A Proof of The Triangular Ashbaugh-Benguria-Payne-Pólya-Weinberger Inequality
AA PROOF OF THE TRIANGULARASHBAUGH–BENGURIA–PAYNE–P ´OLYA–WEINBERGERINEQUALITY
RYAN ARBON, MOHAMMED MANNAN, MICHAEL PSENKA, AND SEYOON RAGAVAN
Abstract.
In this paper, we show that for all triangles in the plane, theequilateral triangle maximizes the ratio of the first two Dirichlet–Laplacianeigenvalues. This is an extension of work by Siudeja [12], who proved theinequality in the case of acute triangles. The proof utilizes inequalities due toSiudeja and Freitas [5], together with improved variational bounds. Introduction
For triangles in the Euclidean plane, the explicit values for the eigenvalues of theDirichlet–Laplacian problem are only known in the case of the equilateral, 30-60-90, and 45-45-90 triangles. However, it is known that for a given domain D in theplane, the Dirichlet–Laplacian eigenvalues form a non-decreasing sequence, whichwe order as { λ i } i ∈ N . From now on, given a domain D in the plane, we will use thephrase “the eigenvalues of D ” to refer to the Dirichlet–Laplacian eigenvalues of D .The Payne–P´olya–Weinberger (PPW) inequality dates back to 1955, when L.Payne, G. P´olya, and H. Weinberger published a paper [10] proving a bound onthe ratio of the first two eigenvalues λ /λ of a bounded domain D in the plane,namely that λ /λ ≤
3. Payne, P´olya, and Weinberger conjectured that this ratiois maximized when D is the disc, that is:(1.1) λ λ ≤ λ λ (cid:12)(cid:12)(cid:12)(cid:12) disc ≈ . . The original PPW inequality was generalized to dimension n by Thompson in[13], who showed that(1.2) λ λ ≤ n and conjectured that(1.3) λ λ ≤ λ λ (cid:12)(cid:12)(cid:12)(cid:12) n -dimensional ball = (cid:18) j n/ j n/ − (cid:19) , where j m is the first positive zero of the Bessel function of order m . The originalPPW conjecture, along with its n -dimensional generalization by Thompson, wasproven in 1992 by Ashbaugh and Benguria in [2, 3], which led to a natural question:loosely stated, do more regular shapes maximize the ratio λ /λ ? In particular, asstated in [1], the polygonal Ashbaugh-Benguria-PPW Conjecture states that the a r X i v : . [ m a t h . SP ] S e p egular n -gon in the plane maximizes λ /λ in the class of n -gons. More backgroundon the PPW inequality can be found in [6].The purpose of this paper is to show that the ratio λ /λ of eigenvalues of theequilateral triangle is maximized among triangles, as stated below: Theorem 1.1.
For an arbitrary triangle, the following inequality holds: (1.4) λ λ (cid:12)(cid:12)(cid:12)(cid:12) triangle ≤ λ λ (cid:12)(cid:12)(cid:12)(cid:12) equilateral = 73 . This corresponds to the case k = 3 of Conjecture 6.31 in [7] and Open Problem4 of [1], that is, the triangular case of the polygonal Ashbaugh–Benguria–PPWinequality. 2. Proof Outline
In our paper, we prove Theorem 1.1 with simple, explicit proofs split into minimalcase work. We achieve this in multiple ways. Our proof of Theorem 1.1 relies heavilyon work done by Siudeja, who proved in [12] that Theorem 1.1 holds when restrictedto acute triangles. Since the acute case is proven in [12], we restrict our attentionto obtuse and right triangles. We additionally utilize bounds proved by Siudejaand Freitas in [5]. Once we restrict ourselves to the obtuse case and introducenew bounds for the eigenvalues, we are able to finish the proof of Theorem 1.1with only four simple cases, illustrated in Figure 1, using mostly simple univariateoptimization problems and