid:13)(cid:13) L ∞ ( ∂ Ω) Z ∂ Ω f ( x ) dσ ( x ) . (7)Note that we can get arbitrarily close to equality in the above by taking f harmonic and f | ∂ Ω vanishing away from a neighbourhood of where the modulus of the normal derivativeachieves its maximum.As a consequence Theorem 1.1 follows as a corollary of the following result: Theorem 1.2.
Let Ω ⊂ R d , d ≥ , be a bounded convex domain and u Ω solve (6) , then k∇ u Ω k L ∞ (Ω) < d | Ω || ∂ Ω | . (8) Moreover, if the constant d in the right-hand side were replaced by something smaller thereexists a bounded convex domain Ω ⊂ R d for which the inequality fails.Remark . A couple of remarks:(1) Firstly, since u Ω vanishes on the boundary and∆ |∇ u Ω | = 2 d X i,j =1 (cid:16) ∂ u Ω ∂x i ∂x j (cid:17) ≥ (cid:13)(cid:13) ∂u Ω ∂ν (cid:13)(cid:13) L ∞ ( ∂ Ω) = k∇ u Ω k L ∞ (Ω) . Thus any boundfor c d (Ω) implies a corresponding bound for k∇ u Ω k L ∞ (Ω) .(2) Secondly, an application of Greens identity yields a matching lower bound: k∇ u Ω k L ∞ (Ω) = (cid:13)(cid:13)(cid:13) ∂u Ω ∂ν (cid:13)(cid:13)(cid:13) L ∞ ( ∂ Ω) ≥ | ∂ Ω | Z ∂ Ω (cid:12)(cid:12)(cid:12) ∂u Ω ∂ν (cid:12)(cid:12)(cid:12) dσ ( x ) = | Ω || ∂ Ω | , (9)where equality holds if and only if Ω is a ball, by the classical overdetermined Serrinproblem (see [13, 17]).1.3. A family of Hermite–Hadamard-type inequalities.
In Section 4 we consider aone-parameter family of inequalities containing the Hermite–Hadamard inequality (1) as aspecial case. Specifically, we consider for α ≤ d the inequality Z Ω f ( x ) dx ≤ c d,α (Ω) | Ω | αd | ∂ Ω | − αd − Z ∂ Ω f ( x ) dσ ( x ) , (10)where as before Ω ⊂ R d is a convex domain and f : Ω → R is a non-negative subharmonicfunction. When α = d the inequality (10) is nothing but (1), while for α = 1 it reduces toan inequality studied in [2, 6, 10, 19]. Again our main interest is towards upper bounds for c d,α (Ω). We note that for α > d no uniform bound can hold since by taking f ≡ S. LARSON such a bound would imply a reverse isoperimetric inequality, which is a contradiction.However, for α < d a uniform bound can easily be deduced from the end-point case α = d and the isoperimetric inequality. In fact, as we shall see in Section 4 any upper bound for c d,α (Ω) combined with the isoperimetric inequality implies an upper bound for c d,α (Ω) forall α < α . In particular, the bounds for α = 1 in [2, 6, 10, 19] imply bounds for all α < c d,α = sup { c d,α (Ω) : Ω ⊂ R d , convex and bounded } . (11)However, what we find interesting is that the dependence of c d,α (Ω) on the geometryappears very different for α < d compared to the end-point case α = d . Indeed, we shallprove that when α < d the constant c d,α (Ω) becomes small if Ω has high eccentricity. Weemphasize that this is fundamentally different form the behaviour we expect in the case α = d ≥
3, and by (3) know to be true when α = d = 2.As in the case α = d our results can equivalently be phrased in terms of bounds for k∇ u Ω k L ∞ (Ω) . Indeed, arguing as for α = d one concludes that c d,α (Ω) = | Ω | − αd | ∂ Ω | α − d − k∇ u Ω k L ∞ (Ω) . (12)Our main result in this direction is the following: Theorem 1.4.
Let Ω ⊂ R d , d ≥ , be a bounded convex domain and u Ω solve (6) . Then,for any α ≤ d , k∇ u Ω k L ∞ (Ω) ≤ ˜ c d,α (cid:16) r (Ω) D (Ω) (cid:17) d − αd ( d − | Ω | αd | ∂ Ω | − αd − , (13) with ˜ c d,α > depending only on d, α . Moreover, the power of r (Ω) /D (Ω) is optimal. While Theorem 1.2 tells us that for α = d the supremum (11) is not attained and weexpect any sequence of { Ω k } k ≥ , with | Ω k | = 1, satisfyinglim k →∞ | ∂ Ω k |k∇ u Ω k k L ∞ (Ω k ) = d to become unbounded in the limit, Theorem 1.4 tells us that the situation for α < d isdifferent: Corollary 1.5.
For Ω ⊂ R d , d ≥ , denote by u Ω the solution of (6) . For α < d anysequence of convex domains { Ω k } k ≥ ⊂ R d , with | Ω k | = 1 , satisfying lim inf k →∞ | ∂ Ω k | α − d − k∇ u Ω k k L ∞ (Ω k ) > , is uniformly bounded in the Hausdorff metric. In particular, up to translation any suchsequence contains a subsequence converging with respect to the Hausdorff metric.Remark . As a consequence of Corollary 1.5 we find that if { Ω k } k ≥ is a maximizingsequence for c d,α there exists a subsequence which, after translation, converges to a convexdomain Ω ∗ . Naturally, it is tempting to claim that the limit Ω ∗ realizes the supremum, SHARP HERMITE–HADAMARD INEQUALITY 5 c d,α = c d,α (Ω ∗ ). However, although it is not very difficult to conclude that, up to passingto a subsequence, ∇ u Ω k → ∇ u Ω ∗ in L p ( R d ) , for any p < ∞ , we are at this point unable to deduce thatlim k →∞ k∇ u Ω k k L ∞ (Ω k ) = k∇ u Ω ∗ k L ∞ (Ω ∗ ) . Nevertheless, it is an interesting question to understand the shape of such limiting domains,and how their geometry depends on α . The fact that makes this question particularlyintriguing is that Ω ∗ is expected to be quite different from a ball. In fact, by arguing asin (9) and using the isoperimetric inequality one finds that balls are the unique minimizersof c d,α (Ω) for all α ≤ d , c d (Ω) = | Ω | − αd | ∂ Ω | α − d − k∇ u Ω k L ∞ (Ω) ≥ (cid:20) | Ω | d | ∂ Ω | d − (cid:21) d − α ≥ d α − dd − ω α − dd ( d − d , where equality holds if and only if Ω is a ball and ω d denotes the volume of the d -dimensionalunit ball. For the case d = 2 and α = 1 candidates for maximizing domains were obtainedin [6].Before we move on we note that there are three cases of the shape optimization problemassociated to (11) which appear particularly natural:(1) α = d : Corresponding to the inequality k∇ u Ω k L ∞ (Ω) ≤ c d | Ω || ∂ Ω | . This case is ofparticular interest as the end-point and strongest inequality in the range, indeedfor any α < α ′ ≤ d we have c d,α ≤ c d,α ′ (cid:0) dω /dd (cid:1) α − α ′ d − (see (24) below).(2) α = 1: Corresponding to the inequality k∇ u Ω k L ∞ (Ω) ≤ c d, | Ω | d , and by scaling theshape optimization problem of maximizing k∇ u Ω k L ∞ (Ω) with a measure constraint.