other elementary techniques. This approach stands outamong previous publications on this problem, which tend to involve many cases,each of which involves complicated inequalities in multiple variables that are provenwith the assistance of a computer.In addition for our proof, we build new variational bounds on λ from thoseprovided in [12] that are tighter for moderately obtuse triangles, and we apply asimple monotonicity argument to obtain a bound that is effective for very obtusetriangles. We describe this in detail in Section 3.We will use d to denote the diameter of the triangle, which we normalize to 1.We consider triangles in the Euclidean plane with vertices at (0 , , p, q )without loss of generality. To be right-angled or obtuse at ( p, q ), the third vertex( p, q ) must belong on the boundary of or inside the circle ( p − / + q = 1 / p ≥ / q ≥
0. This region is shown in Figure 1.Hence, our triangles have shortest height h equal to q and area A equal to q .Moreover, we use θ to denote the smallest angle of the triangle which will be at(0 , θ = tan − ( q/p ).Our primary strategy is to combine the following estimates for λ from [5]:(2.1) λ ≥ π (1 /d + 1 /h ) and(2.2) λ ≥ θj π/θ A , in combination with new bounds on λ . These inequalities are equations (2.6) and(2.8) in [5], respectively.We obtain new bounds on λ using a variational approach with test functionsbased on known eigenfunctions for the 45-45-90 and 30-60-90 triangles. For very igure 1. Illustration for single vertex ( p, q ) of obtuse trianglesin cases I, II, III, and IV. The other two vertices are always (0 , , p, q ). Note that as λ /λ is invariant under scaling andrigid motions, we can restrict to this quarter semi-circle withoutloss of generality.flat triangles, we enclose a rectangle within the triangle. We will refer to thesebounds as “45-45-90 λ bound,” “30-60-90 λ bound,” and “rectangle λ bound.”As seen in Figure 1, we divide this region into four areas which we addressindividually. Area I employs the 45-45-90 bound and bound (2.1), Area II usesthe 30-60-90 bound and bound (2.1), Area III uses the rectangle bound and bound(2.2), and finally Area IV employs the rectangle bound and bound (2.1).We now make these λ estimates precise before going into casework.3. Upper Bounds on λ Mathematica code reproducing all computations for this section and Section 4is available on GitHub .3.1. Variational Bounds.
For these bounds on λ we use the variational charac-terization λ | T = inf f ,f sup α (cid:82) T |∇ ( αf + f ) | (cid:82) T ( αf + f ) = inf f ,f sup α Aα + 2 Bα + CDα + 2 Eα + F , https://github.com/sragavan99/triangle-ppw-inequality here A = (cid:90) T |∇ f | , B = (cid:90) T ∇ f · ∇ f , C = (cid:90) T |∇ f | ,D = (cid:90) T f , E = (cid:90) T f f , F = (cid:90) T f . (3.1)As usual, f , f must be linearly independent and vanish at the boundary of T .To choose test functions f , f , we use the idea of “transplanting eigenfunctions”used in [5, 8, 11, 12]. We take the first two eigenfunctions of a 45-45-90 or 30-60-90 triangle and transplant them onto T with a suitable affine transformation.These bounds can also be found in [12], but the affine transformations used theresignificantly distort the triangle when it is right or obtuse. We thus obtain betterbounds for the obtuse and right cases by choosing different affine transformationsthat have smaller distortion for right/obtuse triangles; we will point out thesedifferences.3.1.1. We take our 30-60-90 triangle to have vertices at (0 , / , / , √ / z = π (2 x −
1) and t = π (1 − y √ ): φ , ( x, y ) = sin(4 z ) sin(2 t ) − sin(5 z ) sin( t ) − sin( z ) sin(3 t ) ,φ , ( x, y ) = sin(5 z ) sin(3 t ) − sin(2 z ) sin(4 t ) − sin(7 z ) sin( t ) . Let L be the affine mapping sending (0 ,
0) to (1 / , √ / p, q ) to (1 / , ,
0) to (0 , , , (1 , , (0 , √
3) and considers an affine mapping preserving (0 , ,
0) and sending ( p, q ) to (0 , √ p, q ) is mapped into the 30 ◦ angle at(0 , √ φ , ◦ L and φ , ◦ L . We can then evaluatecoefficients given by (3.1) using these test functions: ( p, q ) = − π + 4 p (1245184 − p + 100800( − p ) π )345600 q + − q + 806400 π q q ,B ( p, q ) = − p ( − p ) + 2486372 q q ,C ( p, q ) = − p (6209536 − π ) + 28 p ( − π )1058400 q + − q + 1146600 π (3 + 4 q )1058400 q ,D ( p, q ) = 3 q , E ( p, q ) = 0 , F ( p, q ) = 3 q . (3.2)As expected, E ( p, q ) = 0 since φ , and φ , are orthogonal on the original30-60-90 triangle and this will be preserved by an affine transformation. Thus ourfinal bound for λ is(3.3) λ ≤ sup α A ( p, q ) α + 2 B ( p, q ) α + C ( p, q ) D ( p, q ) α + F ( p, q ) . We take our 45-45-90 triangle to be that with vertices at(0 , , (1 , , (0 , φ , ( x, y ) = sin(2 πx ) sin( πy ) + sin( πx ) sin(2 πy ) ,φ , ( x, y ) = sin(3 πx ) sin( πy ) − sin( πx ) sin(3 πy ) . These can be derived by noting that eigenfunctions of this triangle can be re-flected over the line y = 1 − x to obtain an eigenfunction of the unit square thatvanishes along this diagonal. We define L to be the affine mapping sending ( p, q )to (0 , ,
0) to (1 , ,
0) to (0 , p, q ) to the right angle at (0 , ,
0) and (1 ,
0) and sending ( p, q ) to (0 , p, q ) is sent to the 45 ◦ angleat (0 , φ , ◦ L and φ , ◦ L . From these testfunctions, we obtain the following coefficients: ( p, q ) = p (256 − π ) + p ( −
256 + 90 π ) − q + 45 π (1 + 2 q )72 q ,B ( p, q ) = 512(1 − p )175 q ,C ( p, q ) = 5 π (1 − p + 2 p + 2 q )4 q ,D ( p, q ) = q , E ( p, q ) = 0 , F ( p, q ) = q . (3.4)Once again, it can be seen without doing any integration that E ( p, q ) = 0.This gives us the following bound:(3.5) λ ≤ sup α A ( p, q ) α + 2 B ( p, q ) α + C ( p, q ) D ( p, q ) α + F ( p, q ) . Rectangle Bound.
When our triangle is very obtuse (i.e. q is very small),the bounds on λ described so far are insufficient. This is not surprising, since inthis region our affine transformations are still quite distortive. Thus, we addressthis case with a different approach. As stated in Section 2, this bound is obtained byenclosing a rectangle inside the triangle, with one side coinciding with the triangle’sdiameter. A visualization is given in Figure 2. As the triangle becomes more obtuse,it becomes closer to the enclosed rectangle in shape, so we expect this estimate tobe more effective. It is straightforward to see that if such a rectangle R has height qt for t ∈ (0 , − t .Let us take t = √ q . This clearly is in (0 , q < ⇔ q < (cid:112) q ⇔
11 + q >
11 + (cid:112) q ⇔ > (1 + q ) t ⇔ − t > qt. Then by monotonicity of Dirichlet eigenvalues, we obtain the following inequality: λ ≤ λ ( R ) = π ( 4(1 − t ) + 1( qt ) ) = π (1 + (cid:112) q ) q . This then yields the rectangle bound:(3.7) λ ≤ π (1 + (cid:112) q ) q . Proofs in Each Area
We will now begin to prove Theorem 1.1 by splitting into the four cases indicatedby Figure 2. igure 2. For very obtuse triangles, we bound λ by enclosing arectangle within the triangle and using monotonicity.4.1. Area I.
In this region, we have q ≥ .
156 and p ≤ .