(3) α = 0: Corresponding to the inequality k∇ u Ω k L ∞ (Ω) ≤ c d, | ∂ Ω | d − , and by scal-ing the shape optimization problem of maximizing k∇ u Ω k L ∞ (Ω) with a perimeterconstraint.Although the maximum of the gradient of the torsion function is a classical quantity inthe Saint Venant theory of elasticity (the maximum shear stress), there is to the authorsknowledge little known concerning these shape optimization problems. Apart from α = d and α = 1, which have recently been considered in the context of Hermite–Hadamard-typeinequalities [2, 6], we are unaware of results in this direction. In particular, the problemwhen α = 0 appears not to have been studied. That being said there is a wide range ofbounds for k∇ u Ω k L ∞ (Ω) under various assumptions on Ω available in the literature, werefer to [8] and references therein.In the proof of our main result we shall see that the problem for α = d is closely relatedto maximizing k∇ u Ω k L ∞ (Ω) with a constraint on the inradius of Ω. In fact, our proofof Theorem 1.2 relies on showing that this shape optimization problem is solved by theinfinite slab, which follows by combining an inequality of Sperb with a result of Ba˜nuelosand Kr¨oger [1, 18]. Although the results obtained in this paper essentially settle the shape S. LARSON optimization problem when α = d it would be interesting to obtain quantitative resultssimilar in spirit to (3) also when d ≥ Proof of Theorems 1.1 & 1.2
By the argument in the previous section Theorem 1.1 follows as a consequence of Theo-rem 1.2. Indeed, the inequality (5) follows from Theorem 1.2 and (7) as does the sharpnessof the constant. Moreover, since (8) is strict, equality holds in (5) if and only if f ( x ) ≡ width as Ω, an application of a result of Ba˜nuelosand Kr¨oger reduces the problem to a slab of the same inradius as Ω. This allows us toremove the extra factor ∼ √ d in the result of [2] which arose as a consequence of usingSteinhagen’s inequality [20] to bound the width w (Ω) in terms of the inradius r (Ω). Theprecise differences in the proofs will be explained in greater detail below. Proof of Theorem 1.2.
We split the proof in two steps. In the first step the bound c d (Ω) < d is established, while in the second step the sharpness of the bound is proved by explicitconstruction of a family of bounded convex sets { Ω η } η satisfying lim η →∞ c d (Ω η ) = d . Step 1: (Proof of the upper bound c d (Ω) < d ) Recall that for any bounded convex domainΩ ⊂ R d we have | Ω || ∂ Ω | ≤ r (Ω) ≤ d | Ω || ∂ Ω | , (14)see for instance [9, eq. (13)]. Equality in the upper bound of (14) holds if and only ifΩ is tangential to a ball (i.e. all its regular supporting hyperplanes are tangent to thesame inscribed ball [16]). Also the lower bound is seen to be sharp by considering the sets( − , × ( − R, R ) d − as R → ∞ .By (14) we have that | ∂ Ω || Ω | k∇ u Ω k L ∞ (Ω) ≤ dr (Ω) − k∇ u Ω k L ∞ (Ω) . (15)We wish to maximize the right-hand side with respect to Ω. We aim to show thatsup { r (Ω) − k∇ u Ω k L ∞ (Ω) : Ω ⊂ R d , convex and bounded } = 1 , (16)and that the supremum is not achieved. Note that by scaling this is equivalent to maxi-mizing k∇ u Ω k L ∞ (Ω) among all convex domains of a given inradius.A classical inequality of Sperb allows us to