65, and we employ bounds(3.5) and (2.1). We wish to show that(4.1) A ( p, q ) α + 2 B ( p, q ) α + C ( p, q ) D ( p, q ) α + F ( p, q ) ≤ π (cid:18) q (cid:19) for all real α (hence the bound holds for sup α ). Clearing denominators and rear-ranging, we can equivalently show that the following inequality holds:256 α (288 − p − − p ) p + q ) α ) − π (1 + q ) (1 + α )+7875 π (1 + 2( − p ) p + 2 q )(2 + α ) ≤ . (4.2)For fixed q and α , this is a quadratic in p with leading coefficient given by31500 π − α + 15750 π α . We know that31500 π − α + 15750 π α > π > p and α , this is a quadratic in q with leadingcoefficient 350(69 π − α + 24 π α ). We also know that350(69 π − α + 24 π α ) > π > p and q are extremalassuming the other one is fixed. Thus, we only need to check at the points(0 . , . , (0 . , . p, (cid:112) / − ( p − / ) where p ∈ [0 . , . p, q ) = (0 . , . α − π (70267 + 327457 α ))1250 ≤ . This clearly holds since 1805312 < π · p, q ) = (0 . , . α ( − α ) − π (112318 + 1225453 α )5000 ≤ . he LHS is a quadratic in α that attains a maximum of approximately − . < α ≈ − . / (cid:112) / − q , q ) for q ∈ [ √ / , / − (cid:112) − q α − π ( − − α + 14 q (2 + q )(1 + α )) ≤ . For fixed q this is a quadratic in α , so it suffices to show that the leading coef-ficient and the discriminant are both negative. The leading coefficient of (4.3) isgiven by − π ( − q + 14 q ) , which on the interval [ √ , .
5] achieves a maximum of ≈ − . < q = √ .Hence, the leading coefficient is negative. The discriminant of (4.3) is given by:(4.4) 72( − − q ) − π ( − q (2 + q ))( − q (2 + q ))) . This is a quartic in q , so it may be maximized explicitly; however, we providea simpler argument here. We claim that it is decreasing in q over the interval[ √ / , / q is given by: − π q − π q − π ) q + 524790000 π . This is clearly decreasing for q >
0, and is hence at most its value at q = √ / ≈ − . · < q = √ /
20 with value ≈ − . · <
0. This implies that (4.3)holds, completing our proof for Area I.4.2.
Area II.
In this area, we have q ≥ max(0 . , . p − .
38) and we utilizeequations (3.3) and (2.1). That is, we wish to show that(4.5) A ( p, q ) α + 2 B ( p, q ) α + C ( p, q ) D ( p, q ) α + F ( p, q ) ≤ π (cid:18) q (cid:19) for all real α . Upon clearing denominators and rearranging, the problem is to showthat the following inequality holds:3(( − p ( − π ) + 28 p ( − π ) − q ) + 54331200 π (3 + q ( − q ))) α + ( − p − p − q ) α − − p ( − π ) + 28 p ( − π ) − q + 7761600 π (57 − q + 83 q )) ≤ . (4.6)This proof proceeds similarly to our proof in Area I. For fixed q and α , (4.6) isa quadratic in p with leading coefficient given by − − α (13319850 + 7851173 α ) + 2217600 π (13 + 7 α )) , and at fixed p and α , (4.6) is a quadratic in q with leading coefficient given by(4.8) 3( − − α − α + 7761600 π (83 + 35 α )) . As in the proof for Area I, we now show that these leading coefficients are positivefor all real α . We again check that the discriminant of each is negative, guaranteeingno real roots, which in combination with having a positive leading coefficient implieseach is positive for all real α . We can bound (4.7) in the following way:84( − · · π ) = 84( − · π ) > − + 2217600 · ·
7) = 84( − + 139708800) > − + 10 ) > . Similarly, we can bound (4.8) in the following way:3( − · π ) > − + 2 · ) > − · · − · · ( − · · π ) · ( − · π ) ≈ − . · < , and the discriminant corresponding to (4.8) is given by(3 · − − · · ( − · π ) · ( − · π ) ≈ − . · < . Therefore, both discriminants of the above polynomials with positive leadingcoefficients are negative, and they are therefore both always positive for all real α . Hence, as in Area I, it suffices to show (4.6) at points where both p and q are extremal assuming the other one is fixed. Thus we only need to check at thepoint (0 . , . p, . p − .
38) as p ranges over[ , √ ], and the semicircular arc covering points ( p, (cid:112) / − ( p − / )as p ranges over [0 . , √ ].We first address the point (0 . , . α (1414193259450 + 1768358569901 α ))62500+ 3( − π (312007 + 977515 α ))62500 ≤ . The LHS is a quadratic in α with leading coefficient ≈ − . · < ≈ − . · < q = 1 . p − .
38. Plugging this into (4.6), wewish to show that:
625 (528( − p − p + 11025(682563+ 5 p ( − p )) α + 1617( − p − p + 126000(10401 + 5 p ( − p )) π ) α ) ≤ . (4.9)For fixed p , this is a quadratic in α , so we only need to check that its leadingcoefficient and discriminant are both negative. Firstly, its leading coefficient is:1617625 ( − p − p + 126000(10401 + 5 p ( − p )) π ) . This is a quadratic in p , and over the interval of interest it is maximized at p = , where its value is ≈ − . · <
0, so indeed the leading coefficientof (4.9) is negative. Next, the discriminant of (4.9) is given by:1764390625 (5625(4620157398 + 5 p ( − p )) − − p − p + 126000(10401+ 5 p ( − p )) π )( − p − p + 11025(682563 + 5 p ( − p )) π )) . (4.10)We now show that (4.10) is negative over our interval [ , √ ]. Again,we could do this explicitly but we provide a simpler proof. The second derivativeof (4.10) is a quadratic in p that is minimized at , achieving a minimum of1 . · >
0, so this quartic is convex over this interval. Hence to show thatit is negative it suffices to check that it is negative at the endpoints of our interval.At p = , this discriminant simplifies to the following: − · ( − π ( − π ))) ≈ − . · < p = √ this discriminant simplifies to: − · · ( − · ( − √ π · ( − · ( − √ · ( − √ · π )) ≈ − . · < . Thus, (4.10) is indeed negative. This proves (4.9), achieving the desired result alongthe line segment q = 1 . p − . inally, we show (4.6) along the semi-circular boundary arc; in this domain, weare restricted to q = (cid:112) / − ( p − / and p ∈ [0 . , √ ]. For conve-nience, let p = 0 .
65 and p = √ . Plugging in q = (cid:112) / − ( p − / to(4 . − α )(8 + 77 α ) + 236196 p (22 + 7 α )(8 + 77 α )+ 18110400 p π (1 + α ) − pπ (73 + 49 α ) − π ( −
19+ 14 (cid:112) p (1 − p ) + 7( − (cid:112) p (1 − p )) α ) ≤ . (4.11)Once again, for fixed p this is a quadratic in α . Its leading coefficient is27( − p − pπ + 18110400 p π − − (cid:112) (1 − p ) p ) π ):= 27 f ( p ) , and its constant coefficient is27( − p − pπ + 18110400 p π − −
19 + 14 (cid:112) − ( − p ) p ) π ):= 27 g ( p ) . We show that f ( p ) is negative in the interval [ p , p ] by showing that f ( p ) is strictlyincreasing in [ p , p ] and that f ( p ) <
0. Note that the derivative f (cid:48) ( p ) is given bythe following: f (cid:48) ( p ) = (127309644 − π ) + 2 · π · p + 18110400 π p − (cid:112) (1 − p ) p . Since [ p , p ] ⊂ (0 . , p − √ (1 − p ) p are increas-ing on [ p , p ]. Since both are also positive on [ p , p ], it follows that p − √ (1 − p ) p isincreasing on [ p , p ]. We can then conclude that f (cid:48) ( p ) is increasing on [ p , p ], andthus: f (cid:48) ( p ) ≥ f (cid:48) ( p )= (127309644 − π ) + 2 · π · .
65 + 18110400 π . √ . ≈ . · > . Hence f is increasing, and evaluating f ( p ) gives the following: − √ − √ π + 724416(1423 + 10 √ π − − (cid:115) (1423 + 10 √ − − √ )1945 ) π ≈ − . · < , showing that f ( p ) is negative for p ∈ [ p , p ], and thus the leading coefficient of(4.11) is negative. It remains to show that its discriminant is also negative. imilarly, it is easy to see that g ( p ) is negative for p ∈ [ p , p ]. By considering thederivative of g , we may computationally verify that g has exactly one minimum in( p , p ) and no other local extrema. In fact, the minimizing value of g lies between p = 0 .
81 and p = 0 .
82, and thus the function g has the additional property thatthere exists 0 . < p < .