bound the maximum of the gradient of u Ω in terms of the function itself: k∇ u Ω k L ∞ (Ω) ≤ k u Ω k L ∞ . (17)When d = 2 this is [18, eq. (6.12)] while for d ≥ − ∆ u = 2 resulting in a different SHARP HERMITE–HADAMARD INEQUALITY 7 constant (see also [15] for non-smooth Ω). It is easily checked by explicit calculation thatequality in (17) holds if Ω is the infinite slab ( − , × R d − .Let p Ω ( t, x, y ) denote the heat kernel of the Dirichlet Laplacian on Ω, then u Ω ( x ) = Z ∞ Z Ω p Ω ( t, x, y ) dydt . By integrating the bound of [1, Theorem 1] with respect to t , for any x ∈ Ω, u Ω ( x ) = Z ∞ Z Ω p Ω ( t, x, y ) dydt ≤ Z ∞ Z S r (Ω) p S r (Ω) ( t, , y ) dydt = u S r (Ω) (0) , (18)where S r (Ω) = ( − r (Ω) , r (Ω)) × R d − and equality holds if and only if Ω is an infinite slaband dist( x, ∂ Ω) = r (Ω). Since u S r (Ω) ( x ) = r (Ω) − x , it follows that k u Ω k L ∞ (Ω) < r (Ω) ⊂ R d . For d = 2 and with non-strict inequality this bound wasproved by Sperb [18, eq. (6.13)].Combining (17) and (19) we find r (Ω) − k∇ u Ω k L ∞ (Ω) ≤ √ r (Ω) − k u Ω k / L ∞ (Ω) < , which proves that the supremum in (16) is at most 1 and that this value is not attained.In view of (15), this completes the proof of the inequality in Theorem 1.2.Before moving on to the sharpness of the result we for the sake of comparison explainhow the proof above differs from that given in [2]. Specifically, the difference appears atthe point in the proof where we appeal to the result of Ba˜nuelos and Kr¨oger. In [2] themaximum principle and that, in appropriately chosen coordinates,Ω ⊂ S w (Ω) / = (cid:0) − w (Ω)2 , w (Ω)2 (cid:1) × R d − was used to bound k u Ω k L ∞ (Ω) ≤ k u S w (Ω) / k L ∞ ( S w (Ω) / ) = w (Ω) . (20)As r (Ω) ≤ w (Ω) / w (Ω) ≤ ( √ d r (Ω) if d is odd2 d +1 √ d +2 r (Ω) if d is even , (21)completes the proof of the upper bound in (4). While Sperb’s inequality (17) is sharp forthe infinite slab, equality in Steinhagen’s inequality (21) holds only for the regular ( d + 1)-simplex. These competing facts are what leads to the superfluous factor ∼ √ d in the boundfrom [2]. S. LARSON
Step 2: (Proof of sharpness) We shall construct a family of bounded convex sets { Ω η } η suchthat lim η →∞ c d (Ω η ) = d . To achieve this we need to choose Ω η such that we are arbitrarilyclose to equality in each part of our proof for the upper bound. Namely, we need to beclose to equality in both the upper bound of (14) and almost attain the supremum in (16).As of yet we have not proved that the supremum (16) is not less than 1, but this is nottoo difficult. Indeed, one can consider the sequence Ω R = ( − , × ( − R, R ) d − and showthat equality holds in the limit R → ∞ . However, as we wish to prove that equality can beattained while simultaneously being close to equality in the upper bound of (14) we needto choose our domains a bit more carefully.Let Ω η be the ( d + 1)-simplex obtained by taking a regular d -simplex of sidelength η ≫ x = 0 centred at the origin and adding a final vertex at (1 , , . . . , η − ε εrC ε,r Ω η Figure 1.