82 such that g is decreasing on [ p , p ] and increasing on[ p , p ]. More details can be found in Section4-2.nb in the GitHub repository.We have that − f and − g are both decreasing and positive on [ p , p ], and hence f g is as well, as is any non-trivial function proportionate to it.Meanwhile, the square of the linear coefficient, 729( − p ) ,is increasing on [ p , p ] and is positive. Thus the discriminant729( − p ) − f ( p ) g ( p )is increasing on [ p , p ]. Since evaluation at p gives ≈ − . · <
0, the discrim-inant is negative over [ p , p ].We now check the region [ p , p ]. In this interval, we have the bounds − p − / + 1 / ≤ (cid:112) (1 − p ) p and 12 − (cid:18) p − (cid:19) − . ≤ (cid:112) (1 − p ) p, which hold for p ∈ [ p , p ].Define the following functions, which are modifications of f and g respectively: f ( p ) := − p − pπ + 18110400 p π − − − (cid:18) p − (cid:19) − . π ,g ( p ) := − p − pπ + 18110400 p π − −
19 + 14( − p − / + 1 / π . Then f ≤ f , and additionally − g is positive in p ∈ [ p , p ], so − f g ≤ − f g . Itis not hard to check that − f is positive in [ p , p ], and as g ≤ g , we similarly have − f g ≤ − f g . So:729( − p ) − f ( p ) g ( p ) ≤ − p ) − f ( p ) g ( p ) . =: r ( p ) . We will now prove that the LHS is negative throughout [ p , p ] by proving thisstatement for r ( p ). To do this, we first want to prove as a lemma that r is concaveup in [ p , p ]; r is a quartic polynomial, meaning that its second derivative r (cid:48)(cid:48) isquadratic. To find the critical point of the quadratic r (cid:48)(cid:48) , we find the root of itsderivative. The derivative is given by: − π p − π + 253038466610012160000 π , which has its root at p = ( − π ) / (24696000 π ) ≈ . ≈ . · >
0. Since r (cid:48)(cid:48) is a quadratic whose leading .70 0.75 0.80 0.85 0.90 0.95 1.000.140.160.180.200.220.24 Figure 3.
Visualization showing that the range of angles θ inArea III is contained in [0 . , . p tan(0 . ≤ q ≤ p tan(0 .
24) (shown inorange).coefficient is − π < r (cid:48)(cid:48) ( p ) is a maximum, and since p ∈ [ p , p ], we merely have to check that both r (cid:48)(cid:48) ( p ) > r (cid:48)(cid:48) ( p ) > r (cid:48)(cid:48) ( p ) > p ∈ [ p , p ]. It suffices to prove that r (cid:48)(cid:48) ( q ) > q > p in place of the statement r (cid:48)(cid:48) ( p ) > r (cid:48)(cid:48) ( p ). At p = 0 .
65 and p = 0 . > p , r (cid:48)(cid:48) evaluates to ≈ . · and ≈ . · , respectively. Therefore, r (cid:48)(cid:48) ( p ) > p ∈ [ p , p ], and r isconcave upwards in this region.Thus it suffices to check that r ( p ) < p = p and p = 0 . > p . At thesepoints, r ( p ) has the values ≈ − . · < ≈ − . · < r ( p ) < p ∈ [ p , p ], implying that the same holds forthe discriminant of (4.11).This completes the proof of Equation (4.6), and hence Theorem 1.1 over AreaII.4.3. Area III.
For this area, we use bounds (3.7) and (2.2). Here we are concernedwith the region where 0 . ≤ q ≤ . p − .