The almost maximizing domain Ω η along with the inscribed box C ε,r . For any ε, r > η sufficiently large so that C ε,r ⊂ Ω η .Since Ω η is a ( d + 1)-simplex and thus tangential to a ball | ∂ Ω η || Ω η | = dr (Ω η ) . In order to complete our proof we need to show that r (Ω η ) − k∇ u Ω k L ∞ (Ω η ) = r (Ω η ) − (cid:13)(cid:13)(cid:13) ∂u Ω η ∂ν (cid:13)(cid:13)(cid:13) L ∞ ( ∂ Ω η ) ≥ o (1) , as η → ∞ . For any ε > r > η large enough so that C ε,r = (cid:8) x ∈ R d : 0 < x < − ε, | x j | < r, j = 2 , . . . , d (cid:9) ⊂ Ω η , see Figure 1. By the maximum principle 0 < u C ε,r ( x ) ≤ u Ω η ( x ) for all x ∈ C ε,r . Since0 ∈ ∂C ε,r ∩ ∂ Ω η and u Ω η (0) = u C ε,r (0) = 0 it holds that (cid:12)(cid:12)(cid:12) ∂u C ε,r (0) ∂ν (cid:12)(cid:12)(cid:12) ≤ (cid:12)(cid:12)(cid:12) ∂u Ω η (0) ∂ν (cid:12)(cid:12)(cid:12) . Consequently, for all ε > , r > η large enough so that r (Ω η ) − (cid:13)(cid:13)(cid:13) ∂u Ω η ∂ν (cid:13)(cid:13)(cid:13) L ∞ ( ∂ Ω η ) ≥ r (Ω η ) − (cid:12)(cid:12)(cid:12) ∂u C ε,r (0) ∂ν (cid:12)(cid:12)(cid:12) ≥ (cid:12)(cid:12)(cid:12) ∂u C ε,r (0) ∂ν (cid:12)(cid:12)(cid:12) , where we used r (Ω η ) ≤ / SHARP HERMITE–HADAMARD INEQUALITY 9 As r → ∞ and ε → u C ε,r converges to u S ( x ) = x (1 − x )2 uniformly on anycompact, where S = { x ∈ R d : 0 < x < } . Moreover, since v = u S − u C ε,r is harmonic in C ε,r and vanishes when x = 0 we find that ˜ v defined as the reflection of v through x = 0,that is for x = ( x , x ′ ) ∈ R d we set˜ v ( x ) = sgn( x ) v ( | x | , x ′ ) , is harmonic in ˜ C ε,r = { x ∈ R d : | x | < − ε, | x j | < r, j = 2 , . . . , d } Consequently, ∂ j ˜ v ( x ) is harmonic for each j = 1 , . . . , d and thus the mean value principleand the divergence theorem yields, for any 0 < ρ < / |∇ ˜ v (0) | = 1 ω d ρ d (cid:12)(cid:12)(cid:12)(cid:12)Z B ρ (0) ∇ ˜ v ( x ) dx (cid:12)(cid:12)(cid:12)(cid:12) = 1 ω d ρ d (cid:12)(cid:12)(cid:12)(cid:12)Z ∂B ρ (0) ˜ v ( x ) xρ dσ ( x ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ ω d ρ d Z ∂B ρ (0) | ˜ v ( x ) | dσ ( x ) . Multiplying both sides by ρ d − and integrating from 0 to 1 / ρ we findthat |∇ ˜ v (0) | ≤ C d k ˜ v k L ( B / (0)) . Since ˜ v converges to zero as r → ∞ and ε → ε → r →∞ (cid:12)(cid:12)(cid:12) ∂u C ε,r (0) ∂ν (cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12) ∂u S (0) ∂ν (cid:12)(cid:12)(cid:12) = 12 . By combining the above we find c d (Ω η ) = | ∂ Ω η || Ω η | k∇ u Ω η k L ∞ (Ω η ) = dr (Ω η ) − k∇ u Ω η k L ∞ (Ω η ) ≥ d + o (1) , as η → ∞ , which proves that c d ≥ d and therefore completes the proof of Theorem 1.2. (cid:3) A quantitative improvement when d = 2In the two-dimensional case a result of M´endez-Hern´andez [11] actually allows one tostrengthen both Theorem 1.1 and Theorem 1.2. Namely, integrating the inequality of [11,Theorem 5.2] with respect to t implies that the supremum of the torsion function of abounded convex domain Ω ⊂ R is bounded not only by that of the strip of same inradiusbut by the supremum of the torsion function in the truncated strip (( − r (Ω) , r (Ω)) × R ) ∩ B D (Ω) − r (Ω) (0).Setting R (Ω) = ( − r (Ω) , r (Ω)) × ( − D (Ω) + r (Ω) , D (Ω) − r (Ω)) and using the maximumprinciple one concludes that k u Ω k L ∞ ≤ k u R (Ω) k L ∞ = u R (Ω) (0 , . (22)The torsion function of a rectangle can be explicitly computed. Indeed, for R l =( − / , / × ( − l, l ), u R l ( x , x ) = 1 − x − π X n ≥ − ( − n n cosh( nπl ) cosh( nπx ) sin( nπ ( x + 1 / . Since u R (Ω) ( x ) = 4 r (Ω) u R D (Ω) − r (Ω)2 r (Ω) ( x/ (2 r (Ω)))we find k u R (Ω) k L ∞ = 4 r (Ω) k u R D (Ω) − r (Ω)2 r (Ω) k L ∞ = 4 r (Ω) u R D (Ω) − r (Ω)2 r (Ω) (0 , r (Ω) − r (Ω) π X n ≥ − ( − n n cosh (cid:0) nπ D (Ω) − r (Ω)2 r (Ω) (cid:1) sin( nπ/ r (Ω) − r (Ω) π X k ≥ ( − k (2 k + 1) cosh (cid:0) (2 k + 1) π D (Ω) − r (Ω)2 r (Ω) (cid:1) ≤ r (Ω) − r (Ω) π (cid:20) (cid:0) π D (Ω) − r (Ω)2 r (Ω) (cid:1) −
19 cosh (cid:0) π D (Ω) − r (Ω)2 r (Ω) (cid:1) (cid:21) , where we in the last step used the fact that x x cosh( x ) , x > , is decreasing.Since cosh( x ), x >
0, is strictly increasing and satisfies x ) ≤ e − x ,1cosh (cid:0) π D (Ω) − r (Ω)2 r (Ω) (cid:1) −
19 cosh (cid:0) π D (Ω) − r (Ω)2 r (Ω) (cid:1) >
89 cosh (cid:0) π D (Ω) − r (Ω)2 r (Ω) (cid:1) ≥ e − π D (Ω) − r (Ω) r (Ω) . We conclude that k u R (Ω) k L ∞ < r (Ω) h − ce − π D (Ω) − r (Ω) r (Ω) i , for any c ≤ π .Consequently, the inequalities in Theorems 1.1 and 1.2 can for d = 2 be strengthenedto: For Ω ⊂ R bounded and convex k∇ u Ω k L ∞ (Ω) < c (Ω) | Ω || ∂ Ω | and 1 | Ω | Z Ω f ( x ) dx ≤ c (Ω) | ∂ Ω | Z ∂ Ω f ( x ) dσ ( x ) , for all non-negative subharmonic functions f : Ω → R , with c (Ω) < q − ce − π/ D (Ω) − r (Ω) r (Ω) for some constant c >
0. 4.
Upper bounds for c d,α (Ω)We now turn our attention to the case α < d . As mentioned in the introduction ourresult for α = d together with the isoperimetric inequality implies that c d,α < ∞ . Indeed, SHARP HERMITE–HADAMARD INEQUALITY 11 by (12), Theorem 1.2, and the isoperimetric inequality c d,α (Ω) = (cid:20) | Ω | d | ∂ Ω | d − (cid:21) d − α | ∂ Ω || Ω | k∇ u Ω k L ∞ (Ω) ≤ d α − d − ω α − dd ( d − d . (23)Note that this bound cannot possibly be sharp, indeed we have equality in the isoperimetricinequality if and only if Ω is a ball in which case we are far from equality in Theorem 1.2.Utilizing a bound for c d, proved in [2] allows us to do better by a negative power of d : Lemma 4.1.
Let Ω ⊂ R d , d ≥ , be a bounded convex domain and u Ω solve (6) . Then,for ≤ α ≤ d , c d,α ≤ d α − d − − d − α d − ω α − dd ( d − d , while for α ≤ c d,α ≤ d α − d − − ω α − dd ( d − d . Remark . For d = 2 both bounds can be improved by utilizing the better bound for c , obtained in [6]. Proof of Lemma 4.1.