38. Note that in this region the rangeof angles θ achieved is contained in [0 . , . p, q , andis shown visually in Figure 3. Detailed code for producing this figure and theother computations in this section can be found in Section4-3.nb in the GitHubrepository. oting that A = q , we can then rearrange our target inequality to the following: π √ q ) q θj π/θ A ≤ ⇔ π (1 + (cid:112) q ) q ≤ θj π/θ A ⇔ π (1 + (cid:112) q ) q ≤ θj π/θ . Call the LHS f ( q ) and the RHS g ( θ ). We claim that these are decreasing func-tions of q and θ respectively in the region we are interested in. We first check thisfor f ( q ), omitting the constant factor of 3 π : ddq (1 + (cid:112) q ) q = 2 · / (1 + (cid:112) q ) q / − (1 + (cid:112) q ) q = (1 + (cid:112) q ) q (2 · / q / − (1 + (cid:112) q ))= (1 + (cid:112) q ) q ( (cid:112) q − ≤ , since q ≤ / g ( θ ), let t = π/θ ∈ [13 , t is decreasing with respect to θ , wewish to show (omitting the constant factor of 7) that j t t is increasing with respectto t in this region: ddt j t t = t j t dj t dt − j t t . So we want to show that t j t dj t dt − j t ≥ ⇔ j t ≤ tj t dj t dt ⇔ j t ≤ t dj t dt . To do this,we use the following results shown in [4]:(1) dj t dt > t > j ≈ . > .(2) j t is concave as a function of t (this is Corollary 3.3 of [4]).The first point means we just need to show that j t ≤ t . Note that at t = 13we have j t ≤ . <
26 = 2 t so it suffices to show that dj t dt ≤ t ≥ dj t dt is non-increasing so we just need to showthat dj t dt (cid:12)(cid:12)(cid:12) t =13 ≤
2. But this is straightforward; by concavity, the LHS is at most j − j ≈ . < q , θ ) such that f ( q ) ≤ g ( θ ), then whenever q ≥ q and θ ≤ θ we have f ( q ) ≤ f ( q ) ≤ g ( θ ) ≤ g ( θ ). So if we call the set of points inside our quarter circlesatisfying q ≥ q and θ ≤ θ as S q ,θ then it suffices to specify a set of pairs ( q , θ )each satisfying f ( q ) ≤ g ( θ ) such that the sets S q ,θ collectively cover Area III.We take the following three pairs: • q , θ = 0 . , .
2, satisfies f ( q ) /g ( θ ) ≈ . < • q , θ = 0 . , . f ( q ) /g ( θ ) ≈ . < • q , θ = 0 . , .
24, satisfies f ( q ) /g ( θ ) ≈ . < .70 0.75 0.80 0.85 0.90 0.95 1.000.140.160.180.200.220.24 Figure 4.
Final step for the proof in Area III (shown in bluehere). The orange region is the union of three sector-like regions,each of which corresponds to one S q ,θ . As desired, the orangeregion covers Area III.It is straightforward to verify that the three S q ,θ ’s thus defined cover the regionof interest (since all the conditions of interest are linear or quadratic inequalities in p, q ). This is visually shown in Figure 4.4.4. Area IV.
As mentioned earlier, here we use bounds (3.7) and (2.1). We areconcerned with the region q ≤ . π √ q ) q π (1 + q ) ≤ ⇐⇒ (1 + (cid:112) q ) ( q + 1) ≤ ⇐⇒ (cid:112) q ) − q + 1) ≤ . Call the LHS f ( q ). Note that f (0 . ≈ − . < f is non-decreasing on [0 , . x = √ q (note that this is anincreasing function of q ) and differentiate f with respect to x . We want to show thatthis derivative is non-negative on [0 , √ . x (3 · / − x +12 · / x +5 x ) ≥
0. But this is clear since x ≥ x ≤ √ . < · / ⇒ · / − x > Acknowledgements
We wish to thank Javier G´omez-Serrano for introducing this problem to us inhis class and for guiding us while writing this paper. We also thank the PrincetonUniversity Department of Mathematics. eferences Open problems from the Miniconference on Sharp Eigenvalue Estimates for Partial Differen-tial Operators (April 2020) , http://publish.illinois.edu/eigenvalues2020/files/2020/04/Open-problems.pdf , 2020.2. Mark S. Ashbaugh and Rafael D. Benguria, Proof of the Payne-P´olya-Weinberger conjecture ,Bulletin of the American Mathematical Society (1991), 19–29.3. , A sharp bound for the ratio of the first two eigenvalues of Dirichlet Laplacians andextensions , The Annals of Mathematics (1992), no. 3, 601.4. Arpad Elbert, Luigi Gatteschi, and Andrea Laforgia,
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