For any α < α ′ ≤ d the isoperimetric inequality and (12) implies c d,α (Ω) = (cid:20) | Ω | d | ∂ Ω | d − (cid:21) α ′ − α | Ω | − α ′ d | ∂ Ω | α ′− d − k∇ u Ω k L ∞ (Ω) ≤ (cid:0) dω /dd (cid:1) α − α ′ d − c d,α ′ (Ω) . (24)Choosing α ′ = 1 and using the bound c d, (Ω) ≤ ω − /dd d − / [2, Theorem 3] proves thesecond inequality of the lemma.To prove the first inequality we argue similarly. For any α ≤ α ≤ α we have c d,α (Ω) = h | Ω | − α d | ∂ Ω | α − d − k∇ u Ω k L ∞ (Ω) i α − αα − α h | Ω | − α d | ∂ Ω | α − d − k∇ u Ω k L ∞ (Ω) i α − α α − α = c d,α (Ω) α − αα − α c d,α (Ω) α − α α − α . Choosing α = 1, α = d , and using the bounds of Theorem 1.2 and [2, Theorem 3] yieldsthe desired bound, which completes the proof of Lemma 4.1. (cid:3) We turn to the proofs of Theorem 1.4 and Corollary 1.5.
Proof of Theorem 1.4.
Our goal is to prove | Ω | − αd | ∂ Ω | α − d − k∇ u Ω k L ∞ (Ω) = (cid:20) | Ω | d | ∂ Ω | d − (cid:21) d − α | ∂ Ω || Ω | k∇ u Ω k L ∞ (Ω) . d,α (cid:16) r (Ω) D (Ω) (cid:17) d − αd ( d − . (25)By Theorem 1.2 and (9) 1 ≤ | ∂ Ω || Ω | k∇ u Ω k L ∞ (Ω) < d , therefore it suffices to prove | Ω | d | ∂ Ω | d − . d (cid:16) r (Ω) D (Ω) (cid:17) d ( d − . By John’s lemma [7] there exists an ellipsoid E ⊂ R d such that E is contained in Ω andthe dilation of E by a factor d around its centre contains Ω. Let E be such an ellipsoidassociated with Ω and denote by r ≤ r ≤ . . . ≤ r d the lengths of the semi-axes of E .Note that r and r d are comparable to r (Ω) and D (Ω), respectively. Moreover, since | E | ∼ d d Y j =1 r j and | ∂E | ∼ d d Y j =2 r j the monotonicity of perimeter and volume under inclusion of convex sets implies | Ω | d | ∂ Ω | d − ∼ d Q dj =1 r d j Q dj =2 r d − j = d Y j =2 (cid:16) r r j (cid:17) d ( d − ≤ (cid:16) r r d (cid:17) d ( d − ∼ d (cid:16) r (Ω) D (Ω) (cid:17) d ( d − (26)as claimed.Since each step of the proof is sharp up to constants if r = . . . = r d − the optimality ofthe exponent follows. This completes the proof of Theorem 1.4. (cid:3) Proof of Corollary 1.5.
Let { Ω k } k ≥ be as in the statement of the corollary. Since | Ω k | = 1we have that r (Ω k ) ≤ ω − /dd . Thus, by Theorem 1.4, | ∂ Ω k | α − d − k∇ u Ω k k L ∞ (Ω k ) . d,α (cid:16) r (Ω k ) D (Ω k ) (cid:17) d − αd ( d − . d,α D (Ω) − d − αd ( d − and by assumption lim inf k →∞ | ∂ Ω k | α − d − k∇ u Ω k k L ∞ (Ω k ) > . Combining the above we find D (Ω k ) . d,α
1. The existence of a convergent subsequencefollows from the Blaschke selection theorem [16, Theorem 1.8.7]. (cid:3)
Acknowledgements.
The author wishes to thank the anonymous referee for valuablecomments and suggestions which significantly helped improve the quality of the manuscript